Inverse trigonometric functions and their graphs

Class 12 Mathematics, unit 8, book pages 230–250

Partial. These notes cover the textbook up to printed page 250, the end of 8.2.6 (graph of the inverse cosecant function). Printed pages 251 to 261 are still to be added: the rest of section 8.2 with Exercise 8.2, the book's treatment of the addition and subtraction formulae and of real-life applications (angles of elevation and depression, ramps, buildings, bridges), Exercise 8.3 and the review exercise. The sections 'Addition and subtraction formulae' and 'Applications' are written from the unit's learning outcomes and the 2026 paper questions, not from the book text, so they carry no book section numbers.

An inverse trigonometric function takes the value of a ratio and gives back the angle, for example . Because trigonometric functions repeat, each one is first restricted to a principal interval on which it is one-to-one; this fixes the domain and range of the inverse and makes every answer a single principal value. The unit covers the six inverse functions, their graphs as reflections in , the properties that connect them, and their use in finding angles of elevation, depