Fundamental laws of trigonometry

Class 11 Mathematics, unit 8, book pages 195–215

Practise 74 MCQs Take the unit quiz Review 26 cards

Every identity in this unit grows from one result, the fundamental law $cos (α - β) = cos α cos β + sin α sin β$ , which is proved on the unit circle with the distance formula. From it the unit deduces the sine, cosine and tangent of a sum and a difference, the ratios of allied angles, and the form $r sin (θ + ϕ)$ for $a sin θ + b cos θ$ . It then derives the double, half and triple angle identities and the formulae that turn products of sines and cosines into sums and differences and back again. These identities simplify complicated expressions in calculus, the physical sciences and the social sciences.

Learning outcomes (7)

Establish the fundamental law of trigonometry.
Apply the fundamental law and its deductions to derive the trigonometric ratios of allied angles.
Apply the fundamental law and its deductions to derive the double angle identities.
Apply the fundamental law and its deductions to derive the half angle identities.
Apply the fundamental law and its deductions to derive the triple angle identities.
Express the product of sines and cosines as sums and differences of sines and cosines.
Express the sums and differences of sines and cosines as the product of sines and cosines.

8.1Distance formula

The unit begins by recalling how to find the distance between two points. It is needed in Section 8.2, where the fundamental law of trigonometry is proved by equating two distances on the unit circle.

Distance formula: The distance between the points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ in the plane is $d = A B = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}$ .

d = A B = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}

Distance between two points (book's illustration)

A (5, 4)

and

B (3, 2)

are two points in the plane, find the distance between

A

and

B

Here $(x_{1}, y_{1}) = (5, 4)$ and $(x_{2}, y_{2}) = (3, 2)$ . Substitute into the distance formula.
$A B = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} = (5 - 3)^{2} + (4 - 2)^{2}$
Evaluate the differences and square them.
$= (2)^{2} + (2)^{2} = 4 + 4 = 8$
Simplify the root if wanted: $8 = 22 \approx 2.83$ .
$A B = 8 units$

x

y

B (3, 2)

A (5, 4)

5 - 3 = 2

4 - 2 = 2

The segment

A B

is the hypotenuse of a right-angled triangle whose legs are the horizontal difference

5 - 3 = 2

and the vertical difference

4 - 2 = 2

. Pythagoras' theorem gives

A B = 2^{2} + 2^{2} = 8

A B = 8 = 22 units

8.2Fundamental law of trigonometryOften examined

An identity is an equation that is true for every value of the angle for which both sides are defined. This section proves the one identity from which almost every other identity of the unit is deduced.

Fundamental law of trigonometry: $cos (α - β) = cos α cos β + sin α sin β$ , where $α > β$ .

Derivation Proof of the fundamental law of trigonometry

Aim. Prove that $cos (α - β) = cos α cos β + sin α sin β$ for $α > β$ .

Assumptions.

The circle has radius 1 and centre at the origin $O (0, 0)$ .
$α > β$ , both angles in standard position.

−1

x

y

β

α - β

α - β

O (0, 0)

A (1, 0)

B (cos (α - β), sin (α - β))

C (cos β, sin β)

D (cos α, sin α)

The unit circle.

O C

and

O D

are the terminal arms of

β

and

α

, so

∠ C O D = α - β

. The point

B

is placed so that

∠ A O B = α - β

as well. The two dashed chords

C D

and

A B

then have equal length.

Take a unit circle with centre $O$ . Let the terminal arms $O C$ and $O D$ of the angles $β$ and $α$ (in standard position, $α > β$ ) meet the circle at $C$ and $D$ . On the unit circle the point at angle $θ$ has coordinates $(cos θ, sin θ)$ , so we get the coordinates below.
$C (cos β, sin β), D (cos α, sin α)$
The angle between the arms $O C$ and $O D$ is the difference of the two angles.
$∠ C O D = α - β$
In the same circle, construct the angle $∠ A O B$ in standard position with the same measure $α - β$ . Its initial arm meets the circle at $A (1, 0)$ and its terminal arm at $B$ , whose coordinates are therefore those of the angle $α - β$ .
$A (1, 0), B (cos (α - β), sin (α - β))$
The angles $∠ C O D$ and $∠ A O B$ are equal, and the arms $O C, O D, O A, O B$ are all radii of length 1. So the triangles $C O D$ and $A O B$ are congruent, and their third sides are equal.
$C D = A B$
Write both lengths with the distance formula (Section 8.1).
$(cos α - cos β)^{2} + (sin α - sin β)^{2} = [cos (α - β) - 1]^{2} + [sin (α - β) - 0]^{2}$
Square both sides to remove the roots.
$(cos α - cos β)^{2} + (sin α - sin β)^{2} = [cos (α - β) - 1]^{2} + [sin (α - β)]^{2}$
Expand the left side.
$cos^{2} α + cos^{2} β - 2 cos α cos β + sin^{2} α + sin^{2} β - 2 sin α sin β$
Expand the right side.
$cos^{2} (α - β) + 1 - 2 cos (α - β) + sin^{2} (α - β)$
Rearrange both sides so that the squares are together, and use $cos^{2} θ + sin^{2} θ = 1$ on each pair.
$(cos^{2} α + sin^{2} α) + (cos^{2} β + sin^{2} β) - 2 cos α cos β - 2 sin α sin β = (cos^{2} (α - β) + sin^{2} (α - β)) + 1 - 2 cos (α - β)$
Replace each bracket by 1.
$1 + 1 - 2 cos α cos β - 2 sin α sin β = 1 + 1 - 2 cos (α - β)$
Cancel the 2 on each side, then divide both sides by $- 2$ .
$- 2 cos α cos β - 2 sin α sin β = - 2 cos (α - β)$

Result

cos (α - β) = cos α cos β + sin α sin β (1)

This is the fundamental law of trigonometry. The cosine of a difference is not the difference of the cosines: the right side mixes cosines and sines.

Testing the law on

cos (9 0^{\circ} - 6 0^{\circ})

Does the fundamental law give

cos (9 0^{\circ} - 6 0^{\circ})

correctly? Check it.

Put $α = 9 0^{\circ}$ and $β = 6 0^{\circ}$ in the law.
$cos (9 0^{\circ} - 6 0^{\circ}) = cos 9 0^{\circ} cos 6 0^{\circ} + sin 9 0^{\circ} sin 6 0^{\circ}$
Use $cos 9 0^{\circ} = 0$ , $sin 9 0^{\circ} = 1$ , $cos 6 0^{\circ} = 0.5$ and $sin 6 0^{\circ} \approx 0.866$ .
$cos 3 0^{\circ} = 0 \times 0.5 + 1 \times 0.866 = 0.866$
The left side is $cos 3 0^{\circ} \approx 0.866$ as well, so both sides agree.
$0.866 = 0.866$

the law gives the correct value

Cosine of a sum

Derivation Cosine of a sum, identity (2)

Aim. Deduce $cos (α + β)$ from the fundamental law.

The law (1) holds for any angles, so replace $β$ by $- β$ throughout.
$cos (α - (- β)) = cos α cos (- β) + sin α sin (- β)$
Simplify the left side. On the right side use $cos (- β) = cos β$ (proved below in result (i)) and $sin (- β) = - sin β$ (from the Key facts).
$cos (α + β) = cos α cos β - sin α sin β$

Result

cos (α + β) = cos α cos β - sin α sin β (2)

The sign in the middle of the right side is the opposite of the sign inside the bracket on the left.

Important results from the fundamental law

Before proving the other identities, the book deduces five results from the fundamental law.

Derivation Result (i): $cos (- β) = cos β$

Aim. Show that cosine is an even function.

Put $α = 0$ in the fundamental law.
$cos (0 - β) = cos 0 cos β + sin 0 sin β$
Use $cos 0 = 1$ and $sin 0 = 0$ .
$cos (- β) = 1 \times cos β + 0 \times sin β$

Result

cos (- β) = cos β

Derivation Result (ii): $cos (\frac{π}{2} - β) = sin β$

Aim. Show that the cosine of the complement is the sine.

Put $α = \frac{π}{2}$ in the fundamental law.
$cos (\frac{π}{2} - β) = cos \frac{π}{2} cos β + sin \frac{π}{2} sin β$
Use $cos \frac{π}{2} = 0$ and $sin \frac{π}{2} = 1$ .
$cos (\frac{π}{2} - β) = 0 \times cos β + 1 \times sin β$

Result

cos (\frac{π}{2} - β) = sin β

Derivation Result (iii): $cos (\frac{π}{2} + α) = - sin α$

Aim. Find the cosine of $\frac{π}{2} + α$ .

Put $β = - \frac{π}{2}$ in the fundamental law.
$cos (α - (- \frac{π}{2})) = cos α cos (- \frac{π}{2}) + sin α sin (- \frac{π}{2})$
Use $cos (- \frac{π}{2}) = 0$ and $sin (- \frac{π}{2}) = - 1$ .
$cos (α + \frac{π}{2}) = cos α \times 0 + sin α \times (- 1)$

Result

cos (\frac{π}{2} + α) = - sin α

Derivation Result (iv): $sin (\frac{π}{2} + α) = cos α$

Aim. Find the sine of $\frac{π}{2} + α$ .

Result (ii) is true for every angle $β$ . Replace $β$ by $\frac{π}{2} + α$ .
$cos (\frac{π}{2} - (\frac{π}{2} + α)) = sin (\frac{π}{2} + α)$
Simplify the angle on the left, then use result (i): $cos (- α) = cos α$ .
$cos (- α) = sin (\frac{π}{2} + α) ⟹ cos α = sin (\frac{π}{2} + α)$

Result

sin (\frac{π}{2} + α) = cos α

Derivation Result (v): the sine of a sum and of a difference

Aim. Prove $sin (α + β) = sin α cos β + cos α sin β$ and $sin (α - β) = sin α cos β - cos α sin β$ .

In identity (2), replace $α$ by $\frac{π}{2} + α$ .
$cos (\frac{π}{2} + α + β) = cos (\frac{π}{2} + α) cos β - sin (\frac{π}{2} + α) sin β$
Write the angle on the left as $\frac{π}{2} + (α + β)$ and apply result (iii) to it. On the right use results (iii) and (iv).
$- sin (α + β) = - sin α cos β - cos α sin β$
Multiply both sides by $- 1$ .
$sin (α + β) = sin α cos β + cos α sin β (3)$
Now replace $β$ by $- β$ in identity (3), using $cos (- β) = cos β$ and $sin (- β) = - sin β$ .
$sin (α + (- β)) = sin α cos (- β) + cos α sin (- β)$
Simplify.
$sin (α - β) = sin α cos β - cos α sin β (4)$

Result

sin (α \pm β) = sin α cos β \pm cos α sin β

The four addition and subtraction identities of this section, with the book's numbers
Identity	Formula
$cos (α - β)$ (1)	$cos α cos β + sin α sin β$
$cos (α + β)$ (2)	$cos α cos β - sin α sin β$
$sin (α + β)$ (3)	$sin α cos β + cos α sin β$
$sin (α - β)$ (4)	$sin α cos β - cos α sin β$

Example 1

Prove that

tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β}

Write the tangent as sine over cosine and use identities (3) and (2).
$tan (α + β) = \frac{sin ( α + β )}{cos ( α + β )} = \frac{sin α cos β + cos α sin β}{cos α cos β - sin α sin β}$
Divide the numerator and the denominator by $cos α cos β$ .
$= \frac{\frac{sin α cos β}{cos α cos β} + \frac{cos α sin β}{cos α cos β}}{\frac{cos α cos β}{cos α cos β} - \frac{sin α sin β}{cos α cos β}}$
Cancel in each fraction.
$= \frac{\frac{sin α}{cos α} + \frac{sin β}{cos β}}{1 - \frac{sin α}{cos α} \times \frac{sin β}{cos β}}$
Replace each $\frac{sin}{cos}$ by $tan$ .
$= \frac{tan α + tan β}{1 - tan α tan β}$

tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β}

Check points

Check point: complementary and negative angles

(1) Are the equations

sin α = cos (9 0^{\circ} - α)

and

cos α = sin (9 0^{\circ} - α)

true for all real numbers or only for

0 < α < 9 0^{\circ}

? (2) Amna said that, without finding the values on a calculator, she knows

sin 10 0^{\circ} = cos (- 1 0^{\circ})

. Do you agree with her? Explain why or why not.

(1)

Result (ii) was proved for every angle $β$ , because the fundamental law holds for all angles. Put $β = α$ .
$cos (\frac{π}{2} - α) = sin α$
For the second equation put $β = \frac{π}{2} - α$ in result (ii). The angle $\frac{π}{2} - β$ becomes $α$ .
$cos α = sin (\frac{π}{2} - α)$
Nothing in the working needs $α$ to be acute, so both equations are true for all real $α$ , not only for $0 < α < 9 0^{\circ}$ .

(2)

Use $sin α = cos (9 0^{\circ} - α)$ with $α = 10 0^{\circ}$ . It is valid for every angle, even though $10 0^{\circ}$ is not acute.
$sin 10 0^{\circ} = cos (9 0^{\circ} - 10 0^{\circ}) = cos (- 1 0^{\circ})$
So Amna is right. We agree with her. The identity of part (1) holds for all angles, so no calculator is needed.

(1) true for all real α; (2) yes, sin 10 0^{\circ} = cos (- 1 0^{\circ})

Check point: odd properties and

tan (α - β)

Prove that (i)

sin (- α) = - sin α

(ii)

tan (- α) = - tan α

(iii)

tan (α - β) = \frac{tan α - tan β}{1 + tan α tan β}

(i)

Use result (ii), $cos (\frac{π}{2} - β) = sin β$ , with $β = - α$ .
$sin (- α) = cos (\frac{π}{2} - (- α)) = cos (\frac{π}{2} + α)$
Result (iii) says that this equals $- sin α$ .
$sin (- α) = - sin α$

(ii)

Write the tangent as sine over cosine, then use part (i) and $cos (- α) = cos α$ .
$tan (- α) = \frac{sin ( - α )}{cos ( - α )} = \frac{- sin α}{cos α} = - tan α$

(iii)

In the formula of Example 1 replace $β$ by $- β$ .
$tan (α - β) = \frac{tan α + tan ( - β )}{1 - tan α tan ( - β )}$
Use $tan (- β) = - tan β$ from part (ii).
$tan (α - β) = \frac{tan α - tan β}{1 + tan α tan β}$

tan (α - β) = \frac{tan α - tan β}{1 + tan α tan β}

Worked example: using the fundamental law backwards, and a common mistake

(i) Evaluate

cos 8 0^{\circ} cos 2 0^{\circ} + sin 8 0^{\circ} sin 2 0^{\circ}

. (ii) A student writes

cos (9 0^{\circ} - 6 0^{\circ}) = cos 9 0^{\circ} - cos 6 0^{\circ}

. Test this claim and explain the error.

(i) Recognise the pattern

The expression has the shape $cos α cos β + sin α sin β$ , which is the right side of the fundamental law. So it equals $cos (α - β)$ with $α = 8 0^{\circ}$ and $β = 2 0^{\circ}$ .
$cos 8 0^{\circ} cos 2 0^{\circ} + sin 8 0^{\circ} sin 2 0^{\circ} = cos (8 0^{\circ} - 2 0^{\circ})$
Subtract the angles and use the exact value.
$= cos 6 0^{\circ} = \frac{1}{2}$
Check with a calculator: $0.1736 \times 0.9397 + 0.9848 \times 0.3420 = 0.1632 + 0.3368$ .
$= 0.5000 ✓$

(ii) Test the student's claim

Work out the left side directly.
$cos (9 0^{\circ} - 6 0^{\circ}) = cos 3 0^{\circ} = \frac{3}{2} \approx 0.866$
Work out the student's right side.
$cos 9 0^{\circ} - cos 6 0^{\circ} = 0 - \frac{1}{2} = - \frac{1}{2}$
The two sides differ ( $0.866 \neq = - 0.5$ ), so the claim is false. Cosine does not distribute over a difference of angles. The fundamental law has products and a plus sign.
$cos (9 0^{\circ} - 6 0^{\circ}) = cos 9 0^{\circ} cos 6 0^{\circ} + sin 9 0^{\circ} sin 6 0^{\circ} = 0 \cdot \frac{1}{2} + 1 \cdot \frac{3}{2} = \frac{3}{2}$

(i) \frac{1}{2} (ii) the claim is false; the correct value is \frac{3}{2}

8.3Trigonometric ratios of allied anglesOften examined

Angles such as \dfrac{\pi}{2}\pm\theta, \pi\pm\theta, \dfrac{3\pi}{2}\pm\theta and 2\pi\pm\theta are built from a basic angle \theta by adding or subtracting a right angle or a multiple of a right angle. The trigonometric ratios of such angles can be written in terms of the ratios of \theta itself.

Allied angles: Angles connected with a basic angle \theta by a right angle or its multiple. If \theta is the basic angle, the angles \dfrac{\pi}{2}\pm\theta, \pi\pm\theta, \dfrac{3\pi}{2}\pm\theta, 2\pi\pm\theta and so on are its allied angles.

The following ratios can be derived easily with the help of the fundamental theorem of trigonometry and its deductions (Section 8.2). The book prints them in a 3\times5 grid\; here each row is one allied angle.

Ratios of allied angles. The book writes \sin(\pi\mp\theta)=\pm\sin\theta, \cos\left(\dfrac{3\pi}{2}\pm\theta\right)=\pm\sin\theta, \tan\left(\dfrac{3\pi}{2}\pm\theta\right)=\mp\cot\theta and so on, with the upper signs together and the lower signs together\; each row above is one such case.
Allied angle	\sin	\cos	\tan
\dfrac{\pi}{2}-\theta	\cos\theta	\sin\theta	\cot\theta
\dfrac{\pi}{2}+\theta	\cos\theta	-\sin\theta	-\cot\theta
\pi-\theta	\sin\theta	-\cos\theta	-\tan\theta
\pi+\theta	-\sin\theta	-\cos\theta	\tan\theta
\dfrac{3\pi}{2}-\theta	-\cos\theta	-\sin\theta	\cot\theta
\dfrac{3\pi}{2}+\theta	-\cos\theta	\sin\theta	-\cot\theta
2\pi-\theta	-\sin\theta	\cos\theta	-\tan\theta
2\pi+\theta	\sin\theta	\cos\theta	\tan\theta

Example 2

Use (60^\circ-45^\circ)=15^\circ to find the exact value of (i) \cos15^\circ (ii) \sin15^\circ (iii) \tan15^\circ.

Let \alpha=60^\circ and \beta=45^\circ. The values needed are \sin60^\circ=\dfrac{\sqrt3}{2}\approx0.866, \cos60^\circ=\dfrac12=0.5 and \sin45^\circ=\cos45^\circ=\dfrac{1}{\sqrt2}\approx0.707.

(i)

Use the fundamental law \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos15^\circ=\cos(60^\circ-45^\circ)=\cos60^\circ\cos45^\circ+\sin60^\circ\sin45^\circ
Exact values.
=\dfrac12\cdot\dfrac{1}{\sqrt2}+\dfrac{\sqrt3}{2}\cdot\dfrac{1}{\sqrt2}=\dfrac{1+\sqrt3}{2\sqrt2}=\dfrac{\sqrt2+\sqrt6}{4}
Decimal check with the book's values.
\cos15^\circ=0.5\times0.707+0.866\times0.707=0.3535+0.6123=0.966

(ii)

Use \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin15^\circ=\sin(60^\circ-45^\circ)=\sin60^\circ\cos45^\circ-\cos60^\circ\sin45^\circ
Exact values.
=\dfrac{\sqrt3}{2}\cdot\dfrac{1}{\sqrt2}-\dfrac12\cdot\dfrac{1}{\sqrt2}=\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac{\sqrt6-\sqrt2}{4}
Decimal check.
\sin15^\circ=0.866\times0.707-0.5\times0.707=0.6123-0.3535=0.259

(iii)

Divide the sine by the cosine.
\tan15^\circ=\dfrac{\sin15^\circ}{\cos15^\circ}=\dfrac{0.259}{0.966}=0.268
Exactly, using the values above and rationalising.
\tan15^\circ=\dfrac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}=\dfrac{(\sqrt6-\sqrt2)^2}{6-2}=\dfrac{8-4\sqrt3}{4}=2-\sqrt3\approx0.268

\cos15^\circ=\dfrac{\sqrt6+\sqrt2}{4}\approx0.966,\ \sin15^\circ=\dfrac{\sqrt6-\sqrt2}{4}\approx0.259,\ \tan15^\circ=2-\sqrt3\approx0.268

Example 3

If \sin A=\dfrac15 and \sin B=\dfrac13, where the angles A and B are in quadrant II, find \cos(A+B). In which quadrant does the terminal arm of the angle (A+B) lie?

To use \cos(A+B)=\cos A\cos B-\sin A\sin B we need \cos A and \cos B. Use \cos^2\theta=1-\sin^2\theta for A.
\cos^2A=1-\sin^2A=1-\left(\dfrac15\right)^2=1-\dfrac{1}{25}=\dfrac{24}{25}
Take the square root. In quadrant II cosine is negative, so choose the minus sign.
\cos A=-\dfrac{\sqrt{24}}{5}
Do the same for B.
\cos^2B=1-\left(\dfrac13\right)^2=1-\dfrac19=\dfrac89\;\Longrightarrow\;\cos B=-\dfrac{\sqrt8}{3}\quad(\text{quadrant II})
Substitute into the identity for \cos(A+B). The product of two negatives is positive.
\cos(A+B)=\left(-\dfrac{\sqrt{24}}{5}\right)\left(-\dfrac{\sqrt8}{3}\right)-\left(\dfrac15\right)\left(\dfrac13\right)=\dfrac{\sqrt{192}}{15}-\dfrac{1}{15}
Simplify \sqrt{192}=\sqrt{64\times3}=8\sqrt3.
\cos(A+B)=\dfrac{8\sqrt3-1}{15}
The value is positive (8\sqrt3\approx13.86, so \cos(A+B)\approx0.86). Each of A and B is between 90^\circ and 180^\circ, so A+B is between 180^\circ and 360^\circ. Cosine is positive in quadrant IV only within that range, so the terminal arm of A+B lies in the fourth quadrant.

\cos(A+B)=\dfrac{8\sqrt3-1}{15}\;\ \text{terminal arm in quadrant IV}

Check point: \tan y from \tan(y-45^\circ)

Find the exact value of \tan y when \tan(y-45^\circ)=\dfrac13.

Use the tangent of a difference with \tan45^\circ=1.
\tan(y-45^\circ)=\dfrac{\tan y-\tan45^\circ}{1+\tan y\tan45^\circ}=\dfrac{\tan y-1}{1+\tan y}
Put this equal to \dfrac13 and cross-multiply.
\dfrac{\tan y-1}{1+\tan y}=\dfrac13\;\Longrightarrow\;3\tan y-3=1+\tan y
Collect the terms in \tan y.
2\tan y=4\;\Longrightarrow\;\tan y=2

\tan y=2

Example 4

If \alpha,\beta,\gamma are the interior angles of a triangle, prove that \cot\beta\cot\gamma+\cot\alpha\cot\gamma+\cot\alpha\cot\beta=1.

The angles of a triangle add up to 180^\circ, so \alpha+\beta=180^\circ-\gamma.
\alpha+\beta+\gamma=180^\circ\;\Longrightarrow\;\alpha+\beta=180^\circ-\gamma
Take the tangent of both sides. By the table of allied angles, \tan(180^\circ-\gamma)=-\tan\gamma.
\tan(\alpha+\beta)=\tan(180^\circ-\gamma)=-\tan\gamma
Expand the left side with the tangent identity.
\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=-\tan\gamma
Multiply both sides by 1-\tan\alpha\tan\beta.
\tan\alpha+\tan\beta=-\tan\gamma\,(1-\tan\alpha\tan\beta)=-\tan\gamma+\tan\alpha\tan\beta\tan\gamma
Move -\tan\gamma to the left side.
\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma
Divide both sides by \tan\alpha\tan\beta\tan\gamma.
\dfrac{\tan\alpha}{\tan\alpha\tan\beta\tan\gamma}+\dfrac{\tan\beta}{\tan\alpha\tan\beta\tan\gamma}+\dfrac{\tan\gamma}{\tan\alpha\tan\beta\tan\gamma}=\dfrac{\tan\alpha\tan\beta\tan\gamma}{\tan\alpha\tan\beta\tan\gamma}
Cancel, and replace \dfrac{1}{\tan} by \cot.
\dfrac{1}{\tan\beta\tan\gamma}+\dfrac{1}{\tan\alpha\tan\gamma}+\dfrac{1}{\tan\alpha\tan\beta}=1\;\Longrightarrow\;\cot\beta\cot\gamma+\cot\alpha\cot\gamma+\cot\alpha\cot\beta=1

\cot\beta\cot\gamma+\cot\alpha\cot\gamma+\cot\alpha\cot\beta=1

Check point: \cos105^\circ and \cos\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)

Find the exact value of (i) \cos105^\circ (ii) \cos\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right).

(i)

Write 105^\circ=60^\circ+45^\circ and use \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos105^\circ=\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ
Exact values.
=\dfrac12\cdot\dfrac{1}{\sqrt2}-\dfrac{\sqrt3}{2}\cdot\dfrac{1}{\sqrt2}=\dfrac{1-\sqrt3}{2\sqrt2}=\dfrac{\sqrt2-\sqrt6}{4}

(ii)

The angle is \dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}=15^\circ. Use the fundamental law.
\cos\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=\cos\dfrac{\pi}{3}\cos\dfrac{\pi}{4}+\sin\dfrac{\pi}{3}\sin\dfrac{\pi}{4}
Exact values.
=\dfrac12\cdot\dfrac{1}{\sqrt2}+\dfrac{\sqrt3}{2}\cdot\dfrac{1}{\sqrt2}=\dfrac{1+\sqrt3}{2\sqrt2}=\dfrac{\sqrt2+\sqrt6}{4}

\cos105^\circ=\dfrac{\sqrt2-\sqrt6}{4},\qquad \cos\dfrac{\pi}{12}=\dfrac{\sqrt6+\sqrt2}{4}

Worked example: allied angle \pi-\theta on the unit circle

Draw the unit circle for \theta=30^\circ and the allied angle 180^\circ-30^\circ=150^\circ. Use the picture to find \sin150^\circ, \cos150^\circ and \tan150^\circ, and check against the table of allied angles.

The point P' for 180^\circ-30^\circ=150^\circ is the mirror image of P (for 30^\circ) in the y-axis: same height, opposite x.

The point for 30^\circ on the unit circle has coordinates (\cos30^\circ,\sin30^\circ).
P=\left(\dfrac{\sqrt3}{2},\ \dfrac12\right)
The ray for 150^\circ makes the same 30^\circ angle with the x-axis but on the left, so its point is the mirror image of P in the y-axis: the height stays the same and the x-coordinate changes sign.
P'=\left(-\dfrac{\sqrt3}{2},\ \dfrac12\right)
Read the ratios from the coordinates: cosine is the x-coordinate, sine is the y-coordinate and tangent is y/x.
\sin150^\circ=\dfrac12,\qquad \cos150^\circ=-\dfrac{\sqrt3}{2},\qquad \tan150^\circ=\dfrac{1/2}{-\sqrt3/2}=-\dfrac{1}{\sqrt3}
Compare with the table row for \pi-\theta: \sin\theta, -\cos\theta, -\tan\theta. With \theta=30^\circ these are \tfrac12, -\tfrac{\sqrt3}{2}, -\tfrac1{\sqrt3}, the same values.

\sin150^\circ=\dfrac12,\ \cos150^\circ=-\dfrac{\sqrt3}{2},\ \tan150^\circ=-\dfrac{1}{\sqrt3}

8.4Expressing a sin θ + b cos θ in the form r sin(θ + φ)

A sum of a sine term and a cosine term with the same angle, a\sin\theta+b\cos\theta, can be written as a single sine, r\sin(\theta+\phi). This is done by comparing with the expansion of \sin(\theta+\phi).

Derivation Expressing a\sin\theta+b\cos\theta as r\sin(\theta+\phi)

Aim. Write a\sin\theta+b\cos\theta in the form r\sin(\theta+\phi) and find r and \phi.

The point P(a,b) is at distance r from O. The angle that \overrightarrow{OP} makes with the x-axis is \phi, so a=r\cos\phi and b=r\sin\phi.

Let P(a,b) be a point in the coordinate plane. Let r be the length OP and let \phi be the angle that \overrightarrow{OP} makes with the x-axis. From the right-angled triangle in the figure we can write the coordinates of P in terms of r and \phi.
a=r\cos\phi,\qquad b=r\sin\phi
Replace a and b in the expression.
a\sin\theta+b\cos\theta=r\cos\phi\sin\theta+r\sin\phi\cos\theta
Take out the common factor r and reorder the products.
=r(\cos\phi\sin\theta+\sin\phi\cos\theta)=r(\sin\theta\cos\phi+\cos\theta\sin\phi)
The bracket is the expansion of \sin(\theta+\phi) (identity (3) of Section 8.2).
=r\sin(\theta+\phi)
Find r: square a=r\cos\phi and b=r\sin\phi and add. Use \cos^2\phi+\sin^2\phi=1. (This is also Pythagoras' theorem in the triangle.)
a^2+b^2=r^2(\cos^2\phi+\sin^2\phi)=r^2\;\Longrightarrow\;r=\sqrt{a^2+b^2}
Find \phi: divide b=r\sin\phi by a=r\cos\phi.
\dfrac{b}{a}=\tan\phi\;\Longrightarrow\;\phi=\tan^{-1}\left(\dfrac{b}{a}\right)

Resulta\sin\theta+b\cos\theta=r\sin(\theta+\phi),\qquad r=\sqrt{a^2+b^2},\quad \phi=\tan^{-1}\left(\dfrac ba\right)

The expression has the largest value r and is a sine wave shifted by \phi. The numbers a and b are the coordinates of the point P, so r is its distance from the origin and \phi its direction.

Example 5

Express 12\sin\theta+5\cos\theta in the form r\sin(\theta+\phi).

Compare 12\sin\theta+5\cos\theta with r\cos\phi\sin\theta+r\sin\phi\cos\theta.
a=12=r\cos\phi,\qquad b=5=r\sin\phi
Find r.
r=\sqrt{a^2+b^2}=\sqrt{12^2+5^2}=\sqrt{169}=13
Factor 13 out of the expression, so the coefficients become \dfrac{12}{13} and \dfrac{5}{13}, which are \cos\phi and \sin\phi.
12\sin\theta+5\cos\theta=13\left(\dfrac{12}{13}\sin\theta+\dfrac{5}{13}\cos\theta\right)=13(\sin\theta\cos\phi+\cos\theta\sin\phi)
Use the sine of a sum.
=13\sin(\theta+\phi)
Find \phi from \tan\phi=\dfrac{b}{a}. It is acute because a and b are both positive.
\phi=\tan^{-1}\left(\dfrac{5}{12}\right)\approx22.6^\circ

12\sin\theta+5\cos\theta=13\sin(\theta+\phi),\quad \phi=\tan^{-1}\left(\dfrac{5}{12}\right)

Check point: \cos\theta+\sin\theta

Express \cos\theta+\sin\theta in the form r\sin(\theta+\phi).

Write it as 1\cdot\sin\theta+1\cdot\cos\theta, so a=1 and b=1.
a=1,\qquad b=1
Find r.
r=\sqrt{a^2+b^2}=\sqrt{1+1}=\sqrt2
Find \phi. Both a and b are positive, so \phi is acute.
\phi=\tan^{-1}\left(\dfrac11\right)=\tan^{-1}1=\dfrac{\pi}{4}
Check by expanding.
\sqrt2\sin\left(\theta+\dfrac{\pi}{4}\right)=\sqrt2\left(\sin\theta\cdot\dfrac{1}{\sqrt2}+\cos\theta\cdot\dfrac{1}{\sqrt2}\right)=\sin\theta+\cos\theta

\cos\theta+\sin\theta=\sqrt2\sin\left(\theta+\dfrac{\pi}{4}\right)

Worked example: r\sin(\theta+\phi) when a is negative

Express -5\sin\theta+12\cos\theta in the form r\sin(\theta+\phi), and check the result at \theta=0^\circ and \theta=90^\circ.

The point P(-5,12) is in quadrant II, so \phi is measured past 90^\circ and is about 112.6^\circ. The calculator value \tan^{-1}(-12/5)\approx-67.4^\circ points into quadrant IV, which is the wrong quadrant.

Compare with r\cos\phi\sin\theta+r\sin\phi\cos\theta. Here a=-5 is the coefficient of \sin\theta and b=12 is the coefficient of \cos\theta.
r\cos\phi=-5,\qquad r\sin\phi=12
Find r from r^2=a^2+b^2. The sign of a does not matter here.
r=\sqrt{(-5)^2+12^2}=\sqrt{169}=13
Find the quadrant of \phi from the signs: \cos\phi=-\dfrac5{13}<0 and \sin\phi=\dfrac{12}{13}>0, so \phi is in quadrant II (the point P(-5,12) in the figure).
Find the reference angle from the positive values, then place it in quadrant II. Do not use \tan^{-1}(b/a) blindly: it gives -67.4^\circ, which is in quadrant IV.
\tan^{-1}\left(\dfrac{12}{5}\right)\approx67.4^\circ,\qquad \phi=180^\circ-67.4^\circ=112.6^\circ
Write the result.
-5\sin\theta+12\cos\theta=13\sin(\theta+112.6^\circ)
Check at \theta=0^\circ: the left side is 12 and the right side is 13\sin112.6^\circ=13\times0.9231=12.0. Check at \theta=90^\circ: the left side is -5 and the right side is 13\sin202.6^\circ=13\times(-0.3846)=-5.0. Both agree.

-5\sin\theta+12\cos\theta=13\sin(\theta+\phi),\quad \phi=180^\circ-\tan^{-1}\dfrac{12}{5}\approx112.6^\circ

8.5Double, half and triple angle identitiesOften examined

Double angle, half angle and triple angle identities express the trigonometric ratios of 2\alpha, \dfrac{\theta}{2} and 3\alpha in terms of the ratios of \alpha or \theta. All of them come from the sum identities of Section 8.2. The book numbers them (4) to (20) and the numbers are kept here.

(i) Double angle identities

The double angle identities are obtained by putting \beta=\alpha in the sum identities for sine, cosine and tangent:

\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta (1)
\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta (2)
\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} (3)

Derivation Identity (4): \sin2\alpha

Aim. Express \sin2\alpha in terms of \sin\alpha and \cos\alpha.

Put \beta=\alpha in identity (1).
\sin(\alpha+\alpha)=\sin\alpha\cos\alpha+\cos\alpha\sin\alpha
The two terms on the right are equal, so add them.
\sin2\alpha=2\sin\alpha\cos\alpha

Result\sin2\alpha=2\sin\alpha\cos\alpha\qquad(4)

Derivation Identities (5), (6), (7): \cos2\alpha

Aim. Express \cos2\alpha in three forms.

Put \beta=\alpha in identity (2).
\cos(\alpha+\alpha)=\cos\alpha\cos\alpha-\sin\alpha\sin\alpha\;\Longrightarrow\;\cos2\alpha=\cos^2\alpha-\sin^2\alpha\qquad(5)
Replace \sin^2\alpha by 1-\cos^2\alpha (from \cos^2\alpha+\sin^2\alpha=1) to get a form with cosine only.
\cos2\alpha=\cos^2\alpha-(1-\cos^2\alpha)=2\cos^2\alpha-1\qquad(6)
Replace \cos^2\alpha by 1-\sin^2\alpha to get a form with sine only.
\cos2\alpha=(1-\sin^2\alpha)-\sin^2\alpha=1-2\sin^2\alpha\qquad(7)

Result\cos2\alpha=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha

Use whichever form matches what is given: (6) if \cos\alpha is known, (7) if \sin\alpha is known, (5) if both are known.

Derivation Identities (8), (9), (10): squares in terms of \cos2\alpha

Aim. Express \cos^2\alpha, \sin^2\alpha and \tan^2\alpha using \cos2\alpha.

Relations (6) and (7) can be solved for the squares. From (6), add 1 and divide by 2.
2\cos^2\alpha=1+\cos2\alpha\;\Longrightarrow\;\cos^2\alpha=\dfrac{1+\cos2\alpha}{2}\;\Longrightarrow\;\cos\alpha=\pm\sqrt{\dfrac{1+\cos2\alpha}{2}}\qquad(8)
From (7), move 2\sin^2\alpha across and divide by 2.
2\sin^2\alpha=1-\cos2\alpha\;\Longrightarrow\;\sin^2\alpha=\dfrac{1-\cos2\alpha}{2}\;\Longrightarrow\;\sin\alpha=\pm\sqrt{\dfrac{1-\cos2\alpha}{2}}\qquad(9)
Divide the expression for \sin^2\alpha in (9) by the one for \cos^2\alpha in (8). The 2's cancel.
\tan^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}=\dfrac{1-\cos2\alpha}{1+\cos2\alpha}\qquad(10)

Result\cos^2\alpha=\dfrac{1+\cos2\alpha}{2},\quad\sin^2\alpha=\dfrac{1-\cos2\alpha}{2},\quad\tan^2\alpha=\dfrac{1-\cos2\alpha}{1+\cos2\alpha}

These formulas lower the power of a sine or cosine from 2 to 1 at the cost of doubling the angle. They are used again and again to rewrite powers (Example 9 and Exercise 8.2, question 7).

Derivation Identity (11): \tan2\alpha

Aim. Express \tan2\alpha in terms of \tan\alpha.

Put \beta=\alpha in identity (3).
\tan(\alpha+\alpha)=\dfrac{\tan\alpha+\tan\alpha}{1-\tan\alpha\tan\alpha}
Add the two terms in the numerator and write the product in the denominator as a square.
\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}

Result\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}\qquad(11)

Check point: is \cos2\theta=\sin2(90^\circ-\theta)?

Does \cos2\theta=\sin2(90^\circ-\theta)? Justify your answer.

Simplify the angle on the right: 2(90^\circ-\theta)=180^\circ-2\theta.
\sin2(90^\circ-\theta)=\sin(180^\circ-2\theta)
By the allied angle rule \sin(\pi-x)=\sin x (Section 8.3).
\sin(180^\circ-2\theta)=\sin2\theta
So the right side is \sin2\theta, not \cos2\theta. The two are equal only for special angles such as \theta=22.5^\circ (where \tan2\theta=1).
Counter-example: \theta=30^\circ.
\cos60^\circ=0.5\quad\text{but}\quad\sin2(90^\circ-30^\circ)=\sin120^\circ=0.866

\text{No: }\sin2(90^\circ-\theta)=\sin2\theta\neq\cos2\theta\text{ in general}

Check point: \sin120^\circ

Use the double angle formula to find the exact value of \sin120^\circ.

Write 120^\circ=2\times60^\circ and use \sin2\alpha=2\sin\alpha\cos\alpha with \alpha=60^\circ.
\sin120^\circ=2\sin60^\circ\cos60^\circ
Substitute the exact values.
=2\times\dfrac{\sqrt3}{2}\times\dfrac12=\dfrac{\sqrt3}{2}

\sin120^\circ=\dfrac{\sqrt3}{2}

(ii) Half angle identities

Derivation Identities (12) to (17): half angle identities

Aim. Express the ratios of \theta in terms of \dfrac{\theta}{2}, and the ratios of \dfrac{\theta}{2} in terms of \cos\theta.

Substitute \alpha=\dfrac{\theta}{2} in the double angle identities. For sine, use (4); the angle 2\alpha becomes \theta.
\sin\theta=\sin\left(2\times\dfrac{\theta}{2}\right)=2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}\qquad(12)
For cosine, use (5), (6) and (7).
\cos\theta=\cos^2\dfrac{\theta}{2}-\sin^2\dfrac{\theta}{2}=2\cos^2\dfrac{\theta}{2}-1=1-2\sin^2\dfrac{\theta}{2}\qquad(13)
For tangent, use (11).
\tan\theta=\tan\left(2\times\dfrac{\theta}{2}\right)=\dfrac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}\qquad(14)
From (8), with \alpha=\dfrac{\theta}{2}, the angle 2\alpha becomes \theta.
\cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}\qquad(15)
In the same way from (9).
\sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}\qquad(16)
Divide (16) by (15). The 2's cancel.
\tan\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}\qquad(17)

Result\sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}},\quad\cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}},\quad\tan\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}

These identities are useful in simplifying complex trigonometric expressions. The sign \pm is decided by the quadrant in which \dfrac{\theta}{2} lies, not by the quadrant of \theta.

(iii) Triple angle identities

The ratios \sin3\alpha, \cos3\alpha, \tan3\alpha and so on are called triple angle identities. Each is proved by writing 3\alpha=2\alpha+\alpha and using the sum identities together with the double angle identities.

Derivation Identity (18): \sin3\alpha

Aim. Prove that \sin3\alpha=3\sin\alpha-4\sin^3\alpha.

Write 3\alpha=2\alpha+\alpha and use the sine of a sum.
\sin3\alpha=\sin(2\alpha+\alpha)=\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha
Replace \sin2\alpha by (4) and \cos2\alpha by (7), the sine-only form.
=(2\sin\alpha\cos\alpha)\cos\alpha+(1-2\sin^2\alpha)\sin\alpha
Multiply out.
=2\sin\alpha\cos^2\alpha+\sin\alpha-2\sin^3\alpha
Use \cos^2\alpha=1-\sin^2\alpha so that only sines remain.
=2\sin\alpha(1-\sin^2\alpha)+\sin\alpha-2\sin^3\alpha
Multiply out again.
=2\sin\alpha-2\sin^3\alpha+\sin\alpha-2\sin^3\alpha
Collect like terms.
=3\sin\alpha-4\sin^3\alpha

Result\sin3\alpha=3\sin\alpha-4\sin^3\alpha\qquad(18)

Derivation Identity (19): \cos3\alpha

Aim. Prove that \cos3\alpha=4\cos^3\alpha-3\cos\alpha.

Write 3\alpha=2\alpha+\alpha and use the cosine of a sum.
\cos3\alpha=\cos(2\alpha+\alpha)=\cos2\alpha\cos\alpha-\sin2\alpha\sin\alpha
Replace \cos2\alpha by (6), the cosine-only form, and \sin2\alpha by (4).
=(2\cos^2\alpha-1)\cos\alpha-(2\sin\alpha\cos\alpha)\sin\alpha
Multiply out.
=2\cos^3\alpha-\cos\alpha-2\sin^2\alpha\cos\alpha
Use \sin^2\alpha=1-\cos^2\alpha so that only cosines remain.
=2\cos^3\alpha-\cos\alpha-2(1-\cos^2\alpha)\cos\alpha
Multiply out again.
=2\cos^3\alpha-\cos\alpha-2\cos\alpha+2\cos^3\alpha
Collect like terms.
=4\cos^3\alpha-3\cos\alpha

Result\cos3\alpha=4\cos^3\alpha-3\cos\alpha\qquad(19)

Derivation Identity (20): \tan3\alpha

Aim. Prove that \tan3\alpha=\dfrac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}.

Write 3\alpha=2\alpha+\alpha and use the tangent of a sum.
\tan3\alpha=\tan(2\alpha+\alpha)=\dfrac{\tan2\alpha+\tan\alpha}{1-\tan2\alpha\tan\alpha}
Replace \tan2\alpha by (11).
=\dfrac{\dfrac{2\tan\alpha}{1-\tan^2\alpha}+\tan\alpha}{1-\dfrac{2\tan\alpha}{1-\tan^2\alpha}\tan\alpha}
Put the numerator and the denominator over the common denominator 1-\tan^2\alpha.
=\dfrac{\dfrac{2\tan\alpha+\tan\alpha(1-\tan^2\alpha)}{1-\tan^2\alpha}}{\dfrac{1-\tan^2\alpha-2\tan^2\alpha}{1-\tan^2\alpha}}
Cancel the common factor 1-\tan^2\alpha from top and bottom.
=\dfrac{2\tan\alpha+\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}
Collect like terms in the numerator.
=\dfrac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}

Result\tan3\alpha=\dfrac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}\qquad(20)

Double and triple angle identities at a glance
Name	Identity
\sin2\alpha (4)	2\sin\alpha\cos\alpha
\cos2\alpha (5), (6), (7)	\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha
\tan2\alpha (11)	\dfrac{2\tan\alpha}{1-\tan^2\alpha}
\sin3\alpha (18)	3\sin\alpha-4\sin^3\alpha
\cos3\alpha (19)	4\cos^3\alpha-3\cos\alpha
\tan3\alpha (20)	\dfrac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}

Worked examples

Example 6

Given \sin\theta=\dfrac35, where \theta lies in quadrant I, find the values of \sin2\theta, \cos2\theta and \tan2\theta.

First find \cos\theta from \cos^2\theta=1-\sin^2\theta.
\cos^2\theta=1-\left(\dfrac35\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}
Cosine is positive in quadrant I.
\cos\theta=\dfrac45

(i)

Use \sin2\theta=2\sin\theta\cos\theta.
\sin2\theta=2\times\dfrac35\times\dfrac45=\dfrac{24}{25}

(ii)

Use \cos2\theta=1-2\sin^2\theta (the form that uses the given sine).
\cos2\theta=1-2\times\left(\dfrac35\right)^2=1-\dfrac{18}{25}=\dfrac{7}{25}

(iii)

Divide sine by cosine, using parts (i) and (ii). The denominators 25 cancel.
\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=\dfrac{24/25}{7/25}=\dfrac{24}{7}

\sin2\theta=\dfrac{24}{25},\ \cos2\theta=\dfrac{7}{25},\ \tan2\theta=\dfrac{24}{7}

Example 7

Use the half angle identities to find exact values for (a) \sin15^\circ (b) \tan15^\circ.

(a)

Write 15^\circ=\dfrac{30^\circ}{2} and use identity (16). The angle 15^\circ is in quadrant I, so take the plus sign.
\sin15^\circ=\sin\left(\dfrac{30^\circ}{2}\right)=\sqrt{\dfrac{1-\cos30^\circ}{2}}
Substitute \cos30^\circ=\dfrac{\sqrt3}{2}\approx0.866 (the book uses the decimal).
=\sqrt{\dfrac{1-0.866}{2}}=\sqrt{\dfrac{0.134}{2}}=\sqrt{0.067}=0.259
Exact form: \dfrac{1-\frac{\sqrt3}{2}}{2}=\dfrac{2-\sqrt3}{4}, and \sqrt{2-\sqrt3}=\dfrac{\sqrt6-\sqrt2}{2} (square the right side to check).
\sin15^\circ=\dfrac{\sqrt{2-\sqrt3}}{2}=\dfrac{\sqrt6-\sqrt2}{4}

(b)

Use \tan\dfrac{\theta}{2}=\dfrac{1-\cos\theta}{\sin\theta} (the boxed fact in the book, proved in Example 11), with \theta=30^\circ.
\tan15^\circ=\tan\left(\dfrac{30^\circ}{2}\right)=\dfrac{1-\cos30^\circ}{\sin30^\circ}
Substitute \cos30^\circ\approx0.866 and \sin30^\circ=0.5.
=\dfrac{1-0.866}{0.5}=\dfrac{0.134}{0.5}=0.268
Exact form.
\tan15^\circ=\dfrac{1-\frac{\sqrt3}{2}}{\frac12}=2-\sqrt3

\sin15^\circ=\dfrac{\sqrt6-\sqrt2}{4}\approx0.259,\qquad\tan15^\circ=2-\sqrt3\approx0.268

Example 8

For \cos\alpha=-\dfrac{7}{25} and \alpha in quadrant III, find the values of \sin\dfrac{\alpha}{2} and \cos\dfrac{\alpha}{2}.

Quadrant III means \pi<\alpha<\dfrac{3\pi}{2}. Halve every part to locate \dfrac{\alpha}{2}.
\pi<\alpha<\dfrac{3\pi}{2}\;\Longrightarrow\;\dfrac{\pi}{2}<\dfrac{\alpha}{2}<\dfrac{3\pi}{4}
So \dfrac{\alpha}{2} is in quadrant II, where \sin\dfrac{\alpha}{2}>0 and \cos\dfrac{\alpha}{2}<0.
Use (16) with the plus sign.
\sin\dfrac{\alpha}{2}=\sqrt{\dfrac{1-\cos\alpha}{2}}=\sqrt{\dfrac{1-\left(-\frac{7}{25}\right)}{2}}=\sqrt{\dfrac{32/25}{2}}=\sqrt{\dfrac{16}{25}}=\dfrac45
Use (15) with the minus sign.
\cos\dfrac{\alpha}{2}=-\sqrt{\dfrac{1+\cos\alpha}{2}}=-\sqrt{\dfrac{1+\left(-\frac{7}{25}\right)}{2}}=-\sqrt{\dfrac{18/25}{2}}=-\sqrt{\dfrac{9}{25}}=-\dfrac35

\sin\dfrac{\alpha}{2}=\dfrac45,\qquad\cos\dfrac{\alpha}{2}=-\dfrac35

Check point: \sin\dfrac{\theta}{2} from \cos\theta

Find \sin\left(\dfrac{\theta}{2}\right) when \cos\theta=-\dfrac35, where \dfrac{\pi}{2}<\theta<\pi.

Halve the interval to locate \dfrac{\theta}{2}.
\dfrac{\pi}{2}<\theta<\pi\;\Longrightarrow\;\dfrac{\pi}{4}<\dfrac{\theta}{2}<\dfrac{\pi}{2}
So \dfrac{\theta}{2} is in quadrant I and its sine is positive.
Use (16) with the plus sign.
\sin\dfrac{\theta}{2}=\sqrt{\dfrac{1-\cos\theta}{2}}=\sqrt{\dfrac{1+\frac35}{2}}=\sqrt{\dfrac{8/5}{2}}=\sqrt{\dfrac45}=\dfrac{2}{\sqrt5}

\sin\dfrac{\theta}{2}=\dfrac{2}{\sqrt5}=\dfrac{2\sqrt5}{5}

Worked example: half angles from \tan\theta and the quadrant

Given \tan\theta=-\dfrac43 with \theta in quadrant II, find the exact values of \sin\dfrac\theta2, \cos\dfrac\theta2 and \tan\dfrac\theta2.

The angle \theta (to (-3,4), in quadrant II) and the half angle \theta/2 (to (1.5,3), in quadrant I). Half of an angle between 90^\circ and 180^\circ lies between 45^\circ and 90^\circ, so both \sin\frac\theta2 and \cos\frac\theta2 are positive.

Find \sin\theta and \cos\theta first. In quadrant II, \tan\theta=\dfrac{y}{x}=-\dfrac43 means x=-3 and y=4, so r=\sqrt{9+16}=5.
\sin\theta=\dfrac45,\qquad \cos\theta=-\dfrac35
Decide the signs before calculating. 90^\circ<\theta<180^\circ gives 45^\circ<\dfrac\theta2<90^\circ, which is quadrant I, so all three half-angle ratios are positive (take the + root).
Use the half angle identity for sine.
\sin\dfrac\theta2=\sqrt{\dfrac{1-\cos\theta}{2}}=\sqrt{\dfrac{1+\frac35}{2}}=\sqrt{\dfrac45}=\dfrac{2}{\sqrt5}
Use the half angle identity for cosine.
\cos\dfrac\theta2=\sqrt{\dfrac{1+\cos\theta}{2}}=\sqrt{\dfrac{1-\frac35}{2}}=\sqrt{\dfrac15}=\dfrac{1}{\sqrt5}
Divide.
\tan\dfrac\theta2=\dfrac{2/\sqrt5}{1/\sqrt5}=2
Check with the double angle identity for tangent: \tan\theta=\dfrac{2\tan(\theta/2)}{1-\tan^2(\theta/2)}=\dfrac{4}{1-4}=-\dfrac43, as given.

\sin\dfrac\theta2=\dfrac{2}{\sqrt5},\ \cos\dfrac\theta2=\dfrac{1}{\sqrt5},\ \tan\dfrac\theta2=2

Example 9

Express 4\sin^4x in terms of an expression containing only cosines to the power 1.

Write \sin^4x as the square of \sin^2x and replace \sin^2x by identity (9) with \alpha=x.
4\sin^4x=4(\sin^2x)^2=4\left(\dfrac{1-\cos2x}{2}\right)^2
Square the bracket. The 4 cancels with the denominator 2^2.
=4\left(\dfrac{1-2\cos2x+\cos^22x}{4}\right)=1-2\cos2x+\cos^22x
A square of a cosine is still present. Use (8) with \alpha=2x: \cos^22x=\dfrac{1+\cos4x}{2}.
=1-2\cos2x+\dfrac{1+\cos4x}{2}
Put everything over the denominator 2.
=\dfrac{2-4\cos2x+1+\cos4x}{2}=\dfrac{3-4\cos2x+\cos4x}{2}

4\sin^4x=\dfrac{3-4\cos2x+\cos4x}{2}

Example 10

Find the exact value of \sin22.5^\circ\cdot\cos22.5^\circ.

Identity (4) read backwards: \sin2\alpha=2\sin\alpha\cos\alpha gives \sin\alpha\cos\alpha=\dfrac{\sin2\alpha}{2}.
\sin\alpha\cos\alpha=\dfrac{\sin2\alpha}{2}
Put \alpha=22.5^\circ, so 2\alpha=45^\circ.
\sin22.5^\circ\cos22.5^\circ=\dfrac{\sin(2\cdot22.5^\circ)}{2}=\dfrac{\sin45^\circ}{2}
Substitute \sin45^\circ=\dfrac{1}{\sqrt2}\approx0.707.
=\dfrac{1}{2\sqrt2}=\dfrac{\sqrt2}{4}\approx0.354

\sin22.5^\circ\cos22.5^\circ=\dfrac{\sqrt2}{4}\approx0.354

Check point: (\sin\theta-\cos\theta)^2

Verify (\sin\theta-\cos\theta)^2=1-\sin2\theta.

Expand the square.
(\sin\theta-\cos\theta)^2=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta
Group \sin^2\theta+\cos^2\theta=1 and use 2\sin\theta\cos\theta=\sin2\theta.
=(\sin^2\theta+\cos^2\theta)-2\sin\theta\cos\theta=1-\sin2\theta

(\sin\theta-\cos\theta)^2=1-\sin2\theta

Example 11

Prove that \tan\left(\dfrac{\theta}{2}\right)=\pm\left(\dfrac{1-\cos\theta}{\sin\theta}\right).

Start from identity (17).
\tan\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}
Multiply the numerator and the denominator under the root by 1-\cos\theta.
=\pm\sqrt{\dfrac{(1-\cos\theta)(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}}
The denominator is a difference of squares, 1-\cos^2\theta, and this is \sin^2\theta.
=\pm\sqrt{\dfrac{(1-\cos\theta)^2}{1-\cos^2\theta}}=\pm\sqrt{\dfrac{(1-\cos\theta)^2}{\sin^2\theta}}
Take the square root of the numerator and the denominator.
=\pm\left(\dfrac{1-\cos\theta}{\sin\theta}\right)

\tan\dfrac{\theta}{2}=\pm\left(\dfrac{1-\cos\theta}{\sin\theta}\right)

Example 12

Prove that \dfrac{\sin\theta\cos\theta}{0.5\cos2\theta}=\tan2\theta.

Start from the left side. Multiply the top and bottom by 2.
\text{L.H.S.}=\dfrac{\sin\theta\cos\theta}{0.5\cos2\theta}=\dfrac{2\sin\theta\cos\theta}{2\cdot0.5\cos2\theta}
The numerator is \sin2\theta by (4). In the denominator 2\times0.5=1.
=\dfrac{\sin2\theta}{\cos2\theta}
A sine over a cosine is a tangent.
=\tan2\theta=\text{R.H.S.}

\dfrac{\sin\theta\cos\theta}{0.5\cos2\theta}=\tan2\theta

8.6.1Expressing the product of sines and cosines as sums or differencesOften examined

The sum, difference and product identities turn a product of a sine or cosine with another sine or cosine into a sum or difference (this section), and the reverse (next section). They come from adding and subtracting the four sum and difference identities of Section 8.2.

We have already proved the following identities:

\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta (21)
\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta (22)
\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta (23)
\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta (24)

Derivation Identities (25) to (28)

Aim. Obtain formulas for \sin\alpha\cos\beta, \cos\alpha\sin\beta, \cos\alpha\cos\beta and \sin\alpha\sin\beta as sums or differences.

Add (21) and (22). The terms \cos\alpha\sin\beta cancel and the terms \sin\alpha\cos\beta double.
\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta\qquad(25)
Add (23) and (24). The terms \sin\alpha\sin\beta cancel.
\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta\qquad(26)
Subtract (22) from (21). The terms \sin\alpha\cos\beta cancel and the terms \cos\alpha\sin\beta double.
\sin(\alpha+\beta)-\sin(\alpha-\beta)=2\cos\alpha\sin\beta\qquad(27)
Subtract (24) from (23). The terms \cos\alpha\cos\beta cancel and the terms \sin\alpha\sin\beta double with a minus sign.
\cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta\qquad(28)
Read the four results from right to left, so that each product stands alone on the left side.

Result\begin{aligned}2\sin\alpha\cos\beta&=\sin(\alpha+\beta)+\sin(\alpha-\beta)\\2\cos\alpha\sin\beta&=\sin(\alpha+\beta)-\sin(\alpha-\beta)\\2\cos\alpha\cos\beta&=\cos(\alpha+\beta)+\cos(\alpha-\beta)\\-2\sin\alpha\sin\beta&=\cos(\alpha+\beta)-\cos(\alpha-\beta)\end{aligned}

Each product of two ratios has become a sum or difference of single ratios, of the angles \alpha+\beta and \alpha-\beta. These are the product-to-sum formulae.

Example 13

Express the product 2\cos6\theta\sin3\theta as a sum or difference of sine and cosine.

The product has the form 2\cos\alpha\sin\beta with \alpha=6\theta and \beta=3\theta, so use 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta).
2\cos6\theta\sin3\theta=\sin(6\theta+3\theta)-\sin(6\theta-3\theta)
Add and subtract the angles.
=\sin9\theta-\sin3\theta

2\cos6\theta\sin3\theta=\sin9\theta-\sin3\theta

Example 14

Simplify \sin40^\circ\cos20^\circ+\cos40^\circ\sin20^\circ after converting into sum or difference of sine and cosine.

Make each product look like a formula by writing \dfrac12\times2 in front of it.
\sin40^\circ\cos20^\circ+\cos40^\circ\sin20^\circ=\dfrac12(2\sin40^\circ\cos20^\circ)+\dfrac12(2\cos40^\circ\sin20^\circ)
Apply 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta) to the first and 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta) to the second, with \alpha=40^\circ, \beta=20^\circ.
=\dfrac12\{\sin(40^\circ+20^\circ)+\sin(40^\circ-20^\circ)\}+\dfrac12\{\sin(40^\circ+20^\circ)-\sin(40^\circ-20^\circ)\}
Add and subtract the angles.
=\dfrac12\{\sin60^\circ+\sin20^\circ\}+\dfrac12\{\sin60^\circ-\sin20^\circ\}
Combine. The terms \sin20^\circ cancel.
=\dfrac12\{\sin60^\circ+\sin20^\circ+\sin60^\circ-\sin20^\circ\}=\dfrac12\{2\sin60^\circ\}=\sin60^\circ=\dfrac{\sqrt3}{2}

\dfrac{\sqrt3}{2}

Check point: \sin60^\circ\cos30^\circ

Express \sin60^\circ\cos30^\circ as a sum or difference of sine and cosine and simplify.

Use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta) with \alpha=60^\circ, \beta=30^\circ after dividing by 2.
\sin60^\circ\cos30^\circ=\dfrac12\{\sin(60^\circ+30^\circ)+\sin(60^\circ-30^\circ)\}=\dfrac12\{\sin90^\circ+\sin30^\circ\}
Substitute \sin90^\circ=1 and \sin30^\circ=\dfrac12.
=\dfrac12\left\{1+\dfrac12\right\}=\dfrac12\cdot\dfrac32=\dfrac34
Check: \sin60^\circ\cos30^\circ=\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2}=\dfrac34.

\dfrac12(\sin90^\circ+\sin30^\circ)=\dfrac34

Worked example: a product of three sines with angles 60^\circ\pm\theta

Show that \sin\theta\,\sin(60^\circ-\theta)\,\sin(60^\circ+\theta)=\dfrac14\sin3\theta.

Start with the two sines that form a pair and use 2\sin A\sin B=\cos(A-B)-\cos(A+B) with A=60^\circ+\theta and B=60^\circ-\theta, so A-B=2\theta and A+B=120^\circ.
\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\dfrac12\left[\cos2\theta-\cos120^\circ\right]=\dfrac12\cos2\theta+\dfrac14
Multiply by the remaining factor \sin\theta.
\text{LHS}=\dfrac12\sin\theta\cos2\theta+\dfrac14\sin\theta
Convert \sin\theta\cos2\theta with 2\sin A\cos B=\sin(A+B)+\sin(A-B), taking A=\theta and B=2\theta. Note \sin(-\theta)=-\sin\theta.
\sin\theta\cos2\theta=\dfrac12\left[\sin3\theta+\sin(-\theta)\right]=\dfrac12\left(\sin3\theta-\sin\theta\right)
Substitute and simplify. The \sin\theta terms cancel.
\text{LHS}=\dfrac14\sin3\theta-\dfrac14\sin\theta+\dfrac14\sin\theta=\dfrac14\sin3\theta=\text{RHS}
Quick check at \theta=30^\circ: \sin30^\circ\sin30^\circ\sin90^\circ=\dfrac14 and \dfrac14\sin90^\circ=\dfrac14.

The product \sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta) (solid) and \frac14\sin3\theta (dashed) lie on the same curve, which is what the proof shows.

\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\dfrac14\sin3\theta

8.6.2Expressing the sums or differences of sines and cosines as productOften examined

Derivation Sum to product formulae

Aim. Express \sin p+\sin q, \sin p-\sin q, \cos p+\cos q and \cos p-\cos q as products.

Let \alpha=\dfrac{p+q}{2} and \beta=\dfrac{p-q}{2}. Add them, and subtract them, to find \alpha+\beta and \alpha-\beta.
\alpha+\beta=\dfrac{p+q}{2}+\dfrac{p-q}{2}=\dfrac{2p}{2}=p,\qquad\alpha-\beta=\dfrac{p+q}{2}-\dfrac{p-q}{2}=\dfrac{2q}{2}=q
Substitute these values into identity (25), \sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta.
\sin p+\sin q=2\sin\left(\dfrac{p+q}{2}\right)\cos\left(\dfrac{p-q}{2}\right)
Into identity (27), \sin(\alpha+\beta)-\sin(\alpha-\beta)=2\cos\alpha\sin\beta.
\sin p-\sin q=2\cos\left(\dfrac{p+q}{2}\right)\sin\left(\dfrac{p-q}{2}\right)
Into identity (26), \cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta.
\cos p+\cos q=2\cos\left(\dfrac{p+q}{2}\right)\cos\left(\dfrac{p-q}{2}\right)
Into identity (28), \cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta.
\cos p-\cos q=-2\sin\left(\dfrac{p+q}{2}\right)\sin\left(\dfrac{p-q}{2}\right)

Result\begin{aligned}\sin p+\sin q&=2\sin\dfrac{p+q}{2}\cos\dfrac{p-q}{2}\\\sin p-\sin q&=2\cos\dfrac{p+q}{2}\sin\dfrac{p-q}{2}\\\cos p+\cos q&=2\cos\dfrac{p+q}{2}\cos\dfrac{p-q}{2}\\\cos p-\cos q&=-2\sin\dfrac{p+q}{2}\sin\dfrac{p-q}{2}\end{aligned}

These identities are known as the sum to product formulae. Half the sum and half the difference of the two angles appear on the right.

Example 15

Express \cos45^\circ-\cos15^\circ as a product.

Use \cos p-\cos q=-2\sin\dfrac{p+q}{2}\sin\dfrac{p-q}{2} with p=45^\circ and q=15^\circ.
\cos45^\circ-\cos15^\circ=-2\sin\left(\dfrac{45^\circ+15^\circ}{2}\right)\sin\left(\dfrac{45^\circ-15^\circ}{2}\right)
Evaluate the half sum and the half difference.
=-2\sin30^\circ\sin15^\circ

-2\sin30^\circ\sin15^\circ

Example 16

Show that \dfrac{\sin x+\sin y}{\cos x+\cos y}=\tan\dfrac{x+y}{2}.

Write the numerator with \sin p+\sin q=2\sin\dfrac{p+q}{2}\cos\dfrac{p-q}{2} and the denominator with \cos p+\cos q=2\cos\dfrac{p+q}{2}\cos\dfrac{p-q}{2}.
\dfrac{\sin x+\sin y}{\cos x+\cos y}=\dfrac{2\sin\dfrac{x+y}{2}\cos\dfrac{x-y}{2}}{2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}}
Cancel the common factor 2\cos\dfrac{x-y}{2}.
=\dfrac{\sin\dfrac{x+y}{2}}{\cos\dfrac{x+y}{2}}
Sine over cosine is tangent.
=\tan\dfrac{x+y}{2}

\dfrac{\sin x+\sin y}{\cos x+\cos y}=\tan\dfrac{x+y}{2}

Example 17

Show that \cos6\alpha+\cos5\alpha+\cos3\alpha+\cos2\alpha=4\cos(4\alpha)\cos(1.5\alpha)\cos(0.5\alpha).

Group the terms in pairs whose half sum is the same (6\alpha with 2\alpha, and 5\alpha with 3\alpha, both have half sum 4\alpha).
\cos6\alpha+\cos5\alpha+\cos3\alpha+\cos2\alpha=(\cos6\alpha+\cos2\alpha)+(\cos5\alpha+\cos3\alpha)
Apply \cos p+\cos q=2\cos\dfrac{p+q}{2}\cos\dfrac{p-q}{2} to each bracket.
=2\cos\dfrac{6\alpha+2\alpha}{2}\cos\dfrac{6\alpha-2\alpha}{2}+2\cos\dfrac{5\alpha+3\alpha}{2}\cos\dfrac{5\alpha-3\alpha}{2}
Simplify the angles.
=2\cos4\alpha\cos2\alpha+2\cos4\alpha\cos\alpha
Take out the common factor 2\cos4\alpha.
=2\cos4\alpha(\cos2\alpha+\cos\alpha)
Apply the same formula to the bracket: half sum \dfrac{2\alpha+\alpha}{2}=1.5\alpha and half difference \dfrac{2\alpha-\alpha}{2}=0.5\alpha.
=2\cos4\alpha\times2\cos\dfrac{2\alpha+\alpha}{2}\cos\dfrac{2\alpha-\alpha}{2}
Multiply the 2's.
=4\cos(4\alpha)\cos(1.5\alpha)\cos(0.5\alpha)

4\cos(4\alpha)\cos(1.5\alpha)\cos(0.5\alpha)

Example 18

Show that \sin70^\circ\sin30^\circ\cos20^\circ\cos10^\circ=\dfrac14\cos10^\circ+\dfrac{\sqrt3}{16}+\dfrac18\sin40^\circ.

Replace \sin30^\circ by \dfrac12.
\sin70^\circ\sin30^\circ\cos20^\circ\cos10^\circ=\dfrac12(\sin70^\circ\cos20^\circ\cos10^\circ)
Insert a factor 2 so that 2\sin70^\circ\cos20^\circ appears. Then use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta).
=\dfrac14(2\sin70^\circ\cos20^\circ)\cos10^\circ=\dfrac14[\sin(70^\circ+20^\circ)+\sin(70^\circ-20^\circ)]\cos10^\circ
Use \sin90^\circ=1.
=\dfrac14[\sin90^\circ+\sin50^\circ]\cos10^\circ=\dfrac14[1+\sin50^\circ]\cos10^\circ
Multiply out. In the second term insert a factor 2 so that 2\sin50^\circ\cos10^\circ appears.
=\dfrac14\cos10^\circ+\dfrac14\sin50^\circ\cos10^\circ=\dfrac14\cos10^\circ+\dfrac18\times2\sin50^\circ\cos10^\circ
Apply the same product-to-sum formula again.
=\dfrac14\cos10^\circ+\dfrac18[\sin(50^\circ+10^\circ)+\sin(50^\circ-10^\circ)]=\dfrac14\cos10^\circ+\dfrac18[\sin60^\circ+\sin40^\circ]
Use \sin60^\circ=\dfrac{\sqrt3}{2} and multiply by \dfrac18.
=\dfrac14\cos10^\circ+\dfrac18\left(\dfrac{\sqrt3}{2}+\sin40^\circ\right)=\dfrac14\cos10^\circ+\dfrac{\sqrt3}{16}+\dfrac18\sin40^\circ

\dfrac14\cos10^\circ+\dfrac{\sqrt3}{16}+\dfrac18\sin40^\circ

Check point: \cos195^\circ\cos105^\circ by two methods

Provide two different methods of calculating \cos195^\circ\cos105^\circ, one of which uses the product to sum formula. Which method is easier?

Method 1: product to sum

Use 2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta) with \alpha=195^\circ, \beta=105^\circ.
\cos195^\circ\cos105^\circ=\dfrac12[\cos300^\circ+\cos90^\circ]
Use \cos300^\circ=\cos(360^\circ-60^\circ)=\cos60^\circ=\dfrac12 and \cos90^\circ=0.
=\dfrac12\left[\dfrac12+0\right]=\dfrac14

Method 2: allied angles and the double angle formula

Write each angle with an allied angle: \cos(180^\circ+15^\circ)=-\cos15^\circ and \cos(90^\circ+15^\circ)=-\sin15^\circ.
\cos195^\circ\cos105^\circ=(-\cos15^\circ)(-\sin15^\circ)=\sin15^\circ\cos15^\circ
Use \sin\alpha\cos\alpha=\dfrac{\sin2\alpha}{2}.
=\dfrac{\sin30^\circ}{2}=\dfrac14

Both methods give the same value. Method 1 is shorter: it needs one formula and no allied angles. Method 2 needs two allied angle rules and the double angle formula, though every step is simple.

\dfrac14

Formula sheet

Distance formula

d=AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

A(x_1,y_1) and B(x_2,y_2) are two points in the plane

Fundamental law of trigonometry

\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta

Proved for \alpha>\beta and used for all angles

Cosine of a sum

\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta

Sine of a sum and a difference

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

The signs on the two sides are the same

Tangent of a sum and a difference

\tan(\alpha\pm\beta)=\dfrac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}

Valid when \cos\alpha\neq0, \cos\beta\neq0 and \tan(\alpha\pm\beta) is defined

Negative angles

\sin(-\theta)=-\sin\theta,\quad\cos(-\theta)=\cos\theta,\quad\tan(-\theta)=-\tan\theta

Complementary angles

\sin\left(\dfrac{\pi}{2}-\theta\right)=\cos\theta,\quad\cos\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta,\quad\tan\left(\dfrac{\pi}{2}-\theta\right)=\cot\theta

True for all angles \theta

Allied angles at \pi and 2\pi

\sin(\pi\pm\theta)=\mp\sin\theta,\quad\cos(\pi\pm\theta)=-\cos\theta,\quad\tan(\pi\pm\theta)=\pm\tan\theta

Odd multiples of a right angle change a ratio to its co-ratio; even multiples do not. The sign follows the quadrant (Section 8.3 table)

Form r\sin(\theta+\phi)

a\sin\theta+b\cos\theta=r\sin(\theta+\phi),\quad r=\sqrt{a^2+b^2},\ \phi=\tan^{-1}\left(\dfrac ba\right)

a=r\cos\phi and b=r\sin\phi

Double angle identities

\begin{aligned}\sin2\alpha&=2\sin\alpha\cos\alpha\\ \cos2\alpha&=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha\\ \tan2\alpha&=\dfrac{2\tan\alpha}{1-\tan^2\alpha}\end{aligned}

Squares in terms of \cos2\alpha

\cos^2\alpha=\dfrac{1+\cos2\alpha}{2},\quad\sin^2\alpha=\dfrac{1-\cos2\alpha}{2},\quad\tan^2\alpha=\dfrac{1-\cos2\alpha}{1+\cos2\alpha}

Used to lower the power of a sine or cosine

Half angle identities

\sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}},\quad\cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}},\quad\tan\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}

The sign is chosen from the quadrant of \dfrac{\theta}{2}

Tangent of a half angle

\tan\dfrac{\theta}{2}=\dfrac{1-\cos\theta}{\sin\theta}

Proved in Example 11

Triple angle identities

\begin{aligned}\sin3\alpha&=3\sin\alpha-4\sin^3\alpha\\ \cos3\alpha&=4\cos^3\alpha-3\cos\alpha\\ \tan3\alpha&=\dfrac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}\end{aligned}

Product to sum

\begin{aligned}2\sin\alpha\cos\beta&=\sin(\alpha+\beta)+\sin(\alpha-\beta)\\2\cos\alpha\sin\beta&=\sin(\alpha+\beta)-\sin(\alpha-\beta)\\2\cos\alpha\cos\beta&=\cos(\alpha+\beta)+\cos(\alpha-\beta)\\-2\sin\alpha\sin\beta&=\cos(\alpha+\beta)-\cos(\alpha-\beta)\end{aligned}

Sum to product

\begin{aligned}\sin p+\sin q&=2\sin\dfrac{p+q}{2}\cos\dfrac{p-q}{2}\\\sin p-\sin q&=2\cos\dfrac{p+q}{2}\sin\dfrac{p-q}{2}\\\cos p+\cos q&=2\cos\dfrac{p+q}{2}\cos\dfrac{p-q}{2}\\\cos p-\cos q&=-2\sin\dfrac{p+q}{2}\sin\dfrac{p-q}{2}\end{aligned}

Key terms

Fundamental law of trigonometry: The identity \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta, from which the other identities of the unit are deduced.
Trigonometric identity: An equation involving trigonometric ratios that is true for all angles for which both sides are defined.
Distance formula: AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}, the distance between A(x_1,y_1) and B(x_2,y_2).
Allied angles: Angles connected with a basic angle \theta by a right angle or its multiple, such as \dfrac{\pi}{2}\pm\theta, \pi\pm\theta, \dfrac{3\pi}{2}\pm\theta and 2\pi\pm\theta.
Co-ratio: The partner of a ratio under a right-angle change: sine and cosine, tangent and cotangent, secant and cosecant.
Double angle identities: Identities for \sin2\alpha, \cos2\alpha and \tan2\alpha in terms of the ratios of \alpha.
Half angle identities: Identities for the ratios of \dfrac{\theta}{2} in terms of the ratios of \theta.
Triple angle identities: Identities for \sin3\alpha, \cos3\alpha and \tan3\alpha in terms of the ratios of \alpha.
Product to sum formulae: Identities that write 2\sin\alpha\cos\beta, 2\cos\alpha\sin\beta, 2\cos\alpha\cos\beta and -2\sin\alpha\sin\beta as sums or differences.
Sum to product formulae: Identities that write \sin p\pm\sin q and \cos p\pm\cos q as products of ratios of \dfrac{p+q}{2} and \dfrac{p-q}{2}.

Exercise 8.1p. 202–203

The fundamental law \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta and the identities that follow from it: the sum and difference formulas, the allied-angle results and the form r\sin(\theta+\phi). Find any missing sine or cosine from \sin^2\theta+\cos^2\theta=1, choosing the sign from the quadrant, then substitute into the formula.

Find the values of \cos(\alpha\pm\beta), \sin(\alpha\pm\beta) and \tan(\alpha\pm\beta) for each given pair of angles.

Solution

(i) \alpha=180^\circ,\ \beta=60^\circ

Write down the sine and cosine of each angle.
\cos\alpha=-1,\ \sin\alpha=0,\ \cos\beta=\tfrac12,\ \sin\beta=\tfrac{\sqrt3}{2}
Cosine of the sum, from identity (2): \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos(\alpha+\beta)=\left(-1\right)\left(\tfrac12\right)-\left(0\right)\left(\tfrac{\sqrt3}{2}\right)=-\tfrac12
Cosine of the difference, from the fundamental law (1): \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\left(-1\right)\left(\tfrac12\right)+\left(0\right)\left(\tfrac{\sqrt3}{2}\right)=-\tfrac12
Sine of the sum, from identity (3): \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\sin(\alpha+\beta)=\left(0\right)\left(\tfrac12\right)+\left(-1\right)\left(\tfrac{\sqrt3}{2}\right)=-\tfrac{\sqrt3}{2}
Sine of the difference, from identity (4): \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin(\alpha-\beta)=\left(0\right)\left(\tfrac12\right)-\left(-1\right)\left(\tfrac{\sqrt3}{2}\right)=\tfrac{\sqrt3}{2}
Tangent is sine divided by cosine, so use the two results above.
\tan(\alpha+\beta)=\dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}=\sqrt3
Do the same for the difference.
\tan(\alpha-\beta)=\dfrac{\sin(\alpha-\beta)}{\cos(\alpha-\beta)}=-\sqrt3

\begin{aligned}\cos(\alpha\pm\beta)&:\ -\tfrac12,\ -\tfrac12\\ \sin(\alpha\pm\beta)&:\ -\tfrac{\sqrt3}{2},\ \tfrac{\sqrt3}{2}\\ \tan(\alpha\pm\beta)&:\ \sqrt3,\ -\sqrt3\end{aligned}

(ii) \alpha=90^\circ,\ \beta=60^\circ

Write down the sine and cosine of each angle.
\cos\alpha=0,\ \sin\alpha=1,\ \cos\beta=\tfrac12,\ \sin\beta=\tfrac{\sqrt3}{2}
Cosine of the sum, from identity (2): \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos(\alpha+\beta)=\left(0\right)\left(\tfrac12\right)-\left(1\right)\left(\tfrac{\sqrt3}{2}\right)=-\tfrac{\sqrt3}{2}
Cosine of the difference, from the fundamental law (1): \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\left(0\right)\left(\tfrac12\right)+\left(1\right)\left(\tfrac{\sqrt3}{2}\right)=\tfrac{\sqrt3}{2}
Sine of the sum, from identity (3): \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\sin(\alpha+\beta)=\left(1\right)\left(\tfrac12\right)+\left(0\right)\left(\tfrac{\sqrt3}{2}\right)=\tfrac12
Sine of the difference, from identity (4): \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin(\alpha-\beta)=\left(1\right)\left(\tfrac12\right)-\left(0\right)\left(\tfrac{\sqrt3}{2}\right)=\tfrac12
Tangent is sine divided by cosine, so use the two results above.
\tan(\alpha+\beta)=\dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}=-\tfrac{1}{\sqrt3}
Do the same for the difference.
\tan(\alpha-\beta)=\dfrac{\sin(\alpha-\beta)}{\cos(\alpha-\beta)}=\tfrac{1}{\sqrt3}

\begin{aligned}\cos(\alpha\pm\beta)&:\ -\tfrac{\sqrt3}{2},\ \tfrac{\sqrt3}{2}\\ \sin(\alpha\pm\beta)&:\ \tfrac12,\ \tfrac12\\ \tan(\alpha\pm\beta)&:\ -\tfrac{1}{\sqrt3},\ \tfrac{1}{\sqrt3}\end{aligned}

(iii) \alpha=180^\circ,\ \beta=30^\circ

Write down the sine and cosine of each angle.
\cos\alpha=-1,\ \sin\alpha=0,\ \cos\beta=\tfrac{\sqrt3}{2},\ \sin\beta=\tfrac12
Cosine of the sum, from identity (2): \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos(\alpha+\beta)=\left(-1\right)\left(\tfrac{\sqrt3}{2}\right)-\left(0\right)\left(\tfrac12\right)=-\tfrac{\sqrt3}{2}
Cosine of the difference, from the fundamental law (1): \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\left(-1\right)\left(\tfrac{\sqrt3}{2}\right)+\left(0\right)\left(\tfrac12\right)=-\tfrac{\sqrt3}{2}
Sine of the sum, from identity (3): \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\sin(\alpha+\beta)=\left(0\right)\left(\tfrac{\sqrt3}{2}\right)+\left(-1\right)\left(\tfrac12\right)=-\tfrac12
Sine of the difference, from identity (4): \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin(\alpha-\beta)=\left(0\right)\left(\tfrac{\sqrt3}{2}\right)-\left(-1\right)\left(\tfrac12\right)=\tfrac12
Tangent is sine divided by cosine, so use the two results above.
\tan(\alpha+\beta)=\dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}=\tfrac{1}{\sqrt3}
Do the same for the difference.
\tan(\alpha-\beta)=\dfrac{\sin(\alpha-\beta)}{\cos(\alpha-\beta)}=-\tfrac{1}{\sqrt3}

\begin{aligned}\cos(\alpha\pm\beta)&:\ -\tfrac{\sqrt3}{2},\ -\tfrac{\sqrt3}{2}\\ \sin(\alpha\pm\beta)&:\ -\tfrac12,\ \tfrac12\\ \tan(\alpha\pm\beta)&:\ \tfrac{1}{\sqrt3},\ -\tfrac{1}{\sqrt3}\end{aligned}

(iv) \alpha=\pi,\ \beta=\dfrac{\pi}{3}

Write down the sine and cosine of each angle.
\cos\alpha=-1,\ \sin\alpha=0,\ \cos\beta=\tfrac12,\ \sin\beta=\tfrac{\sqrt3}{2}
Cosine of the sum, from identity (2): \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos(\alpha+\beta)=\left(-1\right)\left(\tfrac12\right)-\left(0\right)\left(\tfrac{\sqrt3}{2}\right)=-\tfrac12
Cosine of the difference, from the fundamental law (1): \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\left(-1\right)\left(\tfrac12\right)+\left(0\right)\left(\tfrac{\sqrt3}{2}\right)=-\tfrac12
Sine of the sum, from identity (3): \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\sin(\alpha+\beta)=\left(0\right)\left(\tfrac12\right)+\left(-1\right)\left(\tfrac{\sqrt3}{2}\right)=-\tfrac{\sqrt3}{2}
Sine of the difference, from identity (4): \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin(\alpha-\beta)=\left(0\right)\left(\tfrac12\right)-\left(-1\right)\left(\tfrac{\sqrt3}{2}\right)=\tfrac{\sqrt3}{2}
Tangent is sine divided by cosine, so use the two results above.
\tan(\alpha+\beta)=\dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}=\sqrt3
Do the same for the difference.
\tan(\alpha-\beta)=\dfrac{\sin(\alpha-\beta)}{\cos(\alpha-\beta)}=-\sqrt3

\begin{aligned}\cos(\alpha\pm\beta)&:\ -\tfrac12,\ -\tfrac12\\ \sin(\alpha\pm\beta)&:\ -\tfrac{\sqrt3}{2},\ \tfrac{\sqrt3}{2}\\ \tan(\alpha\pm\beta)&:\ \sqrt3,\ -\sqrt3\end{aligned}

(v) \alpha=\dfrac{\pi}{2},\ \beta=\dfrac{\pi}{6}

Write down the sine and cosine of each angle.
\cos\alpha=0,\ \sin\alpha=1,\ \cos\beta=\tfrac{\sqrt3}{2},\ \sin\beta=\tfrac12
Cosine of the sum, from identity (2): \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos(\alpha+\beta)=\left(0\right)\left(\tfrac{\sqrt3}{2}\right)-\left(1\right)\left(\tfrac12\right)=-\tfrac12
Cosine of the difference, from the fundamental law (1): \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\left(0\right)\left(\tfrac{\sqrt3}{2}\right)+\left(1\right)\left(\tfrac12\right)=\tfrac12
Sine of the sum, from identity (3): \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\sin(\alpha+\beta)=\left(1\right)\left(\tfrac{\sqrt3}{2}\right)+\left(0\right)\left(\tfrac12\right)=\tfrac{\sqrt3}{2}
Sine of the difference, from identity (4): \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin(\alpha-\beta)=\left(1\right)\left(\tfrac{\sqrt3}{2}\right)-\left(0\right)\left(\tfrac12\right)=\tfrac{\sqrt3}{2}
Tangent is sine divided by cosine, so use the two results above.
\tan(\alpha+\beta)=\dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}=-\sqrt3
Do the same for the difference.
\tan(\alpha-\beta)=\dfrac{\sin(\alpha-\beta)}{\cos(\alpha-\beta)}=\sqrt3

\begin{aligned}\cos(\alpha\pm\beta)&:\ -\tfrac12,\ \tfrac12\\ \sin(\alpha\pm\beta)&:\ \tfrac{\sqrt3}{2},\ \tfrac{\sqrt3}{2}\\ \tan(\alpha\pm\beta)&:\ -\sqrt3,\ \sqrt3\end{aligned}

(vi) \alpha=\dfrac{3\pi}{2},\ \beta=\dfrac{\pi}{4}

Write down the sine and cosine of each angle.
\cos\alpha=0,\ \sin\alpha=-1,\ \cos\beta=\tfrac{1}{\sqrt2},\ \sin\beta=\tfrac{1}{\sqrt2}
Cosine of the sum, from identity (2): \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos(\alpha+\beta)=\left(0\right)\left(\tfrac{1}{\sqrt2}\right)-\left(-1\right)\left(\tfrac{1}{\sqrt2}\right)=\tfrac{1}{\sqrt2}
Cosine of the difference, from the fundamental law (1): \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\left(0\right)\left(\tfrac{1}{\sqrt2}\right)+\left(-1\right)\left(\tfrac{1}{\sqrt2}\right)=-\tfrac{1}{\sqrt2}
Sine of the sum, from identity (3): \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\sin(\alpha+\beta)=\left(-1\right)\left(\tfrac{1}{\sqrt2}\right)+\left(0\right)\left(\tfrac{1}{\sqrt2}\right)=-\tfrac{1}{\sqrt2}
Sine of the difference, from identity (4): \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin(\alpha-\beta)=\left(-1\right)\left(\tfrac{1}{\sqrt2}\right)-\left(0\right)\left(\tfrac{1}{\sqrt2}\right)=-\tfrac{1}{\sqrt2}
Tangent is sine divided by cosine, so use the two results above.
\tan(\alpha+\beta)=\dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}=-1
Do the same for the difference.
\tan(\alpha-\beta)=\dfrac{\sin(\alpha-\beta)}{\cos(\alpha-\beta)}=1

\begin{aligned}\cos(\alpha\pm\beta)&:\ \tfrac{1}{\sqrt2},\ -\tfrac{1}{\sqrt2}\\ \sin(\alpha\pm\beta)&:\ -\tfrac{1}{\sqrt2},\ -\tfrac{1}{\sqrt2}\\ \tan(\alpha\pm\beta)&:\ -1,\ 1\end{aligned}

Book answer. The key gives these in allied-angle form: (i) \cos(180^\circ\pm60^\circ)=-\cos60^\circ, \sin(180^\circ+60^\circ)=-\sin60^\circ, \sin(180^\circ-60^\circ)=\sin60^\circ, \tan(180^\circ+60^\circ)=\tan60^\circ, \tan(180^\circ-60^\circ)=-\tan60^\circ; (ii) \cos(90^\circ+60^\circ)=-\sin60^\circ, \cos(90^\circ-60^\circ)=\sin60^\circ, \sin(90^\circ\pm60^\circ)=\cos60^\circ, \tan(90^\circ+60^\circ)=-\cot60^\circ, \tan(90^\circ-60^\circ)=\cot60^\circ; and similarly for (iii) to (vi). Every one agrees with the numerical values found above.

Use the fundamental law to find exact values (parts a to d).

Solution

(a) Find the exact value of \cos15^\circ by using \cos(45^\circ-30^\circ).

Apply the fundamental law \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta with \alpha=45^\circ, \beta=30^\circ.
\cos15^\circ=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ
Substitute the exact values \cos45^\circ=\sin45^\circ=\tfrac{1}{\sqrt2}, \cos30^\circ=\tfrac{\sqrt3}{2}, \sin30^\circ=\tfrac12.
\cos15^\circ=\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2}+\dfrac{1}{\sqrt2}\cdot\dfrac12
Add the two fractions, which have the same denominator.
\cos15^\circ=\dfrac{\sqrt3+1}{2\sqrt2}

\dfrac{1+\sqrt3}{2\sqrt2}

(b) Use the value of \cos15^\circ found in (a) to find \cos165^\circ by using \cos(180^\circ-15^\circ).

Apply the fundamental law with \alpha=180^\circ, \beta=15^\circ.
\cos165^\circ=\cos180^\circ\cos15^\circ+\sin180^\circ\sin15^\circ
Put \cos180^\circ=-1 and \sin180^\circ=0.
\cos165^\circ=(-1)\cos15^\circ+0=-\cos15^\circ
Use the value from (a).
\cos165^\circ=-\dfrac{1+\sqrt3}{2\sqrt2}

-\dfrac{1+\sqrt3}{2\sqrt2}

Apply the fundamental law with \alpha=360^\circ, \beta=15^\circ.
\cos345^\circ=\cos360^\circ\cos15^\circ+\sin360^\circ\sin15^\circ
Put \cos360^\circ=1 and \sin360^\circ=0.
\cos345^\circ=(1)\cos15^\circ+0=\cos15^\circ
Use the value from (a).
\cos345^\circ=\dfrac{1+\sqrt3}{2\sqrt2}

\dfrac{1+\sqrt3}{2\sqrt2}

(d) Find the exact value of \sin75^\circ.

75^\circ=90^\circ-15^\circ. By the allied-angle result \sin\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta (the sine of an angle equals the cosine of its complement):
\sin75^\circ=\sin(90^\circ-15^\circ)=\cos15^\circ
Use the value from (a).
\sin75^\circ=\dfrac{1+\sqrt3}{2\sqrt2}

\dfrac{1+\sqrt3}{2\sqrt2}

Book answer. (a) \cos15^\circ=\dfrac{1+\sqrt3}{2\sqrt2} (b) \cos165^\circ=-\dfrac{1+\sqrt3}{2\sqrt2} (c) \cos345^\circ=\dfrac{1+\sqrt3}{2\sqrt2} (d) \sin75^\circ=\dfrac{1+\sqrt3}{2\sqrt2}

Note. The same number can be written \dfrac{\sqrt6+\sqrt2}{4}\approx0.966 (multiply top and bottom by \sqrt2).

Exact values of \cos120^\circ, \sin120^\circ, \tan120^\circ, \cos75^\circ, \cos105^\circ, \cos285^\circ and \sin15^\circ (parts a to f).

Solution

(a) Find the exact value of \cos120^\circ by using \cos(180^\circ-60^\circ) and \cos(90^\circ+30^\circ).

First way. Fundamental law with \alpha=180^\circ, \beta=60^\circ.
\cos(180^\circ-60^\circ)=\cos180^\circ\cos60^\circ+\sin180^\circ\sin60^\circ=(-1)\left(\tfrac12\right)+0=-\tfrac12
Second way. Identity (2), \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta, with \alpha=90^\circ, \beta=30^\circ.
\cos(90^\circ+30^\circ)=\cos90^\circ\cos30^\circ-\sin90^\circ\sin30^\circ=0-(1)\left(\tfrac12\right)=-\tfrac12
Both ways give the same value, which is a useful check.
\cos120^\circ=-\dfrac12

\cos120^\circ=-\dfrac12

(b) Find the exact value of \sin120^\circ and then \tan120^\circ.

Write 120^\circ=180^\circ-60^\circ and use identity (4), \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin120^\circ=\sin180^\circ\cos60^\circ-\cos180^\circ\sin60^\circ=0-(-1)\left(\tfrac{\sqrt3}{2}\right)=\tfrac{\sqrt3}{2}
Divide the sine by the cosine found in (a).
\tan120^\circ=\dfrac{\sin120^\circ}{\cos120^\circ}=\dfrac{\sqrt3/2}{-1/2}=-\sqrt3

\sin120^\circ=\dfrac{\sqrt3}{2},\qquad\tan120^\circ=-\sqrt3

Fundamental law with \alpha=120^\circ, \beta=45^\circ.
\cos75^\circ=\cos120^\circ\cos45^\circ+\sin120^\circ\sin45^\circ
Substitute the values of (a) and (b).
\cos75^\circ=\left(-\tfrac12\right)\left(\tfrac{1}{\sqrt2}\right)+\left(\tfrac{\sqrt3}{2}\right)\left(\tfrac{1}{\sqrt2}\right)
Combine over the common denominator 2\sqrt2.
\cos75^\circ=\dfrac{\sqrt3-1}{2\sqrt2}

\cos75^\circ=\dfrac{\sqrt3-1}{2\sqrt2}

(d) Use the value of \cos75^\circ found in (c) to find \cos105^\circ by using \cos(180^\circ-75^\circ).

Fundamental law with \alpha=180^\circ, \beta=75^\circ.
\cos105^\circ=\cos180^\circ\cos75^\circ+\sin180^\circ\sin75^\circ=-\cos75^\circ
Use the value from (c).
\cos105^\circ=-\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac{1-\sqrt3}{2\sqrt2}

\cos105^\circ=\dfrac{1-\sqrt3}{2\sqrt2}

(e) Use the value of \cos75^\circ found in (c) to find \cos285^\circ by using \cos(360^\circ-75^\circ).

Fundamental law with \alpha=360^\circ, \beta=75^\circ.
\cos285^\circ=\cos360^\circ\cos75^\circ+\sin360^\circ\sin75^\circ=\cos75^\circ
Use the value from (c).
\cos285^\circ=\dfrac{\sqrt3-1}{2\sqrt2}

\cos285^\circ=\dfrac{\sqrt3-1}{2\sqrt2}

(f) Find the exact value of \sin15^\circ.

15^\circ=90^\circ-75^\circ, so the sine of 15^\circ is the cosine of its complement 75^\circ (allied angle \sin\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta).
\sin15^\circ=\sin(90^\circ-75^\circ)=\cos75^\circ
Use the value from (c).
\sin15^\circ=\dfrac{\sqrt3-1}{2\sqrt2}

\sin15^\circ=\dfrac{\sqrt3-1}{2\sqrt2}

Book answer. (a) \cos120^\circ=-\tfrac12 (b) \sin120^\circ=\tfrac{\sqrt3}{2}, \tan120^\circ=-\sqrt3 (c) \cos75^\circ=\dfrac{\sqrt3-1}{2\sqrt2} (d) \cos105^\circ=\dfrac{1-\sqrt3}{2\sqrt2} (e) \cos285^\circ=\dfrac{\sqrt3-1}{2\sqrt2} (f) \sin15^\circ=\dfrac{\sqrt3-1}{2\sqrt2}

Rewrite as a single expression.

Solution

(i) \cos6\theta\cos3\theta-\sin6\theta\sin3\theta

The expression has the form \cos\alpha\cos\beta-\sin\alpha\sin\beta with \alpha=6\theta, \beta=3\theta, which is \cos(\alpha+\beta) by identity (2).
\cos6\theta\cos3\theta-\sin6\theta\sin3\theta=\cos(6\theta+3\theta)
Add the angles.
=\cos9\theta

\cos9\theta

(ii) \cos7\theta\cos2\theta+\sin7\theta\sin2\theta

The form \cos\alpha\cos\beta+\sin\alpha\sin\beta is \cos(\alpha-\beta) (the fundamental law), here with \alpha=7\theta, \beta=2\theta.
\cos7\theta\cos2\theta+\sin7\theta\sin2\theta=\cos(7\theta-2\theta)
Subtract the angles.
=\cos5\theta

\cos5\theta

(iii) \sin\left(\dfrac{\theta}{3}\right)\cos\left(\dfrac{\theta}{6}\right)+\cos\left(\dfrac{\theta}{3}\right)\sin\left(\dfrac{\theta}{6}\right)

The form \sin\alpha\cos\beta+\cos\alpha\sin\beta is \sin(\alpha+\beta) by identity (3), with \alpha=\tfrac{\theta}{3}, \beta=\tfrac{\theta}{6}.
\sin\dfrac{\theta}{3}\cos\dfrac{\theta}{6}+\cos\dfrac{\theta}{3}\sin\dfrac{\theta}{6}=\sin\left(\dfrac{\theta}{3}+\dfrac{\theta}{6}\right)
Add the fractions: \tfrac{\theta}{3}+\tfrac{\theta}{6}=\tfrac{2\theta+\theta}{6}.
=\sin\dfrac{3\theta}{6}=\sin\dfrac{\theta}{2}

\sin\dfrac{\theta}{2}

(iv) \sin138^\circ\cos46^\circ-\cos138^\circ\sin46^\circ

The form \sin\alpha\cos\beta-\cos\alpha\sin\beta is \sin(\alpha-\beta) by identity (4), with \alpha=138^\circ, \beta=46^\circ.
\sin138^\circ\cos46^\circ-\cos138^\circ\sin46^\circ=\sin(138^\circ-46^\circ)
Subtract the angles.
=\sin92^\circ

\sin92^\circ

(v) \dfrac{\tan75^\circ-\tan45^\circ}{1+\tan75^\circ\tan45^\circ}

The form \dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} is \tan(\alpha-\beta), here with \alpha=75^\circ, \beta=45^\circ.
\dfrac{\tan75^\circ-\tan45^\circ}{1+\tan75^\circ\tan45^\circ}=\tan(75^\circ-45^\circ)
Subtract the angles.
=\tan30^\circ

\tan30^\circ\ \left(=\dfrac{1}{\sqrt3}\right)

(vi) \dfrac{\tan\dfrac{4\pi}{3}+\tan\dfrac{2\pi}{3}}{1-\tan\dfrac{4\pi}{3}\tan\dfrac{2\pi}{3}}

The form \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} is \tan(\alpha+\beta) (Example 1), here with \alpha=\tfrac{4\pi}{3}, \beta=\tfrac{2\pi}{3}.
\dfrac{\tan\frac{4\pi}{3}+\tan\frac{2\pi}{3}}{1-\tan\frac{4\pi}{3}\tan\frac{2\pi}{3}}=\tan\left(\dfrac{4\pi}{3}+\dfrac{2\pi}{3}\right)
Add the angles.
=\tan2\pi

\tan2\pi\ (=0)

Book answer. (i) \cos9\theta (ii) \cos5\theta (iii) \sin\dfrac{\theta}{2} (iv) \sin92^\circ (v) \tan30^\circ (vi) \tan2\pi

Note. Where an exact number is wanted: \tan30^\circ=\dfrac{1}{\sqrt3} and \tan2\pi=0.

For \sin\alpha=\dfrac45, \tan\beta=-\dfrac{5}{12} with terminal sides of angles \alpha and \beta lie quad. II, find \cos(\alpha+\beta) and \cos(\alpha-\beta).

Solution

Find \cos\alpha from \cos^2\alpha=1-\sin^2\alpha. In quadrant II cosine is negative.
\cos^2\alpha=1-\dfrac{16}{25}=\dfrac{9}{25}\ \Rightarrow\ \cos\alpha=-\dfrac35
For \beta, use a right triangle with sides 5 and 12 (hypotenuse \sqrt{25+144}=13). In quadrant II sine is positive and cosine is negative, which gives \tan\beta=\tfrac{\sin\beta}{\cos\beta}=-\tfrac{5}{12}.
\sin\beta=\dfrac{5}{13},\qquad\cos\beta=-\dfrac{12}{13}
Cosine of the sum: \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos(\alpha+\beta)=\left(-\dfrac35\right)\left(-\dfrac{12}{13}\right)-\left(\dfrac45\right)\left(\dfrac{5}{13}\right)=\dfrac{36}{65}-\dfrac{20}{65}=\dfrac{16}{65}
Cosine of the difference: \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\dfrac{36}{65}+\dfrac{20}{65}=\dfrac{56}{65}

\cos(\alpha+\beta)=\dfrac{16}{65},\qquad\cos(\alpha-\beta)=\dfrac{56}{65}

Book answer. \cos(\alpha+\beta)=\dfrac{16}{65}, \cos(\alpha-\beta)=\dfrac{56}{65}

For \cos\alpha=-\dfrac{7}{25} and \cot\beta=\dfrac{15}{8} with terminal sides of \alpha in quad. II and \beta in quad. III find: (i) \sin(\alpha-\beta) (ii) \cos(\alpha-\beta) (iii) \tan(\alpha-\beta)

Solution

Find \sin\alpha. In quadrant II the sine is positive.
\sin^2\alpha=1-\dfrac{49}{625}=\dfrac{576}{625}\ \Rightarrow\ \sin\alpha=\dfrac{24}{25}
From \cot\beta=\tfrac{15}{8} we get \tan\beta=\tfrac{8}{15}; the hypotenuse is \sqrt{15^2+8^2}=17. In quadrant III both sine and cosine are negative.
\sin\beta=-\dfrac{8}{17},\qquad\cos\beta=-\dfrac{15}{17}
(i) \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin(\alpha-\beta)=\dfrac{24}{25}\left(-\dfrac{15}{17}\right)-\left(-\dfrac{7}{25}\right)\left(-\dfrac{8}{17}\right)=-\dfrac{360}{425}-\dfrac{56}{425}=-\dfrac{416}{425}
(ii) \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\left(-\dfrac{7}{25}\right)\left(-\dfrac{15}{17}\right)+\dfrac{24}{25}\left(-\dfrac{8}{17}\right)=\dfrac{105}{425}-\dfrac{192}{425}=-\dfrac{87}{425}
(iii) Divide the sine by the cosine.
\tan(\alpha-\beta)=\dfrac{-416/425}{-87/425}=\dfrac{416}{87}
Check. \sin^2+\cos^2 must equal 1, and the tangent formula must agree. With \tan\alpha=-\tfrac{24}{7} and \tan\beta=\tfrac{8}{15}:
\begin{aligned}\left(\dfrac{416}{425}\right)^2+\left(\dfrac{87}{425}\right)^2&=\dfrac{173056+7569}{180625}\\&=1\\ \dfrac{-\frac{24}{7}-\frac{8}{15}}{1+\left(-\frac{24}{7}\right)\left(\frac{8}{15}\right)}&=\dfrac{-\frac{416}{105}}{-\frac{87}{105}}\\&=\dfrac{416}{87}\ \checkmark\end{aligned}

\sin(\alpha-\beta)=-\dfrac{416}{425},\quad\cos(\alpha-\beta)=-\dfrac{87}{425},\quad\tan(\alpha-\beta)=\dfrac{416}{87}

Book answer. (i) \dfrac{416}{425} (ii) \dfrac{87}{425} (iii) \dfrac{416}{87}

Note. The key prints (i) and (ii) with positive signs. Those values belong to an acute \beta (\sin\beta=\tfrac{8}{17}, \cos\beta=\tfrac{15}{17}). The question puts \beta in quadrant III, where sine and cosine are both negative, and then the signs of (i) and (ii) are negative as found above. Part (iii) agrees with the key because the two signs cancel in the ratio.

Given \alpha and \beta are acute angles with \sin\alpha=\dfrac{12}{13} and \tan\beta=\dfrac43 find: (i) \sin(\alpha+\beta) (ii) \cos(\alpha+\beta) (iii) \tan(\alpha+\beta)

Solution

Acute angles have positive sine and cosine. Find \cos\alpha.
\cos^2\alpha=1-\dfrac{144}{169}=\dfrac{25}{169}\ \Rightarrow\ \cos\alpha=\dfrac{5}{13}
From \tan\beta=\tfrac43 use a 3-4-5 right triangle.
\sin\beta=\dfrac45,\qquad\cos\beta=\dfrac35
(i) \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\sin(\alpha+\beta)=\dfrac{12}{13}\cdot\dfrac35+\dfrac{5}{13}\cdot\dfrac45=\dfrac{36}{65}+\dfrac{20}{65}=\dfrac{56}{65}
(ii) \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos(\alpha+\beta)=\dfrac{5}{13}\cdot\dfrac35-\dfrac{12}{13}\cdot\dfrac45=\dfrac{15}{65}-\dfrac{48}{65}=-\dfrac{33}{65}
(iii) Divide.
\tan(\alpha+\beta)=\dfrac{56/65}{-33/65}=-\dfrac{56}{33}

\sin(\alpha+\beta)=\dfrac{56}{65},\quad\cos(\alpha+\beta)=-\dfrac{33}{65},\quad\tan(\alpha+\beta)=-\dfrac{56}{33}

Book answer. (i) \dfrac{56}{65} (ii) -\dfrac{33}{65} (iii) -\dfrac{56}{33}

Note. \alpha+\beta is obtuse (cosine negative, sine positive), which is possible even though both angles are acute.

If \sin\alpha=\dfrac35, where 0<\alpha<\dfrac{\pi}{2} and \cos\beta=\dfrac{12}{13}, where \dfrac{3\pi}{2}<\beta<2\pi find: (i) \csc(\alpha+\beta) (ii) \sec(\alpha+\beta) (iii) \cot(\alpha+\beta)

Solution

\alpha is in quadrant I, so \cos\alpha>0.
\cos\alpha=\sqrt{1-\dfrac{9}{25}}=\dfrac45
\beta is in quadrant IV, so \sin\beta<0.
\sin\beta=-\sqrt{1-\dfrac{144}{169}}=-\dfrac{5}{13}
Find \sin(\alpha+\beta).
\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\dfrac35\cdot\dfrac{12}{13}+\dfrac45\left(-\dfrac{5}{13}\right)=\dfrac{36}{65}-\dfrac{20}{65}=\dfrac{16}{65}
Find \cos(\alpha+\beta).
\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\dfrac45\cdot\dfrac{12}{13}-\dfrac35\left(-\dfrac{5}{13}\right)=\dfrac{48}{65}+\dfrac{15}{65}=\dfrac{63}{65}
(i) Cosecant is the reciprocal of sine.
\csc(\alpha+\beta)=\dfrac{65}{16}
(ii) Secant is the reciprocal of cosine.
\sec(\alpha+\beta)=\dfrac{65}{63}
(iii) Cotangent is cosine divided by sine.
\cot(\alpha+\beta)=\dfrac{63/65}{16/65}=\dfrac{63}{16}

\csc(\alpha+\beta)=\dfrac{65}{16},\quad\sec(\alpha+\beta)=\dfrac{65}{63},\quad\cot(\alpha+\beta)=\dfrac{63}{16}

Book answer. (i) \dfrac{65}{16} (ii) \dfrac{65}{63} (iii) \dfrac{63}{16}

Given \alpha and \beta are obtuse angles with \sin\alpha=\dfrac{1}{\sqrt2} and \cos\beta=-\dfrac35 find: (i) \sin(\alpha\pm\beta) (ii) \cos(\alpha\pm\beta) (iii) \tan(\alpha\pm\beta)

Solution

An obtuse angle lies in quadrant II: sine positive, cosine negative. Find \cos\alpha.
\cos\alpha=-\sqrt{1-\dfrac12}=-\dfrac{1}{\sqrt2}
Find \sin\beta.
\sin\beta=\sqrt{1-\dfrac{9}{25}}=\dfrac45

(i) \sin(\alpha\pm\beta)

Sum: \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\sin(\alpha+\beta)=\dfrac{1}{\sqrt2}\left(-\dfrac35\right)+\left(-\dfrac{1}{\sqrt2}\right)\dfrac45=-\dfrac{3}{5\sqrt2}-\dfrac{4}{5\sqrt2}=-\dfrac{7}{5\sqrt2}
Difference: \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.
\sin(\alpha-\beta)=-\dfrac{3}{5\sqrt2}+\dfrac{4}{5\sqrt2}=\dfrac{1}{5\sqrt2}

\sin(\alpha+\beta)=-\dfrac{7}{5\sqrt2},\quad\sin(\alpha-\beta)=\dfrac{1}{5\sqrt2}

(ii) \cos(\alpha\pm\beta)

Sum: \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\cos(\alpha+\beta)=\left(-\dfrac{1}{\sqrt2}\right)\left(-\dfrac35\right)-\dfrac{1}{\sqrt2}\cdot\dfrac45=\dfrac{3}{5\sqrt2}-\dfrac{4}{5\sqrt2}=-\dfrac{1}{5\sqrt2}
Difference: \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\dfrac{3}{5\sqrt2}+\dfrac{4}{5\sqrt2}=\dfrac{7}{5\sqrt2}

\cos(\alpha+\beta)=-\dfrac{1}{5\sqrt2},\quad\cos(\alpha-\beta)=\dfrac{7}{5\sqrt2}

(iii) \tan(\alpha\pm\beta)

Divide the sines by the cosines found in (i) and (ii). The common factor \tfrac{1}{5\sqrt2} cancels.
\tan(\alpha+\beta)=\dfrac{-7}{-1}=7,\qquad\tan(\alpha-\beta)=\dfrac{1}{7}

\tan(\alpha+\beta)=7,\quad\tan(\alpha-\beta)=\dfrac17

Book answer. (i) \sin(\alpha+\beta)=-\dfrac{7}{5\sqrt2}, \sin(\alpha-\beta)=\dfrac{1}{5\sqrt2} (ii) \cos(\alpha+\beta)=-\dfrac{1}{5\sqrt2}, \cos(\alpha-\beta)=\dfrac{7}{5\sqrt2} (iii) \tan(\alpha+\beta)=7, \tan(\alpha-\beta)=\dfrac17

Verify:

Solution

(i) \sin\left(\dfrac{\pi}{2}-\alpha\right)=\cos\alpha

Use \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta with \tfrac{\pi}{2} in the place of \alpha and \alpha in the place of \beta.
\sin\left(\dfrac{\pi}{2}-\alpha\right)=\sin\dfrac{\pi}{2}\cos\alpha-\cos\dfrac{\pi}{2}\sin\alpha
Put \sin\tfrac{\pi}{2}=1 and \cos\tfrac{\pi}{2}=0.
=1\cdot\cos\alpha-0\cdot\sin\alpha=\cos\alpha

(ii) \cos(\pi-\alpha)=-\cos\alpha

Fundamental law with \pi and \alpha.
\cos(\pi-\alpha)=\cos\pi\cos\alpha+\sin\pi\sin\alpha
Put \cos\pi=-1 and \sin\pi=0.
=(-1)\cos\alpha+0=-\cos\alpha

(iii) \cos\left(\alpha+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt2}(\cos\alpha-\sin\alpha)

Use \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta with \beta=\tfrac{\pi}{4}.
\cos\left(\alpha+\dfrac{\pi}{4}\right)=\cos\alpha\cos\dfrac{\pi}{4}-\sin\alpha\sin\dfrac{\pi}{4}
Put \cos\tfrac{\pi}{4}=\sin\tfrac{\pi}{4}=\tfrac{1}{\sqrt2} and take out the common factor.
=\dfrac{1}{\sqrt2}\cos\alpha-\dfrac{1}{\sqrt2}\sin\alpha=\dfrac{1}{\sqrt2}(\cos\alpha-\sin\alpha)

(iv) \sin\left(\beta+\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}(\cos\beta+\sin\beta)

Use \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta with the angles \beta and \tfrac{\pi}{4}.
\sin\left(\beta+\dfrac{\pi}{4}\right)=\sin\beta\cos\dfrac{\pi}{4}+\cos\beta\sin\dfrac{\pi}{4}
Put \cos\tfrac{\pi}{4}=\sin\tfrac{\pi}{4}=\tfrac{1}{\sqrt2}=\tfrac{\sqrt2}{2} (multiply top and bottom by \sqrt2).
=\dfrac{\sqrt2}{2}\sin\beta+\dfrac{\sqrt2}{2}\cos\beta=\dfrac{\sqrt2}{2}(\cos\beta+\sin\beta)

(v) \tan\left(\gamma-\dfrac{\pi}{4}\right)=\dfrac{\tan\gamma-1}{\tan\gamma+1}

Use the difference formula for tangent (Check Point after Example 1): \tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}.
\tan\left(\gamma-\dfrac{\pi}{4}\right)=\dfrac{\tan\gamma-\tan\frac{\pi}{4}}{1+\tan\gamma\tan\frac{\pi}{4}}
Put \tan\tfrac{\pi}{4}=1.
=\dfrac{\tan\gamma-1}{1+\tan\gamma}=\dfrac{\tan\gamma-1}{\tan\gamma+1}

(vi) \tan\left(\gamma+\dfrac{\pi}{4}\right)=\dfrac{1+\tan\gamma}{1-\tan\gamma}=\dfrac{\cos\gamma+\sin\gamma}{\cos\gamma-\sin\gamma}

First equality: \tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} with \tan\tfrac{\pi}{4}=1.
\tan\left(\gamma+\dfrac{\pi}{4}\right)=\dfrac{\tan\gamma+1}{1-\tan\gamma\cdot1}=\dfrac{1+\tan\gamma}{1-\tan\gamma}
Second equality: replace \tan\gamma by \dfrac{\sin\gamma}{\cos\gamma} and multiply the top and bottom by \cos\gamma.
\dfrac{1+\frac{\sin\gamma}{\cos\gamma}}{1-\frac{\sin\gamma}{\cos\gamma}}=\dfrac{\cos\gamma+\sin\gamma}{\cos\gamma-\sin\gamma}

(vii) \cos(x+y)+\cos(x-y)=2\cos x\cos y

Expand each term with identity (2) and the fundamental law.
\cos(x+y)+\cos(x-y)=(\cos x\cos y-\sin x\sin y)+(\cos x\cos y+\sin x\sin y)
The \sin x\sin y terms cancel.
=2\cos x\cos y

(viii) \sin(x+y)-\sin(x-y)=2\cos x\sin y

Expand each term with identities (3) and (4).
\sin(x+y)-\sin(x-y)=(\sin x\cos y+\cos x\sin y)-(\sin x\cos y-\cos x\sin y)
The \sin x\cos y terms cancel and the \cos x\sin y terms add.
=2\cos x\sin y

Note. In (iii) and (iv) the two forms of the constant are equal: \dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}.

Show that:

Solution

(i) \dfrac{\sin(180^\circ+\lambda)\cos(270^\circ+\lambda)}{\sin(180^\circ-\lambda)\cos(270^\circ-\lambda)}=1

Reduce each factor with the allied-angle table. 180^\circ\pm\lambda keeps the sine (an even multiple of 90^\circ) and 270^\circ\pm\lambda changes cosine to sine (an odd multiple). The sign follows the quadrant.
\begin{aligned}\sin(180^\circ+\lambda)&=-\sin\lambda\\ \cos(270^\circ+\lambda)&=\sin\lambda\\ \sin(180^\circ-\lambda)&=\sin\lambda\\ \cos(270^\circ-\lambda)&=-\sin\lambda\end{aligned}
Substitute.
\dfrac{(-\sin\lambda)(\sin\lambda)}{(\sin\lambda)(-\sin\lambda)}=\dfrac{-\sin^2\lambda}{-\sin^2\lambda}=1

(ii) \dfrac{\sin(90^\circ+\alpha)-\cos(360^\circ-\alpha)+\cos\alpha}{\sin(180^\circ-\alpha)+\sin(270^\circ-\alpha)+\cos(90^\circ+\alpha)}=-1

Reduce the numerator terms.
\sin(90^\circ+\alpha)=\cos\alpha,\qquad\cos(360^\circ-\alpha)=\cos\alpha
So the numerator is
\cos\alpha-\cos\alpha+\cos\alpha=\cos\alpha
Reduce the denominator terms.
\sin(180^\circ-\alpha)=\sin\alpha,\quad\sin(270^\circ-\alpha)=-\cos\alpha,\quad\cos(90^\circ+\alpha)=-\sin\alpha
So the denominator is
\sin\alpha-\cos\alpha-\sin\alpha=-\cos\alpha
Divide.
\dfrac{\cos\alpha}{-\cos\alpha}=-1

(iii) \tan\alpha+\tan\beta=\dfrac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta}

Write the tangents as sine over cosine and add with a common denominator.
\tan\alpha+\tan\beta=\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\sin\beta}{\cos\beta}=\dfrac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta}
The numerator is \sin(\alpha+\beta) by identity (3).
=\dfrac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta}

(iv) \sin(\alpha+\beta)\sin(\alpha-\beta)=\cos^2\beta-\cos^2\alpha=\sin^2\alpha-\sin^2\beta

Expand both factors.
\sin(\alpha+\beta)\sin(\alpha-\beta)=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)(\sin\alpha\cos\beta-\cos\alpha\sin\beta)
The product is a difference of squares.
=\sin^2\alpha\cos^2\beta-\cos^2\alpha\sin^2\beta
Replace \cos^2\beta=1-\sin^2\beta and \cos^2\alpha=1-\sin^2\alpha.
=\sin^2\alpha(1-\sin^2\beta)-(1-\sin^2\alpha)\sin^2\beta=\sin^2\alpha-\sin^2\alpha\sin^2\beta-\sin^2\beta+\sin^2\alpha\sin^2\beta=\sin^2\alpha-\sin^2\beta
Second equality. Use \sin^2\theta=1-\cos^2\theta on both terms.
\sin^2\alpha-\sin^2\beta=(1-\cos^2\alpha)-(1-\cos^2\beta)=\cos^2\beta-\cos^2\alpha

(v) \dfrac{\tan(x+y)}{\cot(x-y)}=\dfrac{\tan^2x-\tan^2y}{1-\tan^2x\tan^2y}

\cot(x-y)=\dfrac{1}{\tan(x-y)}, so dividing by it means multiplying by \tan(x-y).
\dfrac{\tan(x+y)}{\cot(x-y)}=\tan(x+y)\tan(x-y)
Use the sum and difference formulas for tangent.
=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\cdot\dfrac{\tan x-\tan y}{1+\tan x\tan y}
Numerators and denominators are each a difference of squares.
=\dfrac{\tan^2x-\tan^2y}{1-\tan^2x\tan^2y}

(vi) \dfrac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=\dfrac{1-\tan\alpha\tan\beta}{1+\tan\alpha\tan\beta}

Expand the numerator and denominator.
\dfrac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=\dfrac{\cos\alpha\cos\beta-\sin\alpha\sin\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}
Divide the top and bottom by \cos\alpha\cos\beta.
=\dfrac{1-\frac{\sin\alpha}{\cos\alpha}\cdot\frac{\sin\beta}{\cos\beta}}{1+\frac{\sin\alpha}{\cos\alpha}\cdot\frac{\sin\beta}{\cos\beta}}=\dfrac{1-\tan\alpha\tan\beta}{1+\tan\alpha\tan\beta}

(vii) \cot(\alpha-\beta)=\dfrac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}

Write cotangent as cosine over sine and expand.
\cot(\alpha-\beta)=\dfrac{\cos(\alpha-\beta)}{\sin(\alpha-\beta)}=\dfrac{\cos\alpha\cos\beta+\sin\alpha\sin\beta}{\sin\alpha\cos\beta-\cos\alpha\sin\beta}
Divide the top and bottom by \sin\alpha\sin\beta.
=\dfrac{\frac{\cos\alpha}{\sin\alpha}\cdot\frac{\cos\beta}{\sin\beta}+1}{\frac{\cos\beta}{\sin\beta}-\frac{\cos\alpha}{\sin\alpha}}=\dfrac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}

(viii) \dfrac{\cos4\theta}{\csc\theta}+\dfrac{\sin4\theta}{\sec\theta}=\sin5\theta

Dividing by \csc\theta is multiplying by \sin\theta, and dividing by \sec\theta is multiplying by \cos\theta.
\dfrac{\cos4\theta}{\csc\theta}+\dfrac{\sin4\theta}{\sec\theta}=\cos4\theta\sin\theta+\sin4\theta\cos\theta
This is \sin\alpha\cos\beta+\cos\alpha\sin\beta with \alpha=4\theta, \beta=\theta.
=\sin(4\theta+\theta)=\sin5\theta

If \alpha+\beta+\gamma=180^\circ, prove that:

Solution

(i) \tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma

From the condition, \alpha+\beta=180^\circ-\gamma. Take the tangent of both sides.
\tan(\alpha+\beta)=\tan(180^\circ-\gamma)=-\tan\gamma
Expand the left side with the sum formula for tangent.
\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=-\tan\gamma
Multiply both sides by 1-\tan\alpha\tan\beta.
\tan\alpha+\tan\beta=-\tan\gamma+\tan\alpha\tan\beta\tan\gamma
Move -\tan\gamma to the left side.
\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma

(ii) \cot\dfrac{\alpha}{2}+\cot\dfrac{\beta}{2}+\cot\dfrac{\gamma}{2}=\cot\dfrac{\alpha}{2}\cot\dfrac{\beta}{2}\cot\dfrac{\gamma}{2}

Write A=\tfrac{\alpha}{2}, B=\tfrac{\beta}{2}, C=\tfrac{\gamma}{2}. Halving the condition gives A+B+C=90^\circ, so A+B=90^\circ-C.
\cot(A+B)=\cot(90^\circ-C)=\tan C
Expand \cot(A+B) as cosine over sine, then divide the top and bottom by \sin A\sin B.
\cot(A+B)=\dfrac{\cos A\cos B-\sin A\sin B}{\sin A\cos B+\cos A\sin B}=\dfrac{\cot A\cot B-1}{\cot B+\cot A}
Equate the two forms and use \tan C=\dfrac{1}{\cot C}.
\dfrac{\cot A\cot B-1}{\cot A+\cot B}=\dfrac{1}{\cot C}
Cross-multiply.
\cot C(\cot A\cot B-1)=\cot A+\cot B\ \Rightarrow\ \cot A\cot B\cot C-\cot C=\cot A+\cot B
Move \cot C to the right side.
\cot A+\cot B+\cot C=\cot A\cot B\cot C

(iii) \tan\dfrac{\alpha}{2}\tan\dfrac{\beta}{2}+\tan\dfrac{\beta}{2}\tan\dfrac{\gamma}{2}+\tan\dfrac{\gamma}{2}\tan\dfrac{\alpha}{2}-1=0

With A=\tfrac{\alpha}{2}, B=\tfrac{\beta}{2}, C=\tfrac{\gamma}{2} we have A+B=90^\circ-C.
\tan(A+B)=\tan(90^\circ-C)=\cot C=\dfrac{1}{\tan C}
Expand the left side with the sum formula for tangent.
\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{1}{\tan C}
Cross-multiply.
\tan C(\tan A+\tan B)=1-\tan A\tan B
Multiply out and move everything to the left.
\tan A\tan C+\tan B\tan C+\tan A\tan B-1=0
This is the required result, since \tan A\tan B=\tan\tfrac{\alpha}{2}\tan\tfrac{\beta}{2}, and so on.
\tan\dfrac{\alpha}{2}\tan\dfrac{\beta}{2}+\tan\dfrac{\beta}{2}\tan\dfrac{\gamma}{2}+\tan\dfrac{\gamma}{2}\tan\dfrac{\alpha}{2}-1=0

Note. Part (i) is the book's Example 4 in another form: the same idea of replacing one angle by 180^\circ minus the other two is used in each part.

Express the following in the form of r\sin(\theta+\phi).

Solution

(i) 12\sin\theta-5\cos\theta

Compare with a\sin\theta+b\cos\theta=r\sin(\theta+\phi), where a=r\cos\phi and b=r\sin\phi (section 8.4).
a=12,\qquad b=-5
Find r=\sqrt{a^2+b^2}.
r=\sqrt{12^2+(-5)^2}=\sqrt{169}=13
Divide each term by r so that the coefficients become \cos\phi and \sin\phi.
12\sin\theta-5\cos\theta=13\left(\dfrac{12}{13}\sin\theta-\dfrac{5}{13}\cos\theta\right)=13(\cos\phi\sin\theta+\sin\phi\cos\theta)=13\sin(\theta+\phi)
Find \phi from \tan\phi=\dfrac{b}{a}, taking the quadrant from the signs of \cos\phi and \sin\phi.
\cos\phi=\dfrac{12}{13}>0,\ \sin\phi=-\dfrac{5}{13}<0\ \Rightarrow\ \phi\text{ is in quadrant IV},\ \phi=\tan^{-1}\left(-\dfrac{5}{12}\right)\approx-22.6^\circ

12\sin\theta-5\cos\theta=13\sin(\theta+\phi),\quad\phi=\tan^{-1}\left(-\dfrac{5}{12}\right)

(ii) 3\sin\theta+4\cos\theta

Compare with a\sin\theta+b\cos\theta=r\sin(\theta+\phi), where a=r\cos\phi and b=r\sin\phi (section 8.4).
a=3,\qquad b=4
Find r=\sqrt{a^2+b^2}.
r=\sqrt{3^2+4^2}=\sqrt{25}=5
Divide each term by r so that the coefficients become \cos\phi and \sin\phi.
3\sin\theta+4\cos\theta=5\left(\dfrac35\sin\theta+\dfrac45\cos\theta\right)=5(\cos\phi\sin\theta+\sin\phi\cos\theta)=5\sin(\theta+\phi)
Find \phi from \tan\phi=\dfrac{b}{a}, taking the quadrant from the signs of \cos\phi and \sin\phi.
\cos\phi=\dfrac35>0,\ \sin\phi=\dfrac45>0\ \Rightarrow\ \phi\text{ is acute},\ \phi=\tan^{-1}\left(\dfrac43\right)\approx53.1^\circ

3\sin\theta+4\cos\theta=5\sin(\theta+\phi),\quad\phi=\tan^{-1}\left(\dfrac43\right)

(iii) \sin\theta-\cos\theta

Compare with a\sin\theta+b\cos\theta=r\sin(\theta+\phi), where a=r\cos\phi and b=r\sin\phi (section 8.4).
a=1,\qquad b=-1
Find r=\sqrt{a^2+b^2}.
r=\sqrt{1^2+(-1)^2}=\sqrt2
Divide each term by r so that the coefficients become \cos\phi and \sin\phi.
\sin\theta-\cos\theta=\sqrt2\left(\dfrac{1}{\sqrt2}\sin\theta-\dfrac{1}{\sqrt2}\cos\theta\right)=\sqrt2(\cos\phi\sin\theta+\sin\phi\cos\theta)=\sqrt2\sin(\theta+\phi)
Find \phi from \tan\phi=\dfrac{b}{a}, taking the quadrant from the signs of \cos\phi and \sin\phi.
\cos\phi=\dfrac{1}{\sqrt2}>0,\ \sin\phi=-\dfrac{1}{\sqrt2}<0\ \Rightarrow\ \phi\text{ is in quadrant IV},\ \phi=\tan^{-1}(-1)=-\dfrac{\pi}{4}
Check at \theta=\tfrac{\pi}{2}: the left side is 1-0=1 and the right side is \sqrt2\sin\tfrac{\pi}{4}=\sqrt2\cdot\tfrac{1}{\sqrt2}=1.

\sin\theta-\cos\theta=\sqrt2\sin\left(\theta-\dfrac{\pi}{4}\right)

Book answer. (i) 12\sin\theta-5\cos\theta=r\sin(\theta+\phi) with r=13, \phi=\tan^{-1}\left(-\dfrac{5}{12}\right) (ii) r=5, \phi=\tan^{-1}\left(\dfrac43\right) (iii) the key says \"Do yourself\"

Note. The key leaves (iii) to the student. The answer above agrees with the method of section 8.4 (r=\sqrt2, \phi=\tan^{-1}(-1)=-\tfrac{\pi}{4}).

A telephone pole is braced by two wires that are both fastened to the ground at a point 3\,\mathrm{m} from the base of the pole. The shorter wire is fastened to the pole 3\,\mathrm{m} above the ground and the longer wire 7\,\mathrm{m} above the ground.

Solution

The pole is the vertical line BL. Both wires are fastened to the ground at G, which is 3 m from the base B. The shorter wire GS reaches 3 m up the pole and the longer wire GL reaches 7 m up.

(a) What is the measure, in degrees, of the angle that the shorter wire makes with the ground?

Triangle BGS has a right angle at B. The side opposite the angle at G is BS=3 and the side adjacent is BG=3.
\tan\angle SGB=\dfrac{BS}{BG}=\dfrac33=1
The angle whose tangent is 1 is 45^\circ.
\angle SGB=45^\circ

45^\circ

(b) Let \theta be the measure of the angle that the longer wire makes with the ground. Find \sin\theta and \cos\theta.

In the right triangle BGL the sides are BL=7 and BG=3. Find the longer wire GL by Pythagoras.
GL=\sqrt{7^2+3^2}=\sqrt{58}
Sine is opposite over hypotenuse and cosine is adjacent over hypotenuse.
\sin\theta=\dfrac{7}{\sqrt{58}},\qquad\cos\theta=\dfrac{3}{\sqrt{58}}

\sin\theta=\dfrac{7}{\sqrt{58}},\quad\cos\theta=\dfrac{3}{\sqrt{58}}

Both wires start at G and go up to the pole. The angle between them is the difference of the two angles with the ground, \theta-45^\circ.
\cos(\theta-45^\circ)=\cos\theta\cos45^\circ+\sin\theta\sin45^\circ
Substitute \cos\theta, \sin\theta and \cos45^\circ=\sin45^\circ=\tfrac{1}{\sqrt2}.
=\dfrac{3}{\sqrt{58}}\cdot\dfrac{1}{\sqrt2}+\dfrac{7}{\sqrt{58}}\cdot\dfrac{1}{\sqrt2}=\dfrac{10}{\sqrt{116}}
Simplify, using \sqrt{116}=2\sqrt{29}.
=\dfrac{10}{2\sqrt{29}}=\dfrac{5}{\sqrt{29}}\approx0.9285

\cos(\theta-45^\circ)=\dfrac{5}{\sqrt{29}}\approx0.9285

(d) Find, to the nearest degree, the measure of the angle between the wires.

Find the angle whose cosine is 0.9285 (calculator, inverse cosine).
\theta-45^\circ=\cos^{-1}(0.9285)\approx21.8^\circ
Check. \theta=\tan^{-1}\tfrac73\approx66.8^\circ, and 66.8^\circ-45^\circ=21.8^\circ.
\approx22^\circ

\approx22^\circ

Book answer. (a) 45^\circ (b) \sin\theta=\dfrac{7}{\sqrt{58}}, \cos\theta=\dfrac{3}{\sqrt{58}} (c) 0.9285 (d) 22^\circ

Exercise 8.2p. 208–209

Double-angle, half-angle and triple-angle identities: \sin2\alpha=2\sin\alpha\cos\alpha, \cos2\alpha=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha, \tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}, and the half-angle forms \sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}, \cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}. The sign of a half-angle value comes from the quadrant of \dfrac{\theta}{2}.

Suppose P(-3,4) lies on the terminal side of \theta when \theta is plotted in standard position. Find \cos2\theta and \sin2\theta and determine the quadrant in which the terminal side of the angle 2\theta lies when it is plotted in standard position.

Solution

Find the distance r=OP from the origin to P with the distance formula (section 8.1).
r=\sqrt{(-3-0)^2+(4-0)^2}=\sqrt{9+16}=5
The trigonometric ratios of \theta are \cos\theta=\dfrac{x}{r} and \sin\theta=\dfrac{y}{r}.
\cos\theta=-\dfrac35,\qquad\sin\theta=\dfrac45
Cosine of the double angle: \cos2\theta=\cos^2\theta-\sin^2\theta.
\cos2\theta=\dfrac{9}{25}-\dfrac{16}{25}=-\dfrac{7}{25}
Sine of the double angle: \sin2\theta=2\sin\theta\cos\theta.
\sin2\theta=2\left(\dfrac45\right)\left(-\dfrac35\right)=-\dfrac{24}{25}
Both \cos2\theta and \sin2\theta are negative, which happens only in quadrant III.
\cos2\theta<0,\ \sin2\theta<0\ \Rightarrow\ 2\theta\text{ lies in quadrant III}
Check. \cos^22\theta+\sin^22\theta must equal 1.
\dfrac{49}{625}+\dfrac{576}{625}=1\ \checkmark

OP is the terminal side of \theta (quadrant II). The point Q(5\cos2\theta,\,5\sin2\theta)=(-1.4,-4.8) on the same circle of radius 5 shows the terminal side of 2\theta, which lies in quadrant III.

\cos2\theta=-\dfrac{7}{25},\quad\sin2\theta=-\dfrac{24}{25},\quad2\theta\text{ in quadrant III}

Book answer. \cos2\theta=-\dfrac{7}{25}, \sin2\theta=-\dfrac{24}{25}, III quadrant

If \sin\alpha=y and \alpha lies in quad. II. Find expressions for \sin2\alpha, \cos2\alpha and \tan2\alpha in terms of y.

Solution

Find \cos\alpha in terms of y from \cos^2\alpha=1-\sin^2\alpha. In quadrant II the cosine is negative, so take the negative root.
\cos\alpha=-\sqrt{1-y^2}
Sine of the double angle: \sin2\alpha=2\sin\alpha\cos\alpha.
\sin2\alpha=2y\left(-\sqrt{1-y^2}\right)=-2y\sqrt{1-y^2}
Cosine of the double angle, using the form \cos2\alpha=1-2\sin^2\alpha which contains only \sin\alpha.
\cos2\alpha=1-2y^2
Tangent is sine divided by cosine.
\tan2\alpha=\dfrac{\sin2\alpha}{\cos2\alpha}=\dfrac{-2y\sqrt{1-y^2}}{1-2y^2}

\sin2\alpha=-2y\sqrt{1-y^2},\quad\cos2\alpha=1-2y^2,\quad\tan2\alpha=\dfrac{-2y\sqrt{1-y^2}}{1-2y^2}

Book answer. \sin2\alpha=-2y\sqrt{1-y^2}, \cos2\alpha=1-2y^2, \tan2\alpha=-\dfrac{2y\sqrt{1-y^2}}{1-2y^2}

Note. \tan2\alpha is undefined where 1-2y^2=0, that is y=\dfrac{1}{\sqrt2} (\alpha=135^\circ).

Use a half angle formula to find the exact value of \cos15^\circ.

Solution

Write 15^\circ as half of 30^\circ and use \cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}} with \theta=30^\circ. 15^\circ is in quadrant I, so take the positive root.
\cos15^\circ=\cos\dfrac{30^\circ}{2}=\sqrt{\dfrac{1+\cos30^\circ}{2}}
Put \cos30^\circ=\dfrac{\sqrt3}{2} and simplify the fraction under the root.
=\sqrt{\dfrac{1+\frac{\sqrt3}{2}}{2}}=\sqrt{\dfrac{2+\sqrt3}{4}}=\dfrac{\sqrt{2+\sqrt3}}{2}
Decimal value.
\cos15^\circ\approx\dfrac{\sqrt{3.732}}{2}\approx0.966
Check against the answer of Exercise 8.1 Q2: \left(\dfrac{\sqrt6+\sqrt2}{4}\right)^2=\dfrac{8+2\sqrt{12}}{16}=\dfrac{2+\sqrt3}{4}, so \dfrac{\sqrt{2+\sqrt3}}{2}=\dfrac{\sqrt6+\sqrt2}{4}=\dfrac{1+\sqrt3}{2\sqrt2}.

\cos15^\circ=\sqrt{\dfrac{1+\cos30^\circ}{2}}=\dfrac{\sqrt{2+\sqrt3}}{2}\approx0.966

Book answer. \cos15^\circ=\sqrt{\dfrac{1+\cos30^\circ}{2}}=0.966

Find (a) \sin2\theta (b) \cos2\theta (c) \tan2\theta (d) \sin\dfrac{\theta}{2} (e) \cos\dfrac{\theta}{2} (f) \tan\dfrac{\theta}{2} when:

Solution

(i) \cos\theta=\dfrac35 where 0<\theta<\dfrac{\pi}{2}

\theta is in quadrant I, so \sin\theta>0.
\sin\theta=\sqrt{1-\dfrac{9}{25}}=\dfrac45
(a) Sine of the double angle, \sin2\theta=2\sin\theta\cos\theta.
\sin2\theta=2\left(\dfrac45\right)\left(\dfrac35\right)=\dfrac{24}{25}
(b) Cosine of the double angle, \cos2\theta=\cos^2\theta-\sin^2\theta (or 2\cos^2\theta-1, or 1-2\sin^2\theta).
\cos2\theta=\dfrac{9}{25}-\dfrac{16}{25}=-\dfrac{7}{25}
(c) Tangent of the double angle is sine divided by cosine.
\tan2\theta=\dfrac{24/25}{-7/25}=-\dfrac{24}{7}
Since 0<\theta<\dfrac{\pi}{2}, we have 0<\dfrac{\theta}{2}<\dfrac{\pi}{4}: quadrant I, so sine, cosine and tangent of \dfrac{\theta}{2} are positive.
(d) \sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}.
\sin\dfrac{\theta}{2}=\sqrt{\dfrac{1-\frac35}{2}}=\sqrt{\dfrac15}=\dfrac{1}{\sqrt5}
(e) \cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}.
\cos\dfrac{\theta}{2}=\sqrt{\dfrac{1+\frac35}{2}}=\sqrt{\dfrac45}=\dfrac{2}{\sqrt5}
(f) \tan\dfrac{\theta}{2} is the ratio of the two values just found.
\tan\dfrac{\theta}{2}=\dfrac{1/\sqrt5}{2/\sqrt5}=\dfrac12

\sin2\theta=\dfrac{24}{25},\ \cos2\theta=-\dfrac{7}{25},\ \tan2\theta=-\dfrac{24}{7},\ \sin\dfrac{\theta}{2}=\dfrac{1}{\sqrt5},\ \cos\dfrac{\theta}{2}=\dfrac{2}{\sqrt5},\ \tan\dfrac{\theta}{2}=\dfrac12

(ii) \tan\theta=\dfrac{12}{5} where \pi<\theta<\dfrac{3\pi}{2}

\theta is in quadrant III, where sine and cosine are both negative. From \tan\theta=\dfrac{12}{5} the hypotenuse is \sqrt{25+144}=13.
\sin\theta=-\dfrac{12}{13},\qquad\cos\theta=-\dfrac{5}{13}
(a) Sine of the double angle, \sin2\theta=2\sin\theta\cos\theta.
\sin2\theta=2\left(-\dfrac{12}{13}\right)\left(-\dfrac{5}{13}\right)=\dfrac{120}{169}
(b) Cosine of the double angle, \cos2\theta=\cos^2\theta-\sin^2\theta (or 2\cos^2\theta-1, or 1-2\sin^2\theta).
\cos2\theta=\dfrac{25}{169}-\dfrac{144}{169}=-\dfrac{119}{169}
(c) Tangent of the double angle is sine divided by cosine.
\tan2\theta=\dfrac{120/169}{-119/169}=-\dfrac{120}{119}
Since \pi<\theta<\dfrac{3\pi}{2}, we have \dfrac{\pi}{2}<\dfrac{\theta}{2}<\dfrac{3\pi}{4}: quadrant II, so \sin\dfrac{\theta}{2}>0 and \cos\dfrac{\theta}{2}<0.
(d) \sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}.
\sin\dfrac{\theta}{2}=+\sqrt{\dfrac{1+\frac{5}{13}}{2}}=\sqrt{\dfrac{9}{13}}=\dfrac{3}{\sqrt{13}}
(e) \cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}.
\cos\dfrac{\theta}{2}=-\sqrt{\dfrac{1-\frac{5}{13}}{2}}=-\sqrt{\dfrac{4}{13}}=-\dfrac{2}{\sqrt{13}}
(f) \tan\dfrac{\theta}{2} is the ratio of the two values just found.
\tan\dfrac{\theta}{2}=\dfrac{3/\sqrt{13}}{-2/\sqrt{13}}=-\dfrac32

\sin2\theta=\dfrac{120}{169},\ \cos2\theta=-\dfrac{119}{169},\ \tan2\theta=-\dfrac{120}{119},\ \sin\dfrac{\theta}{2}=\dfrac{3}{\sqrt{13}},\ \cos\dfrac{\theta}{2}=-\dfrac{2}{\sqrt{13}},\ \tan\dfrac{\theta}{2}=-\dfrac32

(iii) \sin\theta=-\dfrac{7}{25} where \dfrac{3\pi}{2}<\theta<2\pi

\theta is in quadrant IV, so \cos\theta>0.
\cos\theta=\sqrt{1-\dfrac{49}{625}}=\sqrt{\dfrac{576}{625}}=\dfrac{24}{25}
(a) Sine of the double angle, \sin2\theta=2\sin\theta\cos\theta.
\sin2\theta=2\left(-\dfrac{7}{25}\right)\left(\dfrac{24}{25}\right)=-\dfrac{336}{625}
(b) Cosine of the double angle, \cos2\theta=\cos^2\theta-\sin^2\theta (or 2\cos^2\theta-1, or 1-2\sin^2\theta).
\cos2\theta=1-2\sin^2\theta=1-\dfrac{98}{625}=\dfrac{527}{625}
(c) Tangent of the double angle is sine divided by cosine.
\tan2\theta=\dfrac{-336/625}{527/625}=-\dfrac{336}{527}
Since \dfrac{3\pi}{2}<\theta<2\pi, we have \dfrac{3\pi}{4}<\dfrac{\theta}{2}<\pi: quadrant II, so \sin\dfrac{\theta}{2}>0 and \cos\dfrac{\theta}{2}<0.
(d) \sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}.
\sin\dfrac{\theta}{2}=+\sqrt{\dfrac{1-\frac{24}{25}}{2}}=\sqrt{\dfrac{1}{50}}=\dfrac{1}{5\sqrt2}
(e) \cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}.
\cos\dfrac{\theta}{2}=-\sqrt{\dfrac{1+\frac{24}{25}}{2}}=-\sqrt{\dfrac{49}{50}}=-\dfrac{7}{5\sqrt2}
(f) \tan\dfrac{\theta}{2} is the ratio of the two values just found.
\tan\dfrac{\theta}{2}=\dfrac{1/(5\sqrt2)}{-7/(5\sqrt2)}=-\dfrac17

\sin2\theta=-\dfrac{336}{625},\ \cos2\theta=\dfrac{527}{625},\ \tan2\theta=-\dfrac{336}{527},\ \sin\dfrac{\theta}{2}=\dfrac{1}{5\sqrt2},\ \cos\dfrac{\theta}{2}=-\dfrac{7}{5\sqrt2},\ \tan\dfrac{\theta}{2}=-\dfrac17

(iv) \sec\theta=\ (value missing in the book) where \dfrac{3\pi}{2}<\theta<2\pi

The value of \sec\theta is missing in the printed question. The key's answers correspond to \sec\theta=\sqrt5, so we take that value.
\cos\theta=\dfrac{1}{\sec\theta}=\dfrac{1}{\sqrt5}
\theta is in quadrant IV, so \sin\theta<0.
\sin\theta=-\sqrt{1-\dfrac15}=-\dfrac{2}{\sqrt5}
(a) Sine of the double angle, \sin2\theta=2\sin\theta\cos\theta.
\sin2\theta=2\left(-\dfrac{2}{\sqrt5}\right)\left(\dfrac{1}{\sqrt5}\right)=-\dfrac45
(b) Cosine of the double angle, \cos2\theta=\cos^2\theta-\sin^2\theta (or 2\cos^2\theta-1, or 1-2\sin^2\theta).
\cos2\theta=2\cos^2\theta-1=\dfrac25-1=-\dfrac35
(c) Tangent of the double angle is sine divided by cosine.
\tan2\theta=\dfrac{-4/5}{-3/5}=\dfrac43
Since \dfrac{3\pi}{2}<\theta<2\pi, we have \dfrac{3\pi}{4}<\dfrac{\theta}{2}<\pi: quadrant II, so \sin\dfrac{\theta}{2}>0 and \cos\dfrac{\theta}{2}<0.
(d) \sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}.
\sin\dfrac{\theta}{2}=+\sqrt{\dfrac{1-\frac{1}{\sqrt5}}{2}}=\sqrt{\dfrac{\sqrt5-1}{2\sqrt5}}
(e) \cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}.
\cos\dfrac{\theta}{2}=-\sqrt{\dfrac{1+\frac{1}{\sqrt5}}{2}}=-\sqrt{\dfrac{\sqrt5+1}{2\sqrt5}}
(f) \tan\dfrac{\theta}{2} is the ratio of the two values just found.
\tan\dfrac{\theta}{2}=\dfrac{\sqrt{(\sqrt5-1)/(2\sqrt5)}}{-\sqrt{(\sqrt5+1)/(2\sqrt5)}}=-\sqrt{\dfrac{\sqrt5-1}{\sqrt5+1}}

\sin2\theta=-\dfrac45,\ \cos2\theta=-\dfrac35,\ \tan2\theta=\dfrac43,\ \sin\dfrac{\theta}{2}=\sqrt{\dfrac{\sqrt5-1}{2\sqrt5}},\ \cos\dfrac{\theta}{2}=-\sqrt{\dfrac{\sqrt5+1}{2\sqrt5}},\ \tan\dfrac{\theta}{2}=-\sqrt{\dfrac{\sqrt5-1}{\sqrt5+1}}

(v) \csc\theta=\sqrt2 where \dfrac{\pi}{2}<\theta<\pi

\sin\theta=\dfrac{1}{\csc\theta}. \theta is in quadrant II, so \cos\theta<0.
\sin\theta=\dfrac{1}{\sqrt2},\qquad\cos\theta=-\sqrt{1-\dfrac12}=-\dfrac{1}{\sqrt2}
(a) Sine of the double angle, \sin2\theta=2\sin\theta\cos\theta.
\sin2\theta=2\left(\dfrac{1}{\sqrt2}\right)\left(-\dfrac{1}{\sqrt2}\right)=-1
(b) Cosine of the double angle, \cos2\theta=\cos^2\theta-\sin^2\theta (or 2\cos^2\theta-1, or 1-2\sin^2\theta).
\cos2\theta=1-2\sin^2\theta=1-2\cdot\dfrac12=0
(c) Tangent of the double angle, \tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}.
\tan2\theta=\dfrac{-1}{0}\ \text{is undefined (here }2\theta=\tfrac{3\pi}{2}\text{)}
Since \dfrac{\pi}{2}<\theta<\pi, we have \dfrac{\pi}{4}<\dfrac{\theta}{2}<\dfrac{\pi}{2}: quadrant I, so all three ratios of \dfrac{\theta}{2} are positive.
(d) \sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}.
\sin\dfrac{\theta}{2}=\sqrt{\dfrac{1+\frac{1}{\sqrt2}}{2}}=\sqrt{\dfrac{\sqrt2+1}{2\sqrt2}}
(e) \cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}.
\cos\dfrac{\theta}{2}=\sqrt{\dfrac{1-\frac{1}{\sqrt2}}{2}}=\sqrt{\dfrac{\sqrt2-1}{2\sqrt2}}
(f) \tan\dfrac{\theta}{2} is the ratio of the two values just found.
\tan\dfrac{\theta}{2}=\sqrt{\dfrac{\sqrt2+1}{\sqrt2-1}}=\sqrt{\dfrac{(\sqrt2+1)^2}{2-1}}=\sqrt2+1

\sin2\theta=-1,\ \cos2\theta=0,\ \tan2\theta\text{ undefined},\ \sin\dfrac{\theta}{2}=\sqrt{\dfrac{\sqrt2+1}{2\sqrt2}},\ \cos\dfrac{\theta}{2}=\sqrt{\dfrac{\sqrt2-1}{2\sqrt2}},\ \tan\dfrac{\theta}{2}=\sqrt{\dfrac{\sqrt2+1}{\sqrt2-1}}=\sqrt2+1

(vi) \cos\theta=-\dfrac12 where \dfrac{\pi}{2}<\theta<\pi

\theta is in quadrant II, so \sin\theta>0.
\sin\theta=\sqrt{1-\dfrac14}=\dfrac{\sqrt3}{2}
(a) Sine of the double angle, \sin2\theta=2\sin\theta\cos\theta.
\sin2\theta=2\left(\dfrac{\sqrt3}{2}\right)\left(-\dfrac12\right)=-\dfrac{\sqrt3}{2}
(b) Cosine of the double angle, \cos2\theta=\cos^2\theta-\sin^2\theta (or 2\cos^2\theta-1, or 1-2\sin^2\theta).
\cos2\theta=2\cos^2\theta-1=\dfrac12-1=-\dfrac12
(c) Tangent of the double angle is sine divided by cosine.
\tan2\theta=\dfrac{-\sqrt3/2}{-1/2}=\sqrt3
Since \dfrac{\pi}{2}<\theta<\pi, we have \dfrac{\pi}{4}<\dfrac{\theta}{2}<\dfrac{\pi}{2}: quadrant I, so all three ratios of \dfrac{\theta}{2} are positive.
(d) \sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}.
\sin\dfrac{\theta}{2}=\sqrt{\dfrac{1+\frac12}{2}}=\sqrt{\dfrac34}=\dfrac{\sqrt3}{2}
(e) \cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}.
\cos\dfrac{\theta}{2}=\sqrt{\dfrac{1-\frac12}{2}}=\sqrt{\dfrac14}=\dfrac12
(f) \tan\dfrac{\theta}{2} is the ratio of the two values just found.
\tan\dfrac{\theta}{2}=\dfrac{\sqrt3/2}{1/2}=\sqrt3

\sin2\theta=-\dfrac{\sqrt3}{2},\ \cos2\theta=-\dfrac12,\ \tan2\theta=\sqrt3,\ \sin\dfrac{\theta}{2}=\dfrac{\sqrt3}{2},\ \cos\dfrac{\theta}{2}=\dfrac12,\ \tan\dfrac{\theta}{2}=\sqrt3

Book answer. (i) \sin2\theta=\dfrac{24}{25}, \cos2\theta=-\dfrac{7}{25}, \tan2\theta=-\dfrac{24}{7}, \sin\dfrac{\theta}{2}=\dfrac{1}{\sqrt5}, \cos\dfrac{\theta}{2}=\dfrac{2}{\sqrt5}, \tan\dfrac{\theta}{2}=\dfrac12. (ii) \sin2\theta=\dfrac{120}{169}, \cos2\theta=-\dfrac{119}{169}, \tan2\theta=\dfrac{120}{119}, \sin\dfrac{\theta}{2}=\dfrac{3}{\sqrt{13}}, \cos\dfrac{\theta}{2}=\dfrac{-2}{\sqrt{13}}, \tan\dfrac{\theta}{2}=-\dfrac32. (iii) -\dfrac{336}{625}, \dfrac{527}{625}, -\dfrac{336}{527}, \dfrac{1}{5\sqrt2}, -\dfrac{7}{5\sqrt2}, -\dfrac17. (iv) -\dfrac45, -\dfrac35, \dfrac43, \sqrt{\dfrac{\sqrt5-1}{2\sqrt5}}, -\sqrt{\dfrac{\sqrt5+1}{2\sqrt5}}, -\sqrt{\dfrac{\sqrt5-1}{\sqrt5+1}}. (v) -1, 0, undefined, \sqrt{\dfrac{\sqrt2+1}{2\sqrt2}}, \sqrt{\dfrac{\sqrt2-1}{2\sqrt2}}, \sqrt{\dfrac{\sqrt2+1}{\sqrt2-1}}. (vi) -\dfrac{\sqrt3}{2}, -\dfrac12, \sqrt3, \dfrac{\sqrt3}{2}, \dfrac12, \sqrt3.

Note. Two misprints. In (iv) the value of \sec\theta is missing from the question; the key's answers fit \sec\theta=\sqrt5. In (ii) the key gives \tan2\theta=\dfrac{120}{119}, but \tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=\dfrac{120/169}{-119/169}=-\dfrac{120}{119} (2θ is in quadrant II, where tangent is negative). In (v) the question prints only the denominator of \dfrac{\pi}{2}<\theta<\pi; the key fits \dfrac{\pi}{2}.

Find exact values for \sin\theta, \cos\theta and \tan\theta using the information given.

Solution

(i) \sin2\theta=\dfrac{24}{25}, 2\theta in quad. II

Find \cos2\theta first. In quadrant II the cosine is negative.
\cos2\theta=-\sqrt{1-\dfrac{576}{625}}=-\dfrac{7}{25}
Take 0<2\theta<2\pi. If 2\theta is in quadrant II then \dfrac{\pi}{2}<2\theta<\pi, so \dfrac{\pi}{4}<\theta<\dfrac{\pi}{2}: \theta is in quadrant I and its sine and cosine are positive.
Use the half-angle form \cos\theta=\pm\sqrt{\dfrac{1+\cos2\theta}{2}}.
\cos\theta=\sqrt{\dfrac{1-\frac{7}{25}}{2}}=\sqrt{\dfrac{9}{25}}=\dfrac35
Use \sin\theta=\pm\sqrt{\dfrac{1-\cos2\theta}{2}}.
\sin\theta=\sqrt{\dfrac{1+\frac{7}{25}}{2}}=\sqrt{\dfrac{16}{25}}=\dfrac45
Tangent is sine divided by cosine.
\tan\theta=\dfrac{4/5}{3/5}=\dfrac43
Check by going forward again.
2\sin\theta\cos\theta=2\cdot\dfrac45\cdot\dfrac35=\dfrac{24}{25}\ \checkmark,\qquad\cos^2\theta-\sin^2\theta=\dfrac{9-16}{25}=-\dfrac{7}{25}\ \checkmark

\sin\theta=\dfrac45,\quad\cos\theta=\dfrac35,\quad\tan\theta=\dfrac43

(ii) \cos2\theta=-\dfrac{7}{25}, 2\theta in quad. III

The cosine is given and 2\theta is in quadrant III, so the sine is negative: \sin2\theta=-\dfrac{24}{25} (not needed below).
If \pi<2\theta<\dfrac{3\pi}{2} then \dfrac{\pi}{2}<\theta<\dfrac{3\pi}{4}: \theta is in quadrant II, so \sin\theta>0 and \cos\theta<0.
Use the half-angle form \cos\theta=\pm\sqrt{\dfrac{1+\cos2\theta}{2}}.
\cos\theta=-\sqrt{\dfrac{1-\frac{7}{25}}{2}}=-\sqrt{\dfrac{9}{25}}=-\dfrac35
Use \sin\theta=\pm\sqrt{\dfrac{1-\cos2\theta}{2}}.
\sin\theta=+\sqrt{\dfrac{1+\frac{7}{25}}{2}}=\sqrt{\dfrac{16}{25}}=\dfrac45
Tangent is sine divided by cosine.
\tan\theta=\dfrac{4/5}{-3/5}=-\dfrac43
Check by going forward again.
\begin{aligned}\cos^2\theta-\sin^2\theta&=\dfrac{9-16}{25}=-\dfrac{7}{25}\ \checkmark\\ 2\sin\theta\cos\theta&=2\cdot\dfrac45\cdot\left(-\dfrac35\right)=-\dfrac{24}{25}\ \checkmark\end{aligned}

\sin\theta=\dfrac45,\quad\cos\theta=-\dfrac35,\quad\tan\theta=-\dfrac43

(iii) \sin2\theta=-\dfrac{240}{289}, 2\theta in quad. III

Find \cos2\theta first. In quadrant III the cosine is negative.
\cos2\theta=-\sqrt{1-\dfrac{57600}{83521}}=-\sqrt{\dfrac{25921}{83521}}=-\dfrac{161}{289}
If \pi<2\theta<\dfrac{3\pi}{2} then \dfrac{\pi}{2}<\theta<\dfrac{3\pi}{4}: \theta is in quadrant II, so \sin\theta>0 and \cos\theta<0.
Use the half-angle form \cos\theta=\pm\sqrt{\dfrac{1+\cos2\theta}{2}}.
\cos\theta=-\sqrt{\dfrac{1-\frac{161}{289}}{2}}=-\sqrt{\dfrac{64}{289}}=-\dfrac{8}{17}
Use \sin\theta=\pm\sqrt{\dfrac{1-\cos2\theta}{2}}.
\sin\theta=+\sqrt{\dfrac{1+\frac{161}{289}}{2}}=\sqrt{\dfrac{225}{289}}=\dfrac{15}{17}
Tangent is sine divided by cosine.
\tan\theta=\dfrac{15/17}{-8/17}=-\dfrac{15}{8}
Check by going forward again.
2\sin\theta\cos\theta=2\cdot\dfrac{15}{17}\cdot\left(-\dfrac{8}{17}\right)=-\dfrac{240}{289}\ \checkmark

\sin\theta=\dfrac{15}{17},\quad\cos\theta=-\dfrac{8}{17},\quad\tan\theta=-\dfrac{15}{8}

(iv) \cos2\theta=\dfrac{120}{169}, 2\theta in quad. IV

The cosine is given and 2\theta is in quadrant IV, so the sine is negative: \sin2\theta=-\sqrt{1-\dfrac{14400}{28561}}=-\dfrac{119}{169} (not needed below).
If \dfrac{3\pi}{2}<2\theta<2\pi then \dfrac{3\pi}{4}<\theta<\pi: \theta is in quadrant II, so \sin\theta>0 and \cos\theta<0.
Use the half-angle form \cos\theta=\pm\sqrt{\dfrac{1+\cos2\theta}{2}}.
\cos\theta=-\sqrt{\dfrac{1+\frac{120}{169}}{2}}=-\sqrt{\dfrac{289}{338}}=-\dfrac{17}{13\sqrt2}
Use \sin\theta=\pm\sqrt{\dfrac{1-\cos2\theta}{2}}.
\sin\theta=+\sqrt{\dfrac{1-\frac{120}{169}}{2}}=\sqrt{\dfrac{49}{338}}=\dfrac{7}{13\sqrt2}
Tangent is sine divided by cosine.
\tan\theta=\dfrac{7/(13\sqrt2)}{-17/(13\sqrt2)}=-\dfrac{7}{17}
Check by going forward again.
2\sin\theta\cos\theta=2\cdot\dfrac{7}{13\sqrt2}\cdot\left(-\dfrac{17}{13\sqrt2}\right)=-\dfrac{119}{169}\ \checkmark

\sin\theta=\dfrac{7}{13\sqrt2},\quad\cos\theta=-\dfrac{17}{13\sqrt2},\quad\tan\theta=-\dfrac{7}{17}

Book answer. (i) \sin\theta=\dfrac35, \cos\theta=\dfrac45, \tan\theta=\dfrac34 (ii) \sin\theta=\dfrac45, \cos\theta=-\dfrac35, \tan\theta=-\dfrac43 (iii) \sin\theta=\dfrac{15}{17}, \cos\theta=-\dfrac{8}{17}, \tan\theta=-\dfrac{15}{8} (iv) \sin\theta=\dfrac{7}{13\sqrt2}, \cos\theta=-\dfrac{17}{13\sqrt2}, \tan\theta=-\dfrac{7}{17}

Note. Parts (ii), (iii) and (iv) agree with the key. In (i) the key's values \sin\theta=\tfrac35, \cos\theta=\tfrac45 give \cos2\theta=\tfrac{16-9}{25}=+\tfrac{7}{25}, which puts 2\theta in quadrant I, not quadrant II as the question states. With 2\theta in quadrant II (\cos2\theta=-\tfrac{7}{25}) the values are \sin\theta=\tfrac45, \cos\theta=\tfrac35, \tan\theta=\tfrac43, and the check above confirms them. The solutions take 0<2\theta<2\pi.

Use a double-angle identity to find exact values for the following expressions.

Solution

(i) \sin15^\circ\cos15^\circ

From \sin2\alpha=2\sin\alpha\cos\alpha we get \sin\alpha\cos\alpha=\dfrac{\sin2\alpha}{2}. Take \alpha=15^\circ.
\sin15^\circ\cos15^\circ=\dfrac{\sin30^\circ}{2}
Put \sin30^\circ=\tfrac12.
=\dfrac{1/2}{2}=\dfrac14

\dfrac14

(ii) \cos^215^\circ-\sin^215^\circ

This is \cos^2\alpha-\sin^2\alpha=\cos2\alpha with \alpha=15^\circ.
\cos^215^\circ-\sin^215^\circ=\cos30^\circ
Exact value.
=\dfrac{\sqrt3}{2}

\dfrac{\sqrt3}{2}

(iii) 1-2\sin^2\left(\dfrac{\pi}{8}\right)

This is 1-2\sin^2\alpha=\cos2\alpha with \alpha=\dfrac{\pi}{8}.
1-2\sin^2\dfrac{\pi}{8}=\cos\dfrac{2\pi}{8}=\cos\dfrac{\pi}{4}
Exact value.
=\dfrac{1}{\sqrt2}

\dfrac{1}{\sqrt2}

(iv) 2\cos^2\left(\dfrac{\pi}{12}\right)-1

This is 2\cos^2\alpha-1=\cos2\alpha with \alpha=\dfrac{\pi}{12}.
2\cos^2\dfrac{\pi}{12}-1=\cos\dfrac{2\pi}{12}=\cos\dfrac{\pi}{6}
Exact value.
=\dfrac{\sqrt3}{2}

\dfrac{\sqrt3}{2}

(v) \dfrac{2\tan\left(\dfrac{\pi}{12}\right)}{1-\tan^2\left(\dfrac{\pi}{12}\right)}

This is \dfrac{2\tan\alpha}{1-\tan^2\alpha}=\tan2\alpha with \alpha=\dfrac{\pi}{12}.
\dfrac{2\tan\frac{\pi}{12}}{1-\tan^2\frac{\pi}{12}}=\tan\dfrac{\pi}{6}
Exact value.
=\dfrac{1}{\sqrt3}

\dfrac{1}{\sqrt3}

Book answer. (i) \dfrac14 (ii) \dfrac{\sqrt3}{2} (iii) \dfrac{1}{\sqrt2} (iv) \dfrac{\sqrt3}{2} (v) \dfrac{1}{\sqrt3}

Rewrite in terms of an expression containing only cosines to the power 1.

Solution

(i) \sin^2\alpha\cos^2\alpha

Combine the two factors into \sin\alpha\cos\alpha=\dfrac{\sin2\alpha}{2}.
\sin^2\alpha\cos^2\alpha=(\sin\alpha\cos\alpha)^2=\left(\dfrac{\sin2\alpha}{2}\right)^2=\dfrac{\sin^22\alpha}{4}
Use \sin^2A=\dfrac{1-\cos2A}{2} (identity 9) with A=2\alpha.
=\dfrac14\cdot\dfrac{1-\cos4\alpha}{2}=\dfrac{1-\cos4\alpha}{8}

\dfrac{1-\cos4\alpha}{8}

(ii) \sin^4\alpha\cos^2\alpha

Split off \sin^2\alpha\cos^2\alpha, whose value is known from (i).
\sin^4\alpha\cos^2\alpha=(\sin^2\alpha\cos^2\alpha)\sin^2\alpha=\dfrac{1-\cos4\alpha}{8}\cdot\dfrac{1-\cos2\alpha}{2}
Multiply out.
=\dfrac{1}{16}\left(1-\cos2\alpha-\cos4\alpha+\cos2\alpha\cos4\alpha\right)
The product \cos2\alpha\cos4\alpha is not a cosine to the power 1. Use 2\cos A\cos B=\cos(A+B)+\cos(A-B).
\cos4\alpha\cos2\alpha=\dfrac12\left(\cos6\alpha+\cos2\alpha\right)
Substitute and collect the terms.
=\dfrac{1}{16}\left(1-\cos2\alpha-\cos4\alpha+\dfrac12\cos6\alpha+\dfrac12\cos2\alpha\right)=\dfrac{1}{32}\left(2-\cos2\alpha-2\cos4\alpha+\cos6\alpha\right)

\dfrac{1}{32}\left(2-\cos2\alpha-2\cos4\alpha+\cos6\alpha\right)

(iii) \sin^4\alpha\cos^4\alpha

Write it as the fourth power of \sin\alpha\cos\alpha=\dfrac{\sin2\alpha}{2}.
\sin^4\alpha\cos^4\alpha=\left(\dfrac{\sin2\alpha}{2}\right)^4=\dfrac{\sin^42\alpha}{16}=\dfrac{1}{16}\left(\sin^22\alpha\right)^2
Use \sin^22\alpha=\dfrac{1-\cos4\alpha}{2}.
=\dfrac{1}{16}\cdot\dfrac{(1-\cos4\alpha)^2}{4}=\dfrac{1}{64}\left(1-2\cos4\alpha+\cos^24\alpha\right)
Reduce the square with \cos^2A=\dfrac{1+\cos2A}{2}, A=4\alpha.
=\dfrac{1}{64}\left(1-2\cos4\alpha+\dfrac{1+\cos8\alpha}{2}\right)=\dfrac{1}{128}\left(2-4\cos4\alpha+1+\cos8\alpha\right)
Collect the terms.
=\dfrac{1}{128}\left(3-4\cos4\alpha+\cos8\alpha\right)

\dfrac{1}{128}\left(3-4\cos4\alpha+\cos8\alpha\right)

Book answer. (i) \dfrac{1-\cos4\alpha}{8} (ii) \dfrac{1}{16}\left[1-7\cos2\alpha-\cos4\alpha+\cos2\alpha\cos4\alpha\right] (iii) \dfrac{1}{128}\left[3-4\cos4\alpha+\cos8\alpha\right]

Note. The key for (ii) has a misprint: with \alpha=0 the left side is 0, but \tfrac{1}{16}[1-7-1+1]=-\tfrac{6}{16}. The expression \tfrac{1}{16}\left[1-\cos2\alpha-\cos4\alpha+\cos2\alpha\cos4\alpha\right] is right (the 7 should be absent), and it equals the answer above once \cos2\alpha\cos4\alpha is turned into a sum so that only cosines to the power 1 remain. Parts (i) and (iii) agree with the key.

Verify the following identities.

Solution

(i) (\sin\theta+\cos\theta)^2=1+\sin2\theta

Expand the square on the left side.
\text{L.H.S.}=\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta
Use \sin^2\theta+\cos^2\theta=1 and 2\sin\theta\cos\theta=\sin2\theta (identity 4).
=1+\sin2\theta=\text{R.H.S.}

(ii) \tan2x=\dfrac{1}{1-\tan x}-\dfrac{1}{1+\tan x}

Take the right side and subtract the fractions over the common denominator (1-\tan x)(1+\tan x).
\text{R.H.S.}=\dfrac{(1+\tan x)-(1-\tan x)}{(1-\tan x)(1+\tan x)}
Simplify the numerator and write the denominator as a difference of squares.
=\dfrac{2\tan x}{1-\tan^2x}
This is \tan2x by identity (11).
=\tan2x=\text{L.H.S.}

(iii) \tan\dfrac{\theta}{2}=\dfrac{\sin\theta}{1+\cos\theta}

Write the numerator with identity (12) and the denominator with identity (13), \cos\theta=2\cos^2\dfrac{\theta}{2}-1.
\text{R.H.S.}=\dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{1+\left(2\cos^2\frac{\theta}{2}-1\right)}=\dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}
Cancel 2\cos\dfrac{\theta}{2}.
=\dfrac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}=\tan\dfrac{\theta}{2}=\text{L.H.S.}

(iv) \csc2\alpha=\dfrac{\tan\alpha+\cot\alpha}{2}

Write the right side in sines and cosines and add with a common denominator.
\text{R.H.S.}=\dfrac12\left(\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}\right)=\dfrac12\cdot\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha}
The numerator is 1. Then use 2\sin\alpha\cos\alpha=\sin2\alpha.
=\dfrac{1}{2\sin\alpha\cos\alpha}=\dfrac{1}{\sin2\alpha}=\csc2\alpha=\text{L.H.S.}

(v) 8\sin^4\theta=3+\cos4\theta-4\cos2\theta

Write \sin^4\theta=(\sin^2\theta)^2 and use identity (9), \sin^2\theta=\dfrac{1-\cos2\theta}{2}.
\text{L.H.S.}=8\left(\dfrac{1-\cos2\theta}{2}\right)^2=2\left(1-2\cos2\theta+\cos^22\theta\right)
Use identity (8) for the square: \cos^22\theta=\dfrac{1+\cos4\theta}{2}.
=2-4\cos2\theta+2\cdot\dfrac{1+\cos4\theta}{2}=2-4\cos2\theta+1+\cos4\theta
Collect terms.
=3+\cos4\theta-4\cos2\theta=\text{R.H.S.}

(vi) \sin4\theta=4\sin\theta\cos^3\theta-4\sin^3\theta\cos\theta

Write 4\theta=2(2\theta) and use \sin2A=2\sin A\cos A with A=2\theta.
\sin4\theta=2\sin2\theta\cos2\theta
Expand \sin2\theta=2\sin\theta\cos\theta and \cos2\theta=\cos^2\theta-\sin^2\theta.
=2(2\sin\theta\cos\theta)(\cos^2\theta-\sin^2\theta)=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)
Multiply out.
=4\sin\theta\cos^3\theta-4\sin^3\theta\cos\theta

(vii) \sin2\theta=2\cot\theta\sin^2\theta

Write \cot\theta=\dfrac{\cos\theta}{\sin\theta} in the right side.
\text{R.H.S.}=2\cdot\dfrac{\cos\theta}{\sin\theta}\cdot\sin^2\theta
Cancel one \sin\theta.
=2\sin\theta\cos\theta=\sin2\theta=\text{L.H.S.}

(viii) \cos^22x+4\sin^2x\cos^2x=1

Write 4\sin^2x\cos^2x=(2\sin x\cos x)^2=\sin^22x.
\text{L.H.S.}=\cos^22x+(2\sin x\cos x)^2=\cos^22x+\sin^22x
Use \cos^2A+\sin^2A=1 with A=2x.
=1=\text{R.H.S.}

(ix) \cos4\theta=8\cos^4\theta-8\cos^2\theta+1

Use \cos2A=2\cos^2A-1 (identity 6) with A=2\theta.
\cos4\theta=2\cos^22\theta-1
Replace \cos2\theta=2\cos^2\theta-1 and expand the square.
=2\left(2\cos^2\theta-1\right)^2-1=2\left(4\cos^4\theta-4\cos^2\theta+1\right)-1
Multiply out and collect.
=8\cos^4\theta-8\cos^2\theta+1

(x) \sec2x=\dfrac{\cos x}{\cos x+\sin x}+\dfrac{\sin x}{\cos x-\sin x}

Add the fractions on the right over the common denominator (\cos x+\sin x)(\cos x-\sin x)=\cos^2x-\sin^2x.
\text{R.H.S.}=\dfrac{\cos x(\cos x-\sin x)+\sin x(\cos x+\sin x)}{\cos^2x-\sin^2x}
Expand the numerator; the terms \pm\sin x\cos x cancel and \cos^2x+\sin^2x=1.
=\dfrac{\cos^2x-\sin x\cos x+\sin x\cos x+\sin^2x}{\cos^2x-\sin^2x}=\dfrac{1}{\cos2x}
The denominator is \cos2x by identity (5).
=\sec2x=\text{L.H.S.}

(xi) \cos^4x-\sin^4x=\cos2x

The left side is a difference of squares, (\cos^2x)^2-(\sin^2x)^2.
\text{L.H.S.}=(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)
Use identity (5) for the first factor and \cos^2x+\sin^2x=1 for the second.
=\cos2x\cdot1=\cos2x=\text{R.H.S.}

(xii) \tan\dfrac{\beta}{2}+\cot\dfrac{\beta}{2}=2\csc\beta

Write both terms in sines and cosines and add.
\text{L.H.S.}=\dfrac{\sin\frac{\beta}{2}}{\cos\frac{\beta}{2}}+\dfrac{\cos\frac{\beta}{2}}{\sin\frac{\beta}{2}}=\dfrac{\sin^2\frac{\beta}{2}+\cos^2\frac{\beta}{2}}{\sin\frac{\beta}{2}\cos\frac{\beta}{2}}=\dfrac{1}{\sin\frac{\beta}{2}\cos\frac{\beta}{2}}
By identity (12), \sin\beta=2\sin\dfrac{\beta}{2}\cos\dfrac{\beta}{2}, so \sin\dfrac{\beta}{2}\cos\dfrac{\beta}{2}=\dfrac{\sin\beta}{2}.
=\dfrac{1}{\frac{\sin\beta}{2}}=\dfrac{2}{\sin\beta}=2\csc\beta=\text{R.H.S.}

(xiii) \csc2\alpha-\cot2\alpha=\tan\alpha

Write both terms with denominator \sin2\alpha.
\text{L.H.S.}=\dfrac{1}{\sin2\alpha}-\dfrac{\cos2\alpha}{\sin2\alpha}=\dfrac{1-\cos2\alpha}{\sin2\alpha}
Use 1-\cos2\alpha=2\sin^2\alpha (identity 7) and \sin2\alpha=2\sin\alpha\cos\alpha.
=\dfrac{2\sin^2\alpha}{2\sin\alpha\cos\alpha}=\dfrac{\sin\alpha}{\cos\alpha}=\tan\alpha=\text{R.H.S.}

(xiv) \dfrac{\cos^3x-\sin^3x}{\cos x-\sin x}=\dfrac{2+\sin2x}{2}

Factorise the numerator as a difference of cubes, a^3-b^3=(a-b)(a^2+ab+b^2).
\text{L.H.S.}=\dfrac{(\cos x-\sin x)(\cos^2x+\cos x\sin x+\sin^2x)}{\cos x-\sin x}
Cancel the common factor \cos x-\sin x and use \cos^2x+\sin^2x=1.
=1+\sin x\cos x
Write \sin x\cos x=\dfrac{\sin2x}{2}.
=1+\dfrac{\sin2x}{2}=\dfrac{2+\sin2x}{2}=\text{R.H.S.}

(xv) \dfrac{\sin3\alpha}{\sin\alpha}-\dfrac{\cos3\alpha}{\cos\alpha}=2

Subtract the fractions over the common denominator \sin\alpha\cos\alpha.
\text{L.H.S.}=\dfrac{\sin3\alpha\cos\alpha-\cos3\alpha\sin\alpha}{\sin\alpha\cos\alpha}
The numerator is \sin(3\alpha-\alpha) by identity (4).
=\dfrac{\sin2\alpha}{\sin\alpha\cos\alpha}=\dfrac{2\sin\alpha\cos\alpha}{\sin\alpha\cos\alpha}=2=\text{R.H.S.}

(xvi) \dfrac{1-\cos^2\beta}{2-2\cos\beta}=\cos^2\dfrac{\beta}{2}

Replace 1-\cos^2\beta=\sin^2\beta=(1-\cos\beta)(1+\cos\beta) and take 2 out of the denominator.
\text{L.H.S.}=\dfrac{(1-\cos\beta)(1+\cos\beta)}{2(1-\cos\beta)}
Cancel 1-\cos\beta, then use identity (8) with \alpha=\dfrac{\beta}{2}.
=\dfrac{1+\cos\beta}{2}=\cos^2\dfrac{\beta}{2}=\text{R.H.S.}

(xvii) \dfrac{\sin\theta}{1+\cos\theta}=\tan\dfrac{\theta}{2}

This is part (iii) read from the other side. Use identity (12) in the numerator and identity (13), 1+\cos\theta=2\cos^2\dfrac{\theta}{2}, in the denominator.
\text{L.H.S.}=\dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}
Cancel 2\cos\dfrac{\theta}{2}.
=\dfrac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}=\tan\dfrac{\theta}{2}=\text{R.H.S.}

(xviii) \dfrac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}=\cos x

Write \tan\dfrac{x}{2}=\dfrac{\sin(x/2)}{\cos(x/2)} and multiply the top and bottom by \cos^2\dfrac{x}{2}.
\text{L.H.S.}=\dfrac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}}
The denominator is 1 and the numerator is \cos x by identity (13).
=\dfrac{\cos x}{1}=\cos x=\text{R.H.S.}

(xix) \dfrac{\sin2\alpha}{\sin\alpha}-\dfrac{\cos2\alpha}{\cos\alpha}=\sec\alpha

Subtract the fractions over the common denominator \sin\alpha\cos\alpha.
\text{L.H.S.}=\dfrac{\sin2\alpha\cos\alpha-\cos2\alpha\sin\alpha}{\sin\alpha\cos\alpha}
The numerator is \sin(2\alpha-\alpha)=\sin\alpha by identity (4).
=\dfrac{\sin\alpha}{\sin\alpha\cos\alpha}=\dfrac{1}{\cos\alpha}=\sec\alpha=\text{R.H.S.}

(xx) 2\sin^2\dfrac{\beta}{2}+\cos\beta=1

By identity (13), \cos\beta=1-2\sin^2\dfrac{\beta}{2}. Substitute.
\text{L.H.S.}=2\sin^2\dfrac{\beta}{2}+1-2\sin^2\dfrac{\beta}{2}
The two sine terms cancel.
=1=\text{R.H.S.}

(xxi) 2\cos y\sec2y=\dfrac{1}{\cos y-\sin y}+\dfrac{1}{\cos y+\sin y}

Add the fractions on the right side.
\text{R.H.S.}=\dfrac{(\cos y+\sin y)+(\cos y-\sin y)}{(\cos y-\sin y)(\cos y+\sin y)}=\dfrac{2\cos y}{\cos^2y-\sin^2y}
The denominator is \cos2y by identity (5).
=\dfrac{2\cos y}{\cos2y}=2\cos y\sec2y=\text{L.H.S.}

(xxii) 2\sin y\sec2y=\dfrac{1}{\cos y-\sin y}-\dfrac{1}{\cos y+\sin y}

Subtract the fractions on the right side.
\text{R.H.S.}=\dfrac{(\cos y+\sin y)-(\cos y-\sin y)}{(\cos y-\sin y)(\cos y+\sin y)}=\dfrac{2\sin y}{\cos^2y-\sin^2y}
The denominator is \cos2y by identity (5).
=\dfrac{2\sin y}{\cos2y}=2\sin y\sec2y=\text{L.H.S.}

Exercise 8.3p. 212–213

Product-to-sum and sum-to-product formulae. To turn a product into a sum use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta), 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta), 2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta) and -2\sin\alpha\sin\beta=\cos(\alpha+\beta)-\cos(\alpha-\beta). To turn a sum or difference into a product put \alpha=\dfrac{p+q}{2}, \beta=\dfrac{p-q}{2} in these.

12026 Annual I, Set B, Q.2 (xii) OR

Use the product-to-sum formula to change the following to sum or difference.

Solution

(i) 4\sin16x\cos10x

Take out a factor 2 so that the product has the form 2\sin\alpha\cos\beta, with \alpha=16x, \beta=10x.
4\sin16x\cos10x=2\left(2\sin16x\cos10x\right)
Use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta).
=2\left[\sin(16x+10x)+\sin(16x-10x)\right]=2\left[\sin26x+\sin6x\right]

2\left[\sin26x+\sin6x\right]

(ii) 10\cos10y\cos6y

Take out a factor 5 so that the product has the form 2\cos\alpha\cos\beta, with \alpha=10y, \beta=6y.
10\cos10y\cos6y=5\left(2\cos10y\cos6y\right)
Use 2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta).
=5\left[\cos(10y+6y)+\cos(10y-6y)\right]=5\left[\cos16y+\cos4y\right]

5\left[\cos16y+\cos4y\right]

(iii) 2\cos5t\sin3t

This is already 2\cos\alpha\sin\beta with \alpha=5t, \beta=3t. Use 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta).
2\cos5t\sin3t=\sin(5t+3t)-\sin(5t-3t)=\sin8t-\sin2t

\sin8t-\sin2t

(iv) 6\cos5x\sin10x

Write it as 3\left(2\cos5x\sin10x\right) and use 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta) with \alpha=5x, \beta=10x.
6\cos5x\sin10x=3\left[\sin(5x+10x)-\sin(5x-10x)\right]=3\left[\sin15x-\sin(-5x)\right]
Since \sin(-5x)=-\sin5x, the minus sign becomes plus.
=3\left[\sin15x+\sin5x\right]

3\left[\sin15x+\sin5x\right]

(v) \sin(-u)\sin5u

Use \sin(-u)=-\sin u, so the product is -\sin u\sin5u=\dfrac12\left(-2\sin u\sin5u\right).
\sin(-u)\sin5u=\dfrac12\left(-2\sin u\sin5u\right)
Use -2\sin\alpha\sin\beta=\cos(\alpha+\beta)-\cos(\alpha-\beta) with \alpha=u, \beta=5u.
=\dfrac12\left[\cos(u+5u)-\cos(u-5u)\right]=\dfrac12\left[\cos6u-\cos(-4u)\right]
Since \cos(-4u)=\cos4u:
=\dfrac12\left[\cos6u-\cos4u\right]

\dfrac12\left[\cos6u-\cos4u\right]

(vi) -2\sin(-100^\circ)\sin(-20^\circ)

Both sines are of negative angles, so \sin(-100^\circ)=-\sin100^\circ and \sin(-20^\circ)=-\sin20^\circ. The two minus signs cancel.
-2\sin(-100^\circ)\sin(-20^\circ)=-2\sin100^\circ\sin20^\circ
Use -2\sin\alpha\sin\beta=\cos(\alpha+\beta)-\cos(\alpha-\beta) with \alpha=100^\circ, \beta=20^\circ.
=\cos(100^\circ+20^\circ)-\cos(100^\circ-20^\circ)=\cos120^\circ-\cos80^\circ

\cos120^\circ-\cos80^\circ

(vii) \cos23^\circ\sin17^\circ

Write it as \dfrac12\left(2\cos23^\circ\sin17^\circ\right) and use 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta) with \alpha=23^\circ, \beta=17^\circ.
\cos23^\circ\sin17^\circ=\dfrac12\left[\sin(23^\circ+17^\circ)-\sin(23^\circ-17^\circ)\right]
Add and subtract the angles.
=\dfrac12\left[\sin40^\circ-\sin6^\circ\right]

\dfrac12\left[\sin40^\circ-\sin6^\circ\right]

(viii) 2\cos56^\circ\sin48^\circ

Use 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta) with \alpha=56^\circ, \beta=48^\circ.
2\cos56^\circ\sin48^\circ=\sin(56^\circ+48^\circ)-\sin(56^\circ-48^\circ)=\sin104^\circ-\sin8^\circ

\sin104^\circ-\sin8^\circ

(ix) 2\sin75^\circ\sin15^\circ

The formula available is for -2\sin\alpha\sin\beta, so write 2\sin75^\circ\sin15^\circ=-\left(-2\sin75^\circ\sin15^\circ\right) and use -2\sin\alpha\sin\beta=\cos(\alpha+\beta)-\cos(\alpha-\beta).
2\sin75^\circ\sin15^\circ=-\left[\cos(75^\circ+15^\circ)-\cos(75^\circ-15^\circ)\right]=-\left[\cos90^\circ-\cos60^\circ\right]
Remove the bracket.
=\cos60^\circ-\cos90^\circ

\cos60^\circ-\cos90^\circ\ \left(=\dfrac12\right)

(x) 4\sin\dfrac{u+v}{2}\cos\dfrac{u-v}{2}

Write it as 2\left(2\sin\alpha\cos\beta\right) with \alpha=\dfrac{u+v}{2}, \beta=\dfrac{u-v}{2}. Then \alpha+\beta=u and \alpha-\beta=v.
\alpha+\beta=\dfrac{u+v+u-v}{2}=u,\qquad\alpha-\beta=\dfrac{u+v-u+v}{2}=v
Use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta).
4\sin\dfrac{u+v}{2}\cos\dfrac{u-v}{2}=2\left[\sin u+\sin v\right]

2\left[\sin u+\sin v\right]

(xi) 2\cos\dfrac{2u+2v}{2}\sin\dfrac{2u-2v}{2}

Simplify the angles first: \dfrac{2u+2v}{2}=u+v and \dfrac{2u-2v}{2}=u-v.
2\cos(u+v)\sin(u-v)
Use 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta) with \alpha=u+v, \beta=u-v. Then \alpha+\beta=2u and \alpha-\beta=2v.
=\sin2u-\sin2v

\sin2u-\sin2v

Book answer. (i) 2[\sin26x+\sin6x] (ii) 5[\cos16y+\cos4y] (iii) \sin8t-\sin2t (iv) 3[\sin15x+\sin5x] (v) \tfrac12[\cos6u-\cos4u] (vi) \cos120^\circ-\cos80^\circ (vii) \tfrac12[\sin40^\circ-\sin6^\circ] (viii) \sin104^\circ-\sin8^\circ (ix) \cos60^\circ-\cos90^\circ (x) 2[\sin u+\sin v] (xi) \sin2u-\sin2v

Rewrite the sum or difference as a product of two functions.

Solution

(i) \sin70^\circ+\sin30^\circ

Use \sin p+\sin q=2\sin\dfrac{p+q}{2}\cos\dfrac{p-q}{2} with p=70^\circ, q=30^\circ.
\sin70^\circ+\sin30^\circ=2\sin\dfrac{70^\circ+30^\circ}{2}\cos\dfrac{70^\circ-30^\circ}{2}
Simplify the angles.
=2\sin50^\circ\cos20^\circ

2\sin50^\circ\cos20^\circ

(ii) \sin76^\circ-\sin14^\circ

Use \sin p-\sin q=2\cos\dfrac{p+q}{2}\sin\dfrac{p-q}{2} with p=76^\circ, q=14^\circ.
\sin76^\circ-\sin14^\circ=2\cos\dfrac{76^\circ+14^\circ}{2}\sin\dfrac{76^\circ-14^\circ}{2}
Simplify the angles.
=2\cos45^\circ\sin31^\circ

2\cos45^\circ\sin31^\circ

(iii) \cos58^\circ-\cos12^\circ

Use \cos p-\cos q=-2\sin\dfrac{p+q}{2}\sin\dfrac{p-q}{2} with p=58^\circ, q=12^\circ.
\cos58^\circ-\cos12^\circ=-2\sin\dfrac{58^\circ+12^\circ}{2}\sin\dfrac{58^\circ-12^\circ}{2}
Simplify the angles.
=-2\sin35^\circ\sin23^\circ
Check with a calculator: \cos58^\circ-\cos12^\circ=0.5299-0.9781=-0.4482 and -2\sin35^\circ\sin23^\circ=-2(0.5736)(0.3907)=-0.4482.

-2\sin35^\circ\sin23^\circ

(iv) \cos\dfrac{p-q}{2}+\cos\dfrac{p+q}{2}

Use \cos p+\cos q=2\cos\dfrac{p+q}{2}\cos\dfrac{p-q}{2} with the two angles A=\dfrac{p+q}{2} and B=\dfrac{p-q}{2} in place of p and q. Then \dfrac{A+B}{2}=\dfrac{p}{2} and \dfrac{A-B}{2}=\dfrac{q}{2}.
\begin{aligned}\dfrac{A+B}{2}&=\dfrac12\cdot\dfrac{(p+q)+(p-q)}{2}=\dfrac p2\\ \dfrac{A-B}{2}&=\dfrac12\cdot\dfrac{(p+q)-(p-q)}{2}=\dfrac q2\end{aligned}
Substitute.
\cos\dfrac{p-q}{2}+\cos\dfrac{p+q}{2}=2\cos\dfrac p2\cos\dfrac q2

2\cos\dfrac p2\cos\dfrac q2

(v) \sin(-10^\circ)+\sin(-20^\circ)

Use \sin p+\sin q=2\sin\dfrac{p+q}{2}\cos\dfrac{p-q}{2} with p=-10^\circ, q=-20^\circ.
\sin(-10^\circ)+\sin(-20^\circ)=2\sin\dfrac{-10^\circ-20^\circ}{2}\cos\dfrac{-10^\circ+20^\circ}{2}=2\sin(-15^\circ)\cos5^\circ
Use \sin(-15^\circ)=-\sin15^\circ.
=-2\sin15^\circ\cos5^\circ

-2\sin15^\circ\cos5^\circ

Book answer. (i) 2\sin50^\circ\cos20^\circ (ii) 2\cos45^\circ\sin31^\circ (iii) 2\cos35^\circ\cos23^\circ (iv) 2\cos\dfrac p2\cos\dfrac q2 (v) -2\sin15^\circ\cos5^\circ

Note. The key's answer to (iii), 2\cos35^\circ\cos23^\circ, is not right for a difference of cosines: it is the form for \cos p+\cos q (and 2\cos35^\circ\cos23^\circ\approx1.50 while \cos58^\circ-\cos12^\circ\approx-0.448). The correct product is -2\sin35^\circ\sin23^\circ, verified numerically above. Parts (i), (ii), (iv) and (v) agree with the key.

Prove the following identities.

Solution

(i) \dfrac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=\dfrac{1-\tan\alpha\tan\beta}{1+\tan\alpha\tan\beta}

Expand the numerator with identity (23) and the denominator with identity (24).
\text{L.H.S.}=\dfrac{\cos\alpha\cos\beta-\sin\alpha\sin\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}
Divide the top and bottom by \cos\alpha\cos\beta.
=\dfrac{1-\frac{\sin\alpha}{\cos\alpha}\cdot\frac{\sin\beta}{\cos\beta}}{1+\frac{\sin\alpha}{\cos\alpha}\cdot\frac{\sin\beta}{\cos\beta}}=\dfrac{1-\tan\alpha\tan\beta}{1+\tan\alpha\tan\beta}=\text{R.H.S.}

(ii) \dfrac{6\cos8u\sin2u}{\sin(-6u)}=\dfrac{-3\sin10u}{\sin6u}+3

Turn the product in the numerator into a difference: 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta) with \alpha=8u, \beta=2u.
6\cos8u\sin2u=3\left(2\cos8u\sin2u\right)=3\left(\sin10u-\sin6u\right)
In the denominator use \sin(-6u)=-\sin6u.
\text{L.H.S.}=\dfrac{3\left(\sin10u-\sin6u\right)}{-\sin6u}
Divide each term of the numerator by -\sin6u.
=-\dfrac{3\sin10u}{\sin6u}+\dfrac{3\sin6u}{\sin6u}=\dfrac{-3\sin10u}{\sin6u}+3=\text{R.H.S.}

(iii) 4\cos4v\sin3v=2\left(\sin7v-\sin v\right)

Write the left side as 2\left(2\cos4v\sin3v\right) and use 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta) with \alpha=4v, \beta=3v.
\text{L.H.S.}=2\left[\sin(4v+3v)-\sin(4v-3v)\right]
Add and subtract the angles.
=2\left(\sin7v-\sin v\right)=\text{R.H.S.}

(iv) \sin3\theta+\sin\theta=4\cos^2\theta\sin\theta

Use \sin p+\sin q=2\sin\dfrac{p+q}{2}\cos\dfrac{p-q}{2} with p=3\theta, q=\theta.
\text{L.H.S.}=2\sin\dfrac{3\theta+\theta}{2}\cos\dfrac{3\theta-\theta}{2}=2\sin2\theta\cos\theta
Expand \sin2\theta=2\sin\theta\cos\theta.
=2(2\sin\theta\cos\theta)\cos\theta=4\cos^2\theta\sin\theta=\text{R.H.S.}

(v) \cos3x+\cos x=2\cos x\left(\cos2x\right)

Use \cos p+\cos q=2\cos\dfrac{p+q}{2}\cos\dfrac{p-q}{2} with p=3x, q=x.
\text{L.H.S.}=2\cos\dfrac{3x+x}{2}\cos\dfrac{3x-x}{2}=2\cos2x\cos x=2\cos x\left(\cos2x\right)=\text{R.H.S.}

(vi) 2\tan y\cos3y=\sec y\left(\sin4y-\sin2y\right)

Work on the right side. Use \sin p-\sin q=2\cos\dfrac{p+q}{2}\sin\dfrac{p-q}{2} with p=4y, q=2y.
\sin4y-\sin2y=2\cos3y\sin y
Multiply by \sec y=\dfrac{1}{\cos y}.
\text{R.H.S.}=\dfrac{1}{\cos y}\cdot2\cos3y\sin y=2\cdot\dfrac{\sin y}{\cos y}\cos3y=2\tan y\cos3y=\text{L.H.S.}

(vii) \dfrac{\sin6\beta+\sin4\beta}{\sin6\beta-\sin4\beta}=\tan5\beta\cot\beta

Use the sum formula for the numerator and the difference formula for the denominator, with p=6\beta, q=4\beta (so \dfrac{p+q}{2}=5\beta and \dfrac{p-q}{2}=\beta).
\text{L.H.S.}=\dfrac{2\sin5\beta\cos\beta}{2\cos5\beta\sin\beta}
Cancel the 2 and group the ratios.
=\dfrac{\sin5\beta}{\cos5\beta}\cdot\dfrac{\cos\beta}{\sin\beta}=\tan5\beta\cot\beta=\text{R.H.S.}

(viii) \dfrac{\cos3\theta+\cos\theta}{\cos3\theta-\cos\theta}=-\cot2\theta\cot\theta

Use the sum formula for cosines in the numerator and the difference formula in the denominator, with p=3\theta, q=\theta (so \dfrac{p+q}{2}=2\theta and \dfrac{p-q}{2}=\theta).
\text{L.H.S.}=\dfrac{2\cos2\theta\cos\theta}{-2\sin2\theta\sin\theta}
Cancel the 2 and group the ratios.
=-\dfrac{\cos2\theta}{\sin2\theta}\cdot\dfrac{\cos\theta}{\sin\theta}=-\cot2\theta\cot\theta=\text{R.H.S.}

(ix) \dfrac{\cos6x+\cos8x}{\sin6x-\sin4x}=\cot x\cos7x\sec5x

Numerator: sum formula with p=8x, q=6x. Denominator: difference formula with p=6x, q=4x.
\text{L.H.S.}=\dfrac{2\cos7x\cos x}{2\cos5x\sin x}
Cancel the 2 and write \dfrac{\cos x}{\sin x}=\cot x and \dfrac{1}{\cos5x}=\sec5x.
=\dfrac{\cos x}{\sin x}\cdot\cos7x\cdot\dfrac{1}{\cos5x}=\cot x\cos7x\sec5x=\text{R.H.S.}

(x) \dfrac{\cos2\alpha-\cos4\alpha}{\sin2\alpha+\sin4\alpha}=\tan\alpha

Numerator: difference formula for cosines with p=2\alpha, q=4\alpha (so \dfrac{p+q}{2}=3\alpha and \dfrac{p-q}{2}=-\alpha). Denominator: sum formula for sines.
\text{L.H.S.}=\dfrac{-2\sin3\alpha\sin(-\alpha)}{2\sin3\alpha\cos\alpha}
Since \sin(-\alpha)=-\sin\alpha, the numerator becomes 2\sin3\alpha\sin\alpha. Cancel 2\sin3\alpha.
=\dfrac{2\sin3\alpha\sin\alpha}{2\sin3\alpha\cos\alpha}=\dfrac{\sin\alpha}{\cos\alpha}=\tan\alpha=\text{R.H.S.}

(xi) 2\cos2u\cos u+\sin2u\sin u=2\cos^3u

Turn each product into a sum or difference: 2\cos2u\cos u=\cos3u+\cos u and 2\sin2u\sin u=\cos u-\cos3u, so \sin2u\sin u=\dfrac12(\cos u-\cos3u).
\text{L.H.S.}=(\cos3u+\cos u)+\dfrac12(\cos u-\cos3u)=\dfrac12\cos3u+\dfrac32\cos u
Use the triple-angle identity (19), \cos3u=4\cos^3u-3\cos u.
=\dfrac12\left(4\cos^3u-3\cos u\right)+\dfrac32\cos u=2\cos^3u-\dfrac32\cos u+\dfrac32\cos u
The terms in \cos u cancel.
=2\cos^3u=\text{R.H.S.}

(xii) 2\sin2y\sin3y=\cos y-\cos5y

Use -2\sin\alpha\sin\beta=\cos(\alpha+\beta)-\cos(\alpha-\beta), which gives 2\sin\alpha\sin\beta=\cos(\alpha-\beta)-\cos(\alpha+\beta). Take \alpha=3y, \beta=2y.
\text{L.H.S.}=\cos(3y-2y)-\cos(3y+2y)=\cos y-\cos5y=\text{R.H.S.}

(xiii) \dfrac{\cos10x+\cos6x}{\cos6x-\cos10x}=\cot2x\cot8x

Numerator: sum formula with p=10x, q=6x (so \dfrac{p+q}{2}=8x, \dfrac{p-q}{2}=2x). Denominator: difference formula with p=6x, q=10x (so \dfrac{p+q}{2}=8x, \dfrac{p-q}{2}=-2x).
\text{L.H.S.}=\dfrac{2\cos8x\cos2x}{-2\sin8x\sin(-2x)}
Since \sin(-2x)=-\sin2x, the two minus signs cancel.
=\dfrac{2\cos8x\cos2x}{2\sin8x\sin2x}=\dfrac{\cos8x}{\sin8x}\cdot\dfrac{\cos2x}{\sin2x}=\cot8x\cot2x=\text{R.H.S.}

Prove that.

Solution

(i) \cos80^\circ\cos60^\circ\cos40^\circ\cos20^\circ=\dfrac{1}{16}

Put \cos60^\circ=\dfrac12 and group the other three factors.
\text{L.H.S.}=\dfrac12\cos80^\circ\cos40^\circ\cos20^\circ
Turn the product \cos80^\circ\cos40^\circ into a sum, using 2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta), and put \cos120^\circ=-\dfrac12.
\cos80^\circ\cos40^\circ=\dfrac12\left[\cos120^\circ+\cos40^\circ\right]=-\dfrac14+\dfrac12\cos40^\circ
Multiply by \cos20^\circ.
\cos80^\circ\cos40^\circ\cos20^\circ=-\dfrac14\cos20^\circ+\dfrac12\cos40^\circ\cos20^\circ
Turn \cos40^\circ\cos20^\circ into a sum: \dfrac12\left[\cos60^\circ+\cos20^\circ\right]=\dfrac14+\dfrac12\cos20^\circ.
=-\dfrac14\cos20^\circ+\dfrac12\left(\dfrac14+\dfrac12\cos20^\circ\right)=-\dfrac14\cos20^\circ+\dfrac18+\dfrac14\cos20^\circ=\dfrac18
Multiply by the factor \dfrac12 taken out in the first step.
\text{L.H.S.}=\dfrac12\cdot\dfrac18=\dfrac{1}{16}=\text{R.H.S.}

(ii) \sin70^\circ\sin50^\circ\sin30^\circ\sin10^\circ=\dfrac{1}{16}

Put \sin30^\circ=\dfrac12 and start with the product \sin50^\circ\sin10^\circ, using -2\sin\alpha\sin\beta=\cos(\alpha+\beta)-\cos(\alpha-\beta).
\sin50^\circ\sin10^\circ=-\dfrac12\left[\cos60^\circ-\cos40^\circ\right]=\dfrac12\cos40^\circ-\dfrac14
Multiply by \sin70^\circ.
\sin70^\circ\sin50^\circ\sin10^\circ=\dfrac12\cos40^\circ\sin70^\circ-\dfrac14\sin70^\circ
Turn \cos40^\circ\sin70^\circ into a sum, using 2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta): \cos40^\circ\sin70^\circ=\dfrac12\left[\sin110^\circ-\sin(-30^\circ)\right]=\dfrac12\left[\sin110^\circ+\dfrac12\right].
=\dfrac14\sin110^\circ+\dfrac18-\dfrac14\sin70^\circ
Since \sin110^\circ=\sin(180^\circ-70^\circ)=\sin70^\circ, the sine terms cancel.
\sin70^\circ\sin50^\circ\sin10^\circ=\dfrac18
Multiply by \sin30^\circ=\dfrac12.
\text{L.H.S.}=\dfrac12\cdot\dfrac18=\dfrac{1}{16}=\text{R.H.S.}

(iii) \sin\dfrac{\pi}{9}\sin\dfrac{2\pi}{9}\sin\dfrac{3\pi}{9}\sin\dfrac{4\pi}{9}=\dfrac{3}{16}

Change to degrees: \dfrac{\pi}{9}=20^\circ, \dfrac{2\pi}{9}=40^\circ, \dfrac{3\pi}{9}=60^\circ, \dfrac{4\pi}{9}=80^\circ. Put \sin60^\circ=\dfrac{\sqrt3}{2}.
\text{L.H.S.}=\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\dfrac{\sqrt3}{2}\sin20^\circ\sin40^\circ\sin80^\circ
Start with \sin40^\circ\sin20^\circ=-\dfrac12\left[\cos60^\circ-\cos20^\circ\right]=\dfrac12\cos20^\circ-\dfrac14 and multiply by \sin80^\circ.
\sin20^\circ\sin40^\circ\sin80^\circ=\dfrac12\cos20^\circ\sin80^\circ-\dfrac14\sin80^\circ
Turn \cos20^\circ\sin80^\circ into a sum: \dfrac12\left[\sin100^\circ+\sin60^\circ\right].
=\dfrac14\sin100^\circ+\dfrac14\sin60^\circ-\dfrac14\sin80^\circ
Since \sin100^\circ=\sin80^\circ, those two terms cancel and \sin60^\circ=\dfrac{\sqrt3}{2}.
=\dfrac14\sin60^\circ=\dfrac{\sqrt3}{8}
Multiply by the factor \dfrac{\sqrt3}{2} taken out in the first step.
\text{L.H.S.}=\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{8}=\dfrac{3}{16}=\text{R.H.S.}

Review Exercise 8p. 213–215

A mixed review of the unit: sum and difference formulae, allied angles, double, half and triple angles, and product-to-sum and sum-to-product formulae. In a proof, change one side step by step until it equals the other, giving the identity used at each step.

1 (i)

\sin(45^\circ-30^\circ)=\dots

A\dfrac{\sqrt6-\sqrt2}{4}Answer
B\dfrac{\sqrt6+\sqrt2}{4}
C\dfrac{\sqrt6-\sqrt2}{2}
D\dfrac{\sqrt3-\sqrt2}{2}

Why

Use \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta with \alpha=45^\circ, \beta=30^\circ.
\sin15^\circ=\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2}\cdot\dfrac12=\dfrac{\sqrt3-1}{2\sqrt2}
Multiply the top and bottom by \sqrt2.
=\dfrac{\sqrt6-\sqrt2}{4}

Option (b) is \sin75^\circ (a plus sign), and (c) is twice the correct value.

Book answer. (a)

1 (ii)

\tan\left(\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=\dots

A\dfrac{\sqrt3-1}{\sqrt3+1}
B\dfrac{\sqrt3+1}{\sqrt3-1}Answer
C\dfrac{\sqrt3+1}{-\sqrt3-1}
D\dfrac{\sqrt3+1}{-\sqrt3+1}

Why

Use \tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} with \tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3} and \tan\dfrac{\pi}{4}=1.
\tan75^\circ=\dfrac{\frac{1}{\sqrt3}+1}{1-\frac{1}{\sqrt3}}
Multiply the top and bottom by \sqrt3.
=\dfrac{1+\sqrt3}{\sqrt3-1}=\dfrac{\sqrt3+1}{\sqrt3-1}

The angle is 75^\circ, in quadrant I, so the value is positive. Options (c) and (d) are negative ((c) equals -1), and (a) is \tan15^\circ.

Book answer. (b)

1 (iii)

\sin22.5^\circ\cos22.5^\circ+\cos22.5^\circ\sin22.5^\circ=\dots

A\dfrac{-1}{\sqrt3}
B\dfrac{1}{\sqrt3}
C\dfrac{1}{\sqrt2}Answer
D\dfrac{-1}{\sqrt2}

Why

The expression has the form \sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta) with \alpha=\beta=22.5^\circ.
\sin(22.5^\circ+22.5^\circ)=\sin45^\circ=\dfrac{1}{\sqrt2}
The same result follows from 2\sin22.5^\circ\cos22.5^\circ=\sin45^\circ (double-angle identity), as in Example 10.

Book answer. (c)

1 (iv)

\cos(\pi-\theta)=\dots

A\sec\theta
B\pm\cos\theta
C\cos\theta
D-\cos\thetaAnswer

Why

By the fundamental law, \cos(\pi-\theta)=\cos\pi\cos\theta+\sin\pi\sin\theta.
\cos(\pi-\theta)=(-1)\cos\theta+0\cdot\sin\theta=-\cos\theta

The terminal side of \pi-\theta is in quadrant II (for acute \theta), where cosine is negative, and the ratio does not change because \pi is an even multiple of a right angle.

Book answer. (d)

1 (v)

\tan\left(\dfrac{\pi}{2}+\theta\right)=\dots

A\cot\theta
B-\cot\thetaAnswer
C\tan\theta
D-\tan\theta

Why

An odd multiple of a right angle changes tangent into cotangent. Use \sin\left(\dfrac{\pi}{2}+\theta\right)=\cos\theta and \cos\left(\dfrac{\pi}{2}+\theta\right)=-\sin\theta.
\tan\left(\dfrac{\pi}{2}+\theta\right)=\dfrac{\cos\theta}{-\sin\theta}=-\cot\theta

Quadrant II (for acute \theta) has negative tangent, which gives the minus sign. Option (a) has the right ratio but the wrong sign.

Book answer. (b)

1 (vi)

2\sin\alpha\cos\alpha=\dots

A\sin(\pi-2\alpha)Answer
B\sin(\pi+2\alpha)
C\sin(-2\alpha)
D\sin2(\pi-\alpha)

Why

By identity (4), 2\sin\alpha\cos\alpha=\sin2\alpha. Now reduce each option.
\begin{aligned}\sin(\pi-2\alpha)&=\sin2\alpha\\ \sin(\pi+2\alpha)&=-\sin2\alpha\\ \sin(-2\alpha)&=-\sin2\alpha\\ \sin2(\pi-\alpha)&=\sin(2\pi-2\alpha)=-\sin2\alpha\end{aligned}
Only option (a) equals \sin2\alpha.

Book answer. (a)

1 (vii)

\dfrac{\sin2\alpha\cos\alpha}{\cos^3\alpha-\cos\alpha\sin^2\alpha}=\dots

A\csc2\alpha
B-\sec2\alpha
C\tan2\alphaAnswer
D-\tan2\alpha

Why

Take \cos\alpha out of the denominator.
\cos^3\alpha-\cos\alpha\sin^2\alpha=\cos\alpha\left(\cos^2\alpha-\sin^2\alpha\right)=\cos\alpha\cos2\alpha
Cancel \cos\alpha from the top and bottom.
\dfrac{\sin2\alpha\cos\alpha}{\cos\alpha\cos2\alpha}=\dfrac{\sin2\alpha}{\cos2\alpha}=\tan2\alpha

Book answer. (c)

1 (viii)

If \sin\beta=\dfrac35, then \cos2\beta=\dots

A\dfrac{-7}{5}
B\dfrac75
C\dfrac{-7}{25}
D\dfrac{7}{25}Answer

Why

Use \cos2\beta=1-2\sin^2\beta (identity 7), since only \sin\beta is given.
\cos2\beta=1-2\left(\dfrac35\right)^2=1-\dfrac{18}{25}=\dfrac{7}{25}

Options (a) and (b) have absolute value greater than 1, which no cosine can have. Option (c) has the wrong sign.

Book answer. (d)

1 (ix)

\cos^23x-\sin^23x=\dots

A\sin6x
B\cos6xAnswer
C-\sin6x
D-\cos6x

Why

Use identity (5), \cos^2A-\sin^2A=\cos2A, with A=3x.
\cos^23x-\sin^23x=\cos(2\cdot3x)=\cos6x

Book answer. (b)

1 (x)

(\sin x-\cos x)^2=\dots

A1+\sin2x
B1-\cos2x
C1-\sin2xAnswer
D1+\cos x\sin2x

Why

Expand the square and use \sin^2x+\cos^2x=1.
(\sin x-\cos x)^2=\sin^2x-2\sin x\cos x+\cos^2x=1-2\sin x\cos x
Use 2\sin x\cos x=\sin2x.
=1-\sin2x

Option (a) has the sign of the middle term wrong; that is the expansion of (\sin x+\cos x)^2.

Book answer. (c)

1 (xi)

\cos(60^\circ-30^\circ)\neq\dots

A\cos30^\circ
B\sin60^\circ
C\sqrt{1-\sin^230^\circ}
D\cos60^\circ-\cos30^\circAnswer

Why

\cos(60^\circ-30^\circ)=\cos30^\circ=\dfrac{\sqrt3}{2}. Check each option against this value.
\text{(a) }\cos30^\circ=\dfrac{\sqrt3}{2},\quad\text{(b) }\sin60^\circ=\dfrac{\sqrt3}{2},\quad\text{(c) }\sqrt{1-\tfrac14}=\dfrac{\sqrt3}{2}
Option (d) is different. The cosine of a difference is not the difference of the cosines.
\cos60^\circ-\cos30^\circ=\dfrac12-\dfrac{\sqrt3}{2}\approx-0.366\neq\dfrac{\sqrt3}{2}

The question asks for the option that is not equal, so the answer is (d).

Book answer. (d)

1 (xii)

\dfrac{1-\cos x}{\sin x}=\dots

A\tan\left(\dfrac x2\right)Answer
B\cot\left(\dfrac x2\right)
C-\tan\left(\dfrac x2\right)
D-\cot\left(\dfrac x2\right)

Why

Use identity (13), \cos x=1-2\sin^2\dfrac x2, and identity (12), \sin x=2\sin\dfrac x2\cos\dfrac x2.
\dfrac{1-\cos x}{\sin x}=\dfrac{2\sin^2\frac x2}{2\sin\frac x2\cos\frac x2}
Cancel 2\sin\dfrac x2.
=\dfrac{\sin\frac x2}{\cos\frac x2}=\tan\dfrac x2

This is the form \tan\dfrac{\theta}{2}=\dfrac{1-\cos\theta}{\sin\theta} quoted in Example 7.

Book answer. (a)

Given that \sin\theta=\dfrac35, \sin\phi=\dfrac{5}{13} where \theta is obtuse and \phi is acute. Find the values of: (i) \sin(\theta-\phi) (ii) \tan(\theta-\phi) (iii) \tan(\theta+\phi)

Solution

\theta is obtuse (quadrant II), so \cos\theta<0.
\cos\theta=-\sqrt{1-\dfrac{9}{25}}=-\dfrac45,\qquad\tan\theta=\dfrac{3/5}{-4/5}=-\dfrac34
\phi is acute (quadrant I), so \cos\phi>0.
\cos\phi=\sqrt{1-\dfrac{25}{169}}=\dfrac{12}{13},\qquad\tan\phi=\dfrac{5/13}{12/13}=\dfrac{5}{12}
(i) \sin(\theta-\phi)=\sin\theta\cos\phi-\cos\theta\sin\phi.
\sin(\theta-\phi)=\dfrac35\cdot\dfrac{12}{13}-\left(-\dfrac45\right)\dfrac{5}{13}=\dfrac{36}{65}+\dfrac{20}{65}=\dfrac{56}{65}
(ii) \tan(\theta-\phi)=\dfrac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi}. The numerator is -\dfrac34-\dfrac{5}{12}=-\dfrac{9}{12}-\dfrac{5}{12}=-\dfrac{14}{12} and the denominator is 1+\left(-\dfrac34\right)\dfrac{5}{12}=\dfrac{48-15}{48}=\dfrac{33}{48}.
\tan(\theta-\phi)=\dfrac{-14/12}{33/48}=-\dfrac{14}{12}\cdot\dfrac{48}{33}=-\dfrac{56}{33}
(iii) \tan(\theta+\phi)=\dfrac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}. The numerator is -\dfrac34+\dfrac{5}{12}=-\dfrac{9}{12}+\dfrac{5}{12}=-\dfrac{4}{12}=-\dfrac13 and the denominator is 1-\left(-\dfrac34\right)\dfrac{5}{12}=\dfrac{48+15}{48}=\dfrac{63}{48}.
\tan(\theta+\phi)=\dfrac{-1/3}{63/48}=-\dfrac13\cdot\dfrac{48}{63}=-\dfrac{16}{63}
Check (iii) with sines and cosines: \sin(\theta+\phi)=\dfrac{36}{65}-\dfrac{20}{65}=\dfrac{16}{65} and \cos(\theta+\phi)=\left(-\dfrac45\right)\dfrac{12}{13}-\dfrac35\cdot\dfrac5{13}=-\dfrac{63}{65}.
\tan(\theta+\phi)=\dfrac{16/65}{-63/65}=-\dfrac{16}{63}\ \checkmark

\sin(\theta-\phi)=\dfrac{56}{65},\quad\tan(\theta-\phi)=-\dfrac{56}{33},\quad\tan(\theta+\phi)=-\dfrac{16}{63}

Book answer. (i) \dfrac{56}{65} (ii) -\dfrac{56}{33} (iii) -\dfrac{16}{33}

Note. Parts (i) and (ii) agree with the key. The key's (iii), -\dfrac{16}{33}, is a misprint: the denominator of \tan(\theta+\phi) is 1-\tan\theta\tan\phi=\dfrac{63}{48}, which gives -\dfrac{16}{63}. The check with \sin(\theta+\phi)=\dfrac{16}{65} and \cos(\theta+\phi)=-\dfrac{63}{65} confirms -\dfrac{16}{63}.

Express the following as single trigonometric ratios.

Solution

(i) \dfrac{1}{\sqrt2}(\sin\beta+\cos\beta)

Write \dfrac{1}{\sqrt2}=\cos45^\circ=\sin45^\circ and distribute.
\dfrac{1}{\sqrt2}(\sin\beta+\cos\beta)=\sin\beta\cos45^\circ+\cos\beta\sin45^\circ
This has the form \sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta) (identity 3).
=\sin(\beta+45^\circ)
The same expression can also be written with a cosine, because \sin\beta\sin45^\circ+\cos\beta\cos45^\circ=\cos(\beta-45^\circ).
=\cos(\beta-45^\circ)

\sin(\beta+45^\circ)\ \text{or}\ \cos(\beta-45^\circ)

(ii) \dfrac{1}{\sqrt2}\sin75^\circ+\dfrac{1}{\sqrt2}\cos75^\circ

Write \dfrac{1}{\sqrt2}=\cos45^\circ=\sin45^\circ.
\sin75^\circ\cos45^\circ+\cos75^\circ\sin45^\circ
This is \sin(75^\circ+45^\circ) by identity (3).
=\sin120^\circ
Since 120^\circ=90^\circ+30^\circ, the allied-angle rule \sin\left(\tfrac{\pi}{2}+\theta\right)=\cos\theta gives \sin120^\circ=\cos30^\circ, and the value is \dfrac{\sqrt3}{2}.
=\cos30^\circ=\dfrac{\sqrt3}{2}

\sin120^\circ\ \text{or}\ \cos30^\circ

Book answer. (i) \sin(\beta+45^\circ) or \cos(\beta-45^\circ) (ii) \sin120^\circ or \cos30^\circ

Find the values of:

Solution

(i) \dfrac{1+\tan15^\circ}{1-\tan15^\circ}

Write 1=\tan45^\circ in the numerator and the denominator.
\dfrac{1+\tan15^\circ}{1-\tan15^\circ}=\dfrac{\tan45^\circ+\tan15^\circ}{1-\tan45^\circ\tan15^\circ}
This is \tan(\alpha+\beta) with \alpha=45^\circ, \beta=15^\circ.
=\tan(45^\circ+15^\circ)=\tan60^\circ=\sqrt3\approx1.732

\tan60^\circ=\sqrt3\approx1.732

(ii) \cos70^\circ\cos20^\circ-\sin70^\circ\sin20^\circ

This is \cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta) with \alpha=70^\circ, \beta=20^\circ.
\cos(70^\circ+20^\circ)=\cos90^\circ=0

\cos90^\circ=0

Book answer. (i) \tan60^\circ=1.732 (ii) \cos90^\circ=0

Find the values of \tan\theta when \tan(\theta-45^\circ)=\dfrac13.

Solution

Expand the left side with the difference formula for tangent. Put \tan\theta=t and \tan45^\circ=1.
\tan(\theta-45^\circ)=\dfrac{\tan\theta-\tan45^\circ}{1+\tan\theta\tan45^\circ}=\dfrac{t-1}{1+t}
Equate to the given value.
\dfrac{t-1}{1+t}=\dfrac13
Cross-multiply.
3(t-1)=1+t\ \Rightarrow\ 3t-3=1+t
Collect the terms in t.
2t=4\ \Rightarrow\ t=2
Check: \tan(\theta-45^\circ)=\dfrac{2-1}{1+2}=\dfrac13.

\tan\theta=2

Book answer. \tan\theta=2

Prove the following.

Solution

(i) If \sin(\alpha+\theta)=2\cos(\alpha-\theta) prove \tan\alpha=\dfrac{2-\tan\theta}{1-2\tan\theta}.

Expand both sides with identities (3) and (24).
\sin\alpha\cos\theta+\cos\alpha\sin\theta=2\left(\cos\alpha\cos\theta+\sin\alpha\sin\theta\right)
Divide every term by \cos\alpha\cos\theta.
\tan\alpha+\tan\theta=2+2\tan\alpha\tan\theta
Collect the terms in \tan\alpha on the left side.
\tan\alpha-2\tan\alpha\tan\theta=2-\tan\theta
Factorise and divide.
\tan\alpha(1-2\tan\theta)=2-\tan\theta\ \Rightarrow\ \tan\alpha=\dfrac{2-\tan\theta}{1-2\tan\theta}

(ii) If \sin(\alpha-\theta)=\cos(\alpha+\theta) prove \tan\alpha=1.

Expand both sides with identities (4) and (2).
\sin\alpha\cos\theta-\cos\alpha\sin\theta=\cos\alpha\cos\theta-\sin\alpha\sin\theta
Bring the \sin\alpha terms to the left and the \cos\alpha terms to the right.
\sin\alpha\cos\theta+\sin\alpha\sin\theta=\cos\alpha\cos\theta+\cos\alpha\sin\theta
Factorise both sides.
\sin\alpha(\cos\theta+\sin\theta)=\cos\alpha(\cos\theta+\sin\theta)
Divide by \cos\theta+\sin\theta (not zero for a general \theta) and then by \cos\alpha.
\sin\alpha=\cos\alpha\ \Rightarrow\ \tan\alpha=1

Show that:

Solution

(i) \dfrac{4\sin^2\theta\cos\theta}{\cos3\theta+\cos\theta}=\tan2\theta\tan\theta

Turn the denominator into a product: \cos p+\cos q=2\cos\dfrac{p+q}{2}\cos\dfrac{p-q}{2} with p=3\theta, q=\theta.
\cos3\theta+\cos\theta=2\cos2\theta\cos\theta
Substitute and cancel 2\cos\theta.
\text{L.H.S.}=\dfrac{4\sin^2\theta\cos\theta}{2\cos2\theta\cos\theta}=\dfrac{2\sin^2\theta}{\cos2\theta}
Write the right side with sines and cosines: \tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta} (identity 11).
\text{R.H.S.}=\dfrac{2\tan\theta}{1-\tan^2\theta}\cdot\tan\theta=\dfrac{2\tan^2\theta}{1-\tan^2\theta}
Multiply the top and bottom by \cos^2\theta.
=\dfrac{2\sin^2\theta}{\cos^2\theta-\sin^2\theta}=\dfrac{2\sin^2\theta}{\cos2\theta}=\text{L.H.S.}

(ii) \dfrac{\sin10\theta-\sin4\theta}{\sin4\theta+\sin2\theta}=\cos7\theta\sec\theta

Numerator: difference formula \sin p-\sin q=2\cos\dfrac{p+q}{2}\sin\dfrac{p-q}{2} with p=10\theta, q=4\theta. Denominator: sum formula \sin p+\sin q=2\sin\dfrac{p+q}{2}\cos\dfrac{p-q}{2} with p=4\theta, q=2\theta.
\text{L.H.S.}=\dfrac{2\cos7\theta\sin3\theta}{2\sin3\theta\cos\theta}
Cancel 2\sin3\theta.
=\dfrac{\cos7\theta}{\cos\theta}=\cos7\theta\sec\theta=\text{R.H.S.}

Prove that:

Solution

(i) \sqrt{\dfrac{\cos(90^\circ+x)\sec(-x)\tan(180^\circ-x)}{\sec(360^\circ-x)\sin(180^\circ+x)\cot(90^\circ-x)}}=i

Reduce each factor with the allied-angle table. For the numerator:
\cos(90^\circ+x)=-\sin x,\qquad\sec(-x)=\sec x,\qquad\tan(180^\circ-x)=-\tan x
Multiply the three factors.
\text{numerator}=(-\sin x)(\sec x)(-\tan x)=\sin x\sec x\tan x=\tan x\cdot\tan x=\tan^2x
For the denominator:
\sec(360^\circ-x)=\sec x,\qquad\sin(180^\circ+x)=-\sin x,\qquad\cot(90^\circ-x)=\tan x
Multiply the three factors.
\text{denominator}=(\sec x)(-\sin x)(\tan x)=-\tan x\cdot\tan x=-\tan^2x
Divide and take the square root; i is the imaginary unit with i^2=-1.
\sqrt{\dfrac{\tan^2x}{-\tan^2x}}=\sqrt{-1}=i

(ii) \dfrac{\tan^2\left(\dfrac{3\pi}{2}-x\right)\sin^2(\pi+x)\sin(2\pi-x)}{\cos^2(\pi-x)\cot\left(\dfrac{3\pi}{2}+x\right)}=\cos x

Reduce the numerator factors (an odd multiple of \tfrac{\pi}{2} changes tangent to cotangent; the sign follows the quadrant).
\tan\left(\dfrac{3\pi}{2}-x\right)=\cot x,\quad\sin(\pi+x)=-\sin x,\quad\sin(2\pi-x)=-\sin x
So the numerator is
\cot^2x\cdot\sin^2x\cdot(-\sin x)=\dfrac{\cos^2x}{\sin^2x}\cdot\sin^2x\cdot(-\sin x)=-\cos^2x\sin x
Reduce the denominator factors.
\cos(\pi-x)=-\cos x,\quad\cot\left(\dfrac{3\pi}{2}+x\right)=-\tan x
So the denominator is
\cos^2x\cdot(-\tan x)=-\cos^2x\tan x
Divide, cancel -\cos^2x and write \tan x=\dfrac{\sin x}{\cos x}.
\dfrac{-\cos^2x\sin x}{-\cos^2x\tan x}=\dfrac{\sin x}{\tan x}=\sin x\cdot\dfrac{\cos x}{\sin x}=\cos x

Note. In (i) the result is i because the ratio under the root is -1. This uses the imaginary unit of the unit on complex numbers.

Simplify \sqrt{\dfrac{\left(1-\tan^2x\cos(-x)\cos(360^\circ-x)\right)\tan45^\circ}{\left\{\sin90^\circ-\sin(180^\circ+x)\right\}\left\{\sin90^\circ-\cos(90^\circ-x)\right\}}}

Solution

Reduce the numerator. \cos(-x)=\cos x, \cos(360^\circ-x)=\cos x and \tan45^\circ=1.
\left(1-\tan^2x\cos x\cos x\right)\cdot1=1-\tan^2x\cos^2x
Write \tan^2x\cos^2x=\sin^2x.
=1-\sin^2x=\cos^2x
Reduce the denominator. \sin90^\circ=1, \sin(180^\circ+x)=-\sin x and \cos(90^\circ-x)=\sin x.
\{1-(-\sin x)\}\{1-\sin x\}=(1+\sin x)(1-\sin x)
The product is a difference of squares.
=1-\sin^2x=\cos^2x
Divide the numerator by the denominator and take the root.
\sqrt{\dfrac{\cos^2x}{\cos^2x}}=\sqrt1=1

Book answer. 1

Prove that:

Solution

(i) \sin(16x)=16\sin(x)\cos(x)\cos(2x)\cos(4x)\cos(8x)

Use \sin2A=2\sin A\cos A with A=8x.
\sin16x=2\sin8x\cos8x
Apply the same identity to \sin8x=2\sin4x\cos4x.
=2\left(2\sin4x\cos4x\right)\cos8x=4\sin4x\cos4x\cos8x
Apply it to \sin4x=2\sin2x\cos2x.
=4\left(2\sin2x\cos2x\right)\cos4x\cos8x=8\sin2x\cos2x\cos4x\cos8x
Apply it to \sin2x=2\sin x\cos x.
=8\left(2\sin x\cos x\right)\cos2x\cos4x\cos8x=16\sin x\cos x\cos2x\cos4x\cos8x

(ii) \dfrac{1+\cos2\theta}{\sin2\theta-\cos\theta}=\dfrac{2\cos\theta}{2\sin\theta-1}

Use 1+\cos2\theta=2\cos^2\theta (identity 6) and \sin2\theta=2\sin\theta\cos\theta (identity 4).
\text{L.H.S.}=\dfrac{2\cos^2\theta}{2\sin\theta\cos\theta-\cos\theta}
Take \cos\theta out of the denominator.
=\dfrac{2\cos^2\theta}{\cos\theta(2\sin\theta-1)}
Cancel one \cos\theta.
=\dfrac{2\cos\theta}{2\sin\theta-1}=\text{R.H.S.}

(iii) \dfrac{2\tan^2\theta}{\tan^2\theta-1}=\dfrac{-2\tan^2\theta}{\sec^2\theta-2\tan^2\theta}

Work on the denominator of the right side. Use \sec^2\theta=1+\tan^2\theta.
\sec^2\theta-2\tan^2\theta=1+\tan^2\theta-2\tan^2\theta=1-\tan^2\theta=-(\tan^2\theta-1)
Substitute into the right side.
\text{R.H.S.}=\dfrac{-2\tan^2\theta}{-(\tan^2\theta-1)}
The two minus signs cancel.
=\dfrac{2\tan^2\theta}{\tan^2\theta-1}=\text{L.H.S.}

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 141 mark

If \tan\theta=2, what is the value of \cos\theta when \theta is in the first quadrant?

A\dfrac{1}{\sqrt{3}}
B\dfrac{2}{\sqrt{5}}
C\sqrt{5}
D\dfrac{1}{\sqrt{5}}Answer

Why

\tan\theta=2=\dfrac21 in the first quadrant: opposite side 2, adjacent side 1, so the hypotenuse is \sqrt{1^2+2^2}=\sqrt5.

\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac1{\sqrt5}

Option B, \dfrac2{\sqrt5}, is \sin\theta (opposite over hypotenuse), the ratio most often confused with cosine here.

2025 Annual I, Set B, MCQ 151 mark

If \sin x=\dfrac{3}{5}, what is the value of \sin 2x? (0^\circ<x<90^\circ)

A\dfrac{12}{25}
B\dfrac{24}{25}Answer
C\dfrac{7}{25}
D\dfrac{25}{25}

Why

\sin x=\dfrac35 with x acute gives \cos x=\sqrt{1-\dfrac9{25}}=\dfrac45.

\sin2x=2\sin x\cos x=2\cdot\dfrac35\cdot\dfrac45=\dfrac{24}{25}

Option A, \dfrac{12}{25}, forgets the factor 2, and C, \dfrac7{25}, is \cos2x=1-2\sin^2x.

2025 Annual I, Set B, MCQ 181 mark

The expression \sin 75^\circ\cos 15^\circ is equivalent to:

A\tfrac{1}{2}(\sin 90^\circ+\sin 60^\circ)Answer
B\tfrac{1}{2}(\cos 90^\circ+\cos 60^\circ)
C\tfrac{1}{2}(\sin 60^\circ-\sin 90^\circ)
D\tfrac{1}{2}(\cos 60^\circ-\cos 90^\circ)

Why

Use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta) with \alpha=75^\circ, \beta=15^\circ.

\sin75^\circ\cos15^\circ=\dfrac12\big[\sin(75^\circ+15^\circ)+\sin(75^\circ-15^\circ)\big]=\dfrac12(\sin90^\circ+\sin60^\circ)

Options B and D are cosine sums, which come from \cos\alpha\cos\beta or \sin\alpha\sin\beta, and C has the wrong sign. Check: \dfrac12(1+0.866)=0.933=0.966\times0.966.

2026 Annual I, Set B, MCQ 191 mark

\cos(A-B)=

A\cos A\cos B-\sin A\sin B
B\cos A\cos B+\sin A\sin BAnswer
C\cos A\sin B+\sin A\cos B
D\cos A\sin B-\sin A\cos B

Why

This is the fundamental law of trigonometry: \cos(A-B)=\cos A\cos B+\sin A\sin B. The sign between the products is plus for the cosine of a difference.

Option A is \cos(A+B), and C and D are sine formulas (C is \sin(A+B) and D is \sin(B-A) with the wrong sign).

2026 Annual I, Set B, MCQ 201 mark

Using product to sum or difference formula \sin 60^\circ\sin 30^\circ=

A\dfrac{\sqrt3}{4}Answer
B\dfrac{\sqrt3}{2}
C1
D-1

Why

For two sines use -2\sin\alpha\sin\beta=\cos(\alpha+\beta)-\cos(\alpha-\beta), that is \sin\alpha\sin\beta=\dfrac12\big[\cos(\alpha-\beta)-\cos(\alpha+\beta)\big].

\sin60^\circ\sin30^\circ=\dfrac12\big[\cos30^\circ-\cos90^\circ\big]=\dfrac12\left[\dfrac{\sqrt3}{2}-0\right]=\dfrac{\sqrt3}{4}

Option B, \dfrac{\sqrt3}{2}, is \sin60^\circ alone and forgets the factor \sin30^\circ=\dfrac12. Check: 0.866\times0.5=0.433=\dfrac{\sqrt3}{4}.

2025 Annual I, Set B, Q.2 (ii)4 marks

Verify that: \sin\theta\cdot\sin\left(\dfrac{\pi}{3}-\theta\right)\cdot\sin\left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{4}\sin 3\theta

Solution

Combine the last two factors. Use -2\sin\alpha\sin\beta=\cos(\alpha+\beta)-\cos(\alpha-\beta) with \alpha=\dfrac\pi3+\theta, \beta=\dfrac\pi3-\theta, so \alpha+\beta=\dfrac{2\pi}{3} and \alpha-\beta=2\theta.
\sin\left(\dfrac\pi3-\theta\right)\sin\left(\dfrac\pi3+\theta\right)=-\dfrac12\left[\cos\dfrac{2\pi}{3}-\cos2\theta\right]=\dfrac12\left[\cos2\theta+\dfrac12\right]
Use \cos\dfrac{2\pi}3=-\dfrac12, then \cos2\theta=1-2\sin^2\theta.
=\dfrac12\left[1-2\sin^2\theta+\dfrac12\right]=\dfrac34-\sin^2\theta
Multiply by \sin\theta and use the triple angle identity \sin3\theta=3\sin\theta-4\sin^3\theta.
\sin\theta\left(\dfrac34-\sin^2\theta\right)=\dfrac{3\sin\theta-4\sin^3\theta}{4}=\dfrac14\sin3\theta=\text{R.H.S.}\ \checkmark

2025 Annual I, Set B, Q.2 (vi) OR4 marks

Verify that: \dfrac{1+\sin 2\theta+\cos 2\theta}{1+\sin 2\theta-\cos 2\theta}=\cot\theta

Solution

Rewrite the numerator and the denominator with the double angle identities \sin2\theta=2\sin\theta\cos\theta, 1+\cos2\theta=2\cos^2\theta and 1-\cos2\theta=2\sin^2\theta.
\text{Numerator}=(1+\cos2\theta)+\sin2\theta=2\cos^2\theta+2\sin\theta\cos\theta
Take out the common factor 2\cos\theta.
=2\cos\theta(\cos\theta+\sin\theta)
In the same way the denominator is
(1-\cos2\theta)+\sin2\theta=2\sin^2\theta+2\sin\theta\cos\theta=2\sin\theta(\sin\theta+\cos\theta)
Divide. The common factor 2(\sin\theta+\cos\theta) cancels.
\text{L.H.S.}=\dfrac{2\cos\theta(\cos\theta+\sin\theta)}{2\sin\theta(\sin\theta+\cos\theta)}=\dfrac{\cos\theta}{\sin\theta}=\cot\theta=\text{R.H.S.}\ \checkmark

2026 Annual I, Set B, Q.2 (vii)4 marks

Verify the identity: \cos 3\alpha=4\cos^3\alpha-3\cos\alpha.

Solution

Write 3\alpha=2\alpha+\alpha and use the cosine of a sum, \cos(A+B)=\cos A\cos B-\sin A\sin B.
\cos3\alpha=\cos(2\alpha+\alpha)=\cos2\alpha\cos\alpha-\sin2\alpha\sin\alpha
Replace \cos2\alpha by 2\cos^2\alpha-1 (the cosine-only form) and \sin2\alpha by 2\sin\alpha\cos\alpha.
=(2\cos^2\alpha-1)\cos\alpha-(2\sin\alpha\cos\alpha)\sin\alpha
Multiply out.
=2\cos^3\alpha-\cos\alpha-2\sin^2\alpha\cos\alpha
Use \sin^2\alpha=1-\cos^2\alpha so that only cosines remain.
=2\cos^3\alpha-\cos\alpha-2(1-\cos^2\alpha)\cos\alpha=2\cos^3\alpha-\cos\alpha-2\cos\alpha+2\cos^3\alpha
Collect like terms.
=4\cos^3\alpha-3\cos\alpha=\text{R.H.S.}\ \checkmark

Note. To verify, start from one side only (here the left side \cos3\alpha) and reach the other side.

2025 Annual I, Set B, Q.2 (vii) OR (a)

If \tan\theta=\dfrac{3}{4} with \pi<\theta<\dfrac{3\pi}{2}, find the exact value of \sin\dfrac{\theta}{2} without using calculator.

Solution

\pi<\theta<\dfrac{3\pi}2 puts \theta in the third quadrant, where \sin\theta<0 and \cos\theta<0. With \tan\theta=\dfrac34 the hypotenuse is \sqrt{3^2+4^2}=5.
\sin\theta=-\dfrac35,\qquad\cos\theta=-\dfrac45
Halve the inequality to locate \dfrac\theta2. It is in the second quadrant, where sine is positive.
\pi<\theta<\dfrac{3\pi}2\ \Rightarrow\ \dfrac\pi2<\dfrac\theta2<\dfrac{3\pi}4
Use the half angle identity \sin\dfrac\theta2=\sqrt{\dfrac{1-\cos\theta}2} with the plus sign.
\sin\dfrac\theta2=\sqrt{\dfrac{1-\left(-\frac45\right)}{2}}=\sqrt{\dfrac{9/5}{2}}=\sqrt{\dfrac9{10}}=\dfrac3{\sqrt{10}}

\sin\dfrac\theta2=\dfrac{3}{\sqrt{10}}=\dfrac{3\sqrt{10}}{10}

Note. Part (b) of the same question asks for \cos\dfrac\theta2.

2025 Annual I, Set B, Q.2 (vii) OR (b)

If \tan\theta=\dfrac{3}{4} with \pi<\theta<\dfrac{3\pi}{2}, find the exact value of \cos\dfrac{\theta}{2} without using calculator.

Solution

As in part (a): \theta is in the third quadrant, \cos\theta=-\dfrac45, and \dfrac\theta2 is in the second quadrant, where cosine is negative.
\dfrac\pi2<\dfrac\theta2<\dfrac{3\pi}4
Use \cos\dfrac\theta2=-\sqrt{\dfrac{1+\cos\theta}2} (minus sign).
\cos\dfrac\theta2=-\sqrt{\dfrac{1+\left(-\frac45\right)}{2}}=-\sqrt{\dfrac{1/5}{2}}=-\sqrt{\dfrac1{10}}=-\dfrac1{\sqrt{10}}
Check with part (a): \sin^2\dfrac\theta2+\cos^2\dfrac\theta2=\dfrac9{10}+\dfrac1{10}=1.

\cos\dfrac\theta2=-\dfrac1{\sqrt{10}}=-\dfrac{\sqrt{10}}{10}

2026 Annual I, Set B, Q.2 (xi) OR4 marks

Verify: \tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}.

Solution

Write the tangent as sine over cosine and use the sum formulas \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta and \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.
\tan(\alpha+\beta)=\dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}=\dfrac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}
Divide the numerator and the denominator by \cos\alpha\cos\beta (this needs \cos\alpha\neq0 and \cos\beta\neq0).
=\dfrac{\dfrac{\sin\alpha\cos\beta}{\cos\alpha\cos\beta}+\dfrac{\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\dfrac{\cos\alpha\cos\beta}{\cos\alpha\cos\beta}-\dfrac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}}
Cancel in each fraction.
=\dfrac{\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\sin\beta}{\cos\beta}}{1-\dfrac{\sin\alpha}{\cos\alpha}\cdot\dfrac{\sin\beta}{\cos\beta}}
Replace each \dfrac{\sin}{\cos} by \tan.
=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\text{R.H.S.}\ \checkmark

Note. This is Example 1 of Section 8.2 of the book.

2026 Annual I, Set B, Q.2 (xii) OR4 marks

Convert \cos\dfrac{2\theta+2\phi}{2}\sin\dfrac{2\theta-2\phi}{2} into sum or difference form.

Solution

Simplify the angles first.
\cos\dfrac{2\theta+2\phi}{2}\sin\dfrac{2\theta-2\phi}{2}=\cos(\theta+\phi)\sin(\theta-\phi)
The product has the form 2\cos A\sin B, so use 2\cos A\sin B=\sin(A+B)-\sin(A-B) with A=\theta+\phi and B=\theta-\phi. The factor 2 is not in the question, so write \cos A\sin B=\dfrac12\big[\sin(A+B)-\sin(A-B)\big].
A+B=2\theta,\qquad A-B=2\phi
Substitute.
\cos(\theta+\phi)\sin(\theta-\phi)=\dfrac12\big[\sin2\theta-\sin2\phi\big]

\dfrac12\left(\sin2\theta-\sin2\phi\right)

Note. This is part (xi) of Exercise 8.3, question 1, which has the factor 2 in front and so gives \sin2u-\sin2v.

2025 Annual I, Set B, Q.4 OR8 marks

Verify the fundamental law of trigonometry: \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta where \alpha,\beta are real angles in standard position and \alpha>\beta.

Solution

Statement. For real angles \alpha>\beta in standard position, \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.

Derivation Proof of the fundamental law of trigonometry

Aim. Prove that \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta for \alpha>\beta.

Assumptions.

The circle has radius 1 and centre at the origin O(0,0).
\alpha>\beta, both angles in standard position.

The unit circle. OC and OD are the terminal arms of \beta and \alpha, so \angle COD=\alpha-\beta. The point B is placed so that \angle AOB=\alpha-\beta as well. The two dashed chords CD and AB then have equal length.

Take a unit circle with centre O. Let the terminal arms \overrightarrow{OC} and \overrightarrow{OD} of the angles \beta and \alpha (in standard position, \alpha>\beta) meet the circle at C and D. On the unit circle the point at angle \theta has coordinates (\cos\theta,\sin\theta), so we get the coordinates below.
C(\cos\beta,\sin\beta),\qquad D(\cos\alpha,\sin\alpha)
The angle between the arms OC and OD is the difference of the two angles.
\angle COD=\alpha-\beta
In the same circle, construct the angle \angle AOB in standard position with the same measure \alpha-\beta. Its initial arm meets the circle at A(1,0) and its terminal arm at B, whose coordinates are therefore those of the angle \alpha-\beta.
A(1,0),\qquad B\big(\cos(\alpha-\beta),\,\sin(\alpha-\beta)\big)
The angles \angle COD and \angle AOB are equal, and the arms OC,OD,OA,OB are all radii of length 1. So the triangles COD and AOB are congruent, and their third sides are equal.
CD=AB
Write both lengths with the distance formula (Section 8.1).
\sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}=\sqrt{[\cos(\alpha-\beta)-1]^2+[\sin(\alpha-\beta)-0]^2}
Square both sides to remove the roots.
(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2=[\cos(\alpha-\beta)-1]^2+[\sin(\alpha-\beta)]^2
Expand the left side.
\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta+\sin^2\alpha+\sin^2\beta-2\sin\alpha\sin\beta
Expand the right side.
\cos^2(\alpha-\beta)+1-2\cos(\alpha-\beta)+\sin^2(\alpha-\beta)
Rearrange both sides so that the squares are together, and use \cos^2\theta+\sin^2\theta=1 on each pair.
(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=\big(\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta)\big)+1-2\cos(\alpha-\beta)
Replace each bracket by 1.
1+1-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=1+1-2\cos(\alpha-\beta)
Cancel the 2 on each side, then divide both sides by -2.
-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=-2\cos(\alpha-\beta)

Result\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\qquad(1)

This is the fundamental law of trigonometry. The cosine of a difference is not the difference of the cosines: the right side mixes cosines and sines.

2026 Annual I, Set B, Q.5 OR4+4 = 8 marks

If \sin\alpha=\dfrac35, \sin\beta=\dfrac{1}{\sqrt5}, where terminal ray of \alpha and \beta are in 2^{nd} quadrant, then find the value of \cos(\alpha-\beta) and \sin(\alpha+\beta).

Solution

Both angles are in the second quadrant, where sine is positive and cosine is negative. Find the cosines from \cos^2\theta=1-\sin^2\theta.
\cos\alpha=-\sqrt{1-\dfrac9{25}}=-\dfrac45,\qquad\cos\beta=-\sqrt{1-\dfrac15}=-\dfrac{2}{\sqrt5}
**\cos(\alpha-\beta).** Use \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos(\alpha-\beta)=\left(-\dfrac45\right)\left(-\dfrac2{\sqrt5}\right)+\dfrac35\cdot\dfrac1{\sqrt5}=\dfrac{8}{5\sqrt5}+\dfrac{3}{5\sqrt5}=\dfrac{11}{5\sqrt5}
Rationalise: multiply top and bottom by \sqrt5.
\cos(\alpha-\beta)=\dfrac{11\sqrt5}{25}
**\sin(\alpha+\beta).** Use \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\sin(\alpha+\beta)=\dfrac35\left(-\dfrac2{\sqrt5}\right)+\left(-\dfrac45\right)\dfrac1{\sqrt5}=-\dfrac{6}{5\sqrt5}-\dfrac{4}{5\sqrt5}=-\dfrac{10}{5\sqrt5}=-\dfrac2{\sqrt5}
Rationalise.
\sin(\alpha+\beta)=-\dfrac{2\sqrt5}{5}

\cos(\alpha-\beta)=\dfrac{11\sqrt5}{25},\qquad\sin(\alpha+\beta)=-\dfrac{2\sqrt5}{5}

Note. Compare Exercise 8.1, question 5, which uses the same method.

2026 Annual I, Set D, MCQ 141 mark

\cos(\pi+\beta) is equal to:

A\cos\beta
B-\cos\betaAnswer
C\sin\beta
D-\sin\beta

Why

\pi+\beta is an allied angle, and \pi is an even multiple of a right angle, so cosine keeps its name. The angle \pi+\beta is in the third quadrant, where cosine is negative.

\cos(\pi+\beta)=\cos\pi\cos\beta-\sin\pi\sin\beta=(-1)\cos\beta-0=-\cos\beta

Options C and D change cosine into sine, which happens only for \dfrac{\pi}{2} and \dfrac{3\pi}{2}.

2025 Annual I, Set D, MCQ 141 mark

\sin 75^\circ\cos 15^\circ=

A\tfrac{1}{2}(\sin 90^\circ+\sin 60^\circ)Answer
B\tfrac{1}{2}(\cos 90^\circ+\cos 60^\circ)
C\tfrac{1}{2}(\sin 60^\circ-\sin 90^\circ)
D\tfrac{1}{2}(\cos 60^\circ-\cos 90^\circ)

Why

Use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta) with \alpha=75^\circ and \beta=15^\circ.

\sin75^\circ\cos15^\circ=\dfrac12\big[\sin90^\circ+\sin60^\circ\big]

The other options are cosine sums or differences, or have the wrong sign.

2026 Annual I, Set D, MCQ 151 mark

\cos A\cos B is equal to:

A\dfrac12[\cos(A+B)+\cos(A-B)]Answer
B\cos(A+B)
C\cos(A-B)
D\dfrac12[\sin(A+B)+\sin(A-B)]

Why

Add the formulas \cos(A+B)=\cos A\cos B-\sin A\sin B and \cos(A-B)=\cos A\cos B+\sin A\sin B. The sine terms cancel.

\cos(A+B)+\cos(A-B)=2\cos A\cos B\ \Rightarrow\ \cos A\cos B=\dfrac12\big[\cos(A+B)+\cos(A-B)\big]

Option D is the formula for \sin A\cos B, not \cos A\cos B.

2025 Annual I, Set D, MCQ 151 mark

What is value of \cos\left(\dfrac{\theta}{2}\right), if \cos\theta=\dfrac12; (0<\theta<90^\circ)

A\dfrac{\sqrt3}{2}Answer
B\dfrac{1}{2}
C\dfrac{1}{\sqrt2}
D\dfrac{1}{2\sqrt2}

Why

\theta is acute, so \dfrac\theta2 is acute and \cos\dfrac\theta2>0. Use the half angle identity.

\cos\dfrac\theta2=\sqrt{\dfrac{1+\cos\theta}{2}}=\sqrt{\dfrac{1+\frac12}{2}}=\sqrt{\dfrac34}=\dfrac{\sqrt3}2

Option B, \dfrac12, is \cos\theta itself, and C is \cos45^\circ. Check: \cos\theta=\dfrac12 gives \theta=60^\circ, and \cos30^\circ=\dfrac{\sqrt3}2.

2026 Annual I, Set D, Q.2 (ii)4 marks

Verify that: \sin 4\theta=8\sin\theta\cos^3\theta-4\sin\theta\cos\theta.

Solution

Start with the left side and write 4\theta=2(2\theta). Use \sin2x=2\sin x\cos x with x=2\theta.
\sin4\theta=2\sin2\theta\cos2\theta
Replace \sin2\theta by 2\sin\theta\cos\theta and \cos2\theta by 2\cos^2\theta-1.
=2(2\sin\theta\cos\theta)(2\cos^2\theta-1)=4\sin\theta\cos\theta(2\cos^2\theta-1)
Multiply out.
=8\sin\theta\cos^3\theta-4\sin\theta\cos\theta=\text{R.H.S.}\ \checkmark

Note. Exercise 8.2, question 8(vi), proves the related identity \sin4\theta=4\sin\theta\cos^3\theta-4\sin^3\theta\cos\theta.

2025 Annual I, Set D, Q.2 (ii)4 marks

Verify that: \cos\theta\cdot\cos\left(\dfrac{\pi}{3}-\theta\right)\cdot\cos\left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{4}\cos 3\theta.

Solution

Combine the last two factors. Use 2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta) with \alpha=\dfrac\pi3+\theta and \beta=\dfrac\pi3-\theta (so \alpha+\beta=\dfrac{2\pi}3, \alpha-\beta=2\theta).
\cos\left(\dfrac\pi3-\theta\right)\cos\left(\dfrac\pi3+\theta\right)=\dfrac12\left[\cos\dfrac{2\pi}3+\cos2\theta\right]=\dfrac12\left[\cos2\theta-\dfrac12\right]
Use \cos2\theta=2\cos^2\theta-1.
=\dfrac12\left[2\cos^2\theta-\dfrac32\right]=\cos^2\theta-\dfrac34
Multiply by \cos\theta and use the triple angle identity \cos3\theta=4\cos^3\theta-3\cos\theta.
\cos\theta\left(\cos^2\theta-\dfrac34\right)=\dfrac{4\cos^3\theta-3\cos\theta}{4}=\dfrac14\cos3\theta=\text{R.H.S.}\ \checkmark

2026 Annual I, Set D, Q.2 (vi) OR4 marks

Using fundamental law of trigonometry find the value of \cos 15^\circ (do not use calculator).

Solution

Write 15^\circ=60^\circ-45^\circ, whose ratios are known, and use the fundamental law \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.
\cos15^\circ=\cos(60^\circ-45^\circ)=\cos60^\circ\cos45^\circ+\sin60^\circ\sin45^\circ
Substitute the exact values \cos60^\circ=\dfrac12, \sin60^\circ=\dfrac{\sqrt3}{2}, \cos45^\circ=\sin45^\circ=\dfrac1{\sqrt2}.
=\dfrac12\cdot\dfrac1{\sqrt2}+\dfrac{\sqrt3}{2}\cdot\dfrac1{\sqrt2}=\dfrac{1+\sqrt3}{2\sqrt2}
Rationalise: multiply top and bottom by \sqrt2.
\cos15^\circ=\dfrac{\sqrt2+\sqrt6}{4}

\cos15^\circ=\dfrac{\sqrt6+\sqrt2}{4}

Note. This is Example 2(i) of Section 8.3. Decimal check: \dfrac{2.449+1.414}{4}=0.966.

2025 Annual I, Set D, Q.2 (vi) OR4 marks

Verify that: \dfrac{\cos\theta+\cos 4\theta+\cos 7\theta}{\sin\theta+\sin 4\theta+\sin 7\theta}=\cot 4\theta.

Solution

Group the first and last terms in the numerator and in the denominator, because 7\theta and \theta have half sum 4\theta and half difference 3\theta. Use \cos p+\cos q=2\cos\dfrac{p+q}2\cos\dfrac{p-q}2 and \sin p+\sin q=2\sin\dfrac{p+q}2\cos\dfrac{p-q}2.
\cos7\theta+\cos\theta=2\cos4\theta\cos3\theta,\qquad\sin7\theta+\sin\theta=2\sin4\theta\cos3\theta
Substitute and take out the common factors.
\dfrac{\cos\theta+\cos4\theta+\cos7\theta}{\sin\theta+\sin4\theta+\sin7\theta}=\dfrac{2\cos4\theta\cos3\theta+\cos4\theta}{2\sin4\theta\cos3\theta+\sin4\theta}=\dfrac{\cos4\theta\,(2\cos3\theta+1)}{\sin4\theta\,(2\cos3\theta+1)}
Cancel 2\cos3\theta+1.
=\dfrac{\cos4\theta}{\sin4\theta}=\cot4\theta=\text{R.H.S.}\ \checkmark

2026 Annual I, Set D, Q.2 (vii) OR4 marks

Express 2\sin 12\theta\cos 6\theta as a sum or difference.

Solution

The product has the form 2\sin\alpha\cos\beta with \alpha=12\theta and \beta=6\theta, so use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta).
2\sin12\theta\cos6\theta=\sin(12\theta+6\theta)+\sin(12\theta-6\theta)
Add and subtract the angles.
=\sin18\theta+\sin6\theta

\sin18\theta+\sin6\theta

2025 Annual I, Set D, Q.2 (vii) OR (a)

If \sin\theta=-\dfrac{3}{5} and \theta lies in 3rd quadrant, find the exact value of \cos(180^\circ+\theta).

Solution

\theta is in the third quadrant, so \cos\theta<0.
\cos\theta=-\sqrt{1-\sin^2\theta}=-\sqrt{1-\dfrac9{25}}=-\dfrac45
Use the sum formula for cosine; \cos180^\circ=-1 and \sin180^\circ=0, so cosine keeps its name and changes sign.
\cos(180^\circ+\theta)=\cos180^\circ\cos\theta-\sin180^\circ\sin\theta=-\cos\theta
Substitute.
\cos(180^\circ+\theta)=-\left(-\dfrac45\right)=\dfrac45

\dfrac45

2025 Annual I, Set D, Q.2 (vii) OR (b)

If \sin\theta=-\dfrac{3}{5} and \theta lies in 3rd quadrant, find the exact value of \tan(270^\circ-\theta).

Solution

As in part (a), \sin\theta=-\dfrac35 and \cos\theta=-\dfrac45 in the third quadrant.
Expand \sin(270^\circ-\theta) and \cos(270^\circ-\theta) with \sin270^\circ=-1, \cos270^\circ=0.
\begin{aligned}\sin(270^\circ-\theta)&=\sin270^\circ\cos\theta-\cos270^\circ\sin\theta=-\cos\theta\\ \cos(270^\circ-\theta)&=\cos270^\circ\cos\theta+\sin270^\circ\sin\theta=-\sin\theta\end{aligned}
Divide: sine over cosine.
\tan(270^\circ-\theta)=\dfrac{-\cos\theta}{-\sin\theta}=\cot\theta
Substitute the values.
\cot\theta=\dfrac{-\frac45}{-\frac35}=\dfrac43

\dfrac43

Note. The table of allied angles in Section 8.3 gives \tan\left(\dfrac{3\pi}{2}-\theta\right)=\cot\theta directly.

2026 Annual I, Set D, Q.4 OR8 marks

Prove fundamental law of trigonometry: \cos(A-B)=\cos A\cos B+\sin A\sin B. Diagram is necessary.

Solution

Statement. For angles \alpha>\beta in standard position, \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.

Derivation Proof of the fundamental law of trigonometry

Aim. Prove that \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta for \alpha>\beta.

Assumptions.

The circle has radius 1 and centre at the origin O(0,0).
\alpha>\beta, both angles in standard position.

Take a unit circle with centre O. Let the terminal arms \overrightarrow{OC} and \overrightarrow{OD} of the angles \beta and \alpha (in standard position, \alpha>\beta) meet the circle at C and D. On the unit circle the point at angle \theta has coordinates (\cos\theta,\sin\theta), so we get the coordinates below.
C(\cos\beta,\sin\beta),\qquad D(\cos\alpha,\sin\alpha)
The angle between the arms OC and OD is the difference of the two angles.
\angle COD=\alpha-\beta
In the same circle, construct the angle \angle AOB in standard position with the same measure \alpha-\beta. Its initial arm meets the circle at A(1,0) and its terminal arm at B, whose coordinates are therefore those of the angle \alpha-\beta.
A(1,0),\qquad B\big(\cos(\alpha-\beta),\,\sin(\alpha-\beta)\big)
The angles \angle COD and \angle AOB are equal, and the arms OC,OD,OA,OB are all radii of length 1. So the triangles COD and AOB are congruent, and their third sides are equal.
CD=AB
Write both lengths with the distance formula (Section 8.1).
\sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}=\sqrt{[\cos(\alpha-\beta)-1]^2+[\sin(\alpha-\beta)-0]^2}
Square both sides to remove the roots.
(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2=[\cos(\alpha-\beta)-1]^2+[\sin(\alpha-\beta)]^2
Expand the left side.
\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta+\sin^2\alpha+\sin^2\beta-2\sin\alpha\sin\beta
Expand the right side.
\cos^2(\alpha-\beta)+1-2\cos(\alpha-\beta)+\sin^2(\alpha-\beta)
Rearrange both sides so that the squares are together, and use \cos^2\theta+\sin^2\theta=1 on each pair.
(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=\big(\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta)\big)+1-2\cos(\alpha-\beta)
Replace each bracket by 1.
1+1-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=1+1-2\cos(\alpha-\beta)
Cancel the 2 on each side, then divide both sides by -2.
-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=-2\cos(\alpha-\beta)

Result\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\qquad(1)

This is the fundamental law of trigonometry. The cosine of a difference is not the difference of the cosines: the right side mixes cosines and sines.

2025 Annual I, Set D, Q.4 OR8 marks

Verify the fundamental law of trigonometry: \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta where \alpha,\beta are real angles in standard position and \alpha>\beta.

Solution

Statement. For real angles \alpha>\beta in standard position, \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.

Derivation Proof of the fundamental law of trigonometry

Aim. Prove that \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta for \alpha>\beta.

Assumptions.

The circle has radius 1 and centre at the origin O(0,0).
\alpha>\beta, both angles in standard position.

Take a unit circle with centre O. Let the terminal arms \overrightarrow{OC} and \overrightarrow{OD} of the angles \beta and \alpha (in standard position, \alpha>\beta) meet the circle at C and D. On the unit circle the point at angle \theta has coordinates (\cos\theta,\sin\theta), so we get the coordinates below.
C(\cos\beta,\sin\beta),\qquad D(\cos\alpha,\sin\alpha)
The angle between the arms OC and OD is the difference of the two angles.
\angle COD=\alpha-\beta
In the same circle, construct the angle \angle AOB in standard position with the same measure \alpha-\beta. Its initial arm meets the circle at A(1,0) and its terminal arm at B, whose coordinates are therefore those of the angle \alpha-\beta.
A(1,0),\qquad B\big(\cos(\alpha-\beta),\,\sin(\alpha-\beta)\big)
The angles \angle COD and \angle AOB are equal, and the arms OC,OD,OA,OB are all radii of length 1. So the triangles COD and AOB are congruent, and their third sides are equal.
CD=AB
Write both lengths with the distance formula (Section 8.1).
\sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}=\sqrt{[\cos(\alpha-\beta)-1]^2+[\sin(\alpha-\beta)-0]^2}
Square both sides to remove the roots.
(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2=[\cos(\alpha-\beta)-1]^2+[\sin(\alpha-\beta)]^2
Expand the left side.
\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta+\sin^2\alpha+\sin^2\beta-2\sin\alpha\sin\beta
Expand the right side.
\cos^2(\alpha-\beta)+1-2\cos(\alpha-\beta)+\sin^2(\alpha-\beta)
Rearrange both sides so that the squares are together, and use \cos^2\theta+\sin^2\theta=1 on each pair.
(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=\big(\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta)\big)+1-2\cos(\alpha-\beta)
Replace each bracket by 1.
1+1-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=1+1-2\cos(\alpha-\beta)
Cancel the 2 on each side, then divide both sides by -2.
-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=-2\cos(\alpha-\beta)

Result\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\qquad(1)

This is the fundamental law of trigonometry. The cosine of a difference is not the difference of the cosines: the right side mixes cosines and sines.

Exam focus

Prove the fundamental law of trigonometry, \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta, with a diagram. (long)
8-mark Q.4 OR in 2026 Set D and in both 2025 sets (Set B and Set D). Draw the unit circle with A, B, C, D, equate the chords CD and AB with the distance formula and simplify.
Go to answer
Verify a triple or multiple angle identity, such as \cos3\alpha=4\cos^3\alpha-3\cos\alpha or \sin4\theta=8\sin\theta\cos^3\theta-4\sin\theta\cos\theta. (problem)
2026 Set B Q.2 (vii) and 2026 Set D Q.2 (ii), 4 marks each. Write 3\alpha=2\alpha+\alpha (or 4\theta=2\cdot2\theta) and use the sum and double angle identities.
Go to answer
Convert a product of sines and cosines to a sum or difference, for example 2\sin12\theta\cos6\theta, and the reverse. (problem)
2026 Set B Q.2 (xii) OR and 2026 Set D Q.2 (vii) OR (4 marks) and four MCQs: 2026 Set B MCQ 20, Set D MCQ 15, 2025 Set B MCQ 18 and Set D MCQ 14.
Go to answer
Find \cos(\alpha-\beta) and \sin(\alpha+\beta) from \sin\alpha and \sin\beta with the quadrants given. (long)
2026 Set B Q.5 OR (8 marks). Find the cosines with their signs from the quadrant, then use the sum and difference formulas; compare Exercise 8.1, questions 5 to 9.
Go to answer
Verify an identity with angles \dfrac\pi3\pm\theta or with several multiple angles, for example \sin\theta\sin\left(\dfrac\pi3-\theta\right)\sin\left(\dfrac\pi3+\theta\right)=\dfrac14\sin3\theta. (problem)
2025 Set B Q.2 (ii) and (vi) OR, 2025 Set D Q.2 (ii) and (vi) OR (4 marks each). Use product to sum or sum to product, then the double and triple angle identities.
Go to answer
Find exact values of half angles, such as \sin\dfrac\theta2 and \cos\dfrac\theta2 given \tan\theta and the quadrant of \theta. (problem)
2025 Set B Q.2 (vii) OR and 2025 Set D MCQ 15. The sign comes from the quadrant of \dfrac\theta2, not of \theta.
Go to answer
Find an exact value with the fundamental law or allied angles, for example \cos15^\circ or \tan(270^\circ-\theta). (problem)
2026 Set D Q.2 (vi) OR (\cos15^\circ), 2025 Set D Q.2 (vii) OR (allied angles) and 2026 Set D MCQ 14 (\cos(\pi+\beta)).
Go to answer
Verify \tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}. (problem)
2026 Set B Q.2 (xi) OR, 4 marks. It is Example 1 of Section 8.2: divide the sum formulas by \cos\alpha\cos\beta.
Go to answer
Express a\sin\theta+b\cos\theta in the form r\sin(\theta+\phi). (problem)
Not in the 2025 or 2026 paper, but Section 8.4 and Exercise 8.1, question 13, ask it, so it is a likely short question.
Go to answer

8.1Distance formula

8.2Fundamental law of trigonometryOften examined

Derivation Proof of the fundamental law of trigonometry

Cosine of a sum

Derivation Cosine of a sum, identity (2)

Important results from the fundamental law

Derivation Result (i): cos(−β)=cosβ

Derivation Result (ii): cos(2π​−β)=sinβ

Derivation Result (iii): cos(2π​+α)=−sinα

Derivation Result (iv): sin(2π​+α)=cosα

Derivation Result (v): the sine of a sum and of a difference

Check points

(1)

(2)

(i)

(ii)

(iii)

(i) Recognise the pattern

(ii) Test the student's claim

8.3Trigonometric ratios of allied anglesOften examined

(i)

(ii)

(iii)

(i)

(ii)

8.4Expressing a sin θ + b cos θ in the form r sin(θ + φ)

Derivation Expressing a\sin\theta+b\cos\theta as r\sin(\theta+\phi)

8.5Double, half and triple angle identitiesOften examined

(i) Double angle identities

Derivation Identity (4): \sin2\alpha

Derivation Identities (5), (6), (7): \cos2\alpha

Derivation Identities (8), (9), (10): squares in terms of \cos2\alpha

Derivation Identity (11): \tan2\alpha

(ii) Half angle identities

Derivation Identities (12) to (17): half angle identities

(iii) Triple angle identities

Derivation Identity (18): \sin3\alpha

Derivation Identity (19): \cos3\alpha

Derivation Identity (20): \tan3\alpha

Worked examples

(i)

(ii)

(iii)

(a)

(b)

8.6.1Expressing the product of sines and cosines as sums or differencesOften examined

Derivation Identities (25) to (28)

8.6.2Expressing the sums or differences of sines and cosines as productOften examined

Derivation Sum to product formulae

Method 1: product to sum

Method 2: allied angles and the double angle formula

Formula sheet

Key terms

Exercise 8.1p. 202–203

Exercise 8.2p. 208–209

Exercise 8.3p. 212–213

Review Exercise 8p. 213–215

FBISE 2025 and 2026 questions

Derivation Proof of the fundamental law of trigonometry

Derivation Proof of the fundamental law of trigonometry

Derivation Proof of the fundamental law of trigonometry

Exam focus

Derivation Result (i): $cos (- β) = cos β$

Derivation Result (ii): $cos (\frac{π}{2} - β) = sin β$

Derivation Result (iii): $cos (\frac{π}{2} + α) = - sin α$

Derivation Result (iv): $sin (\frac{π}{2} + α) = cos α$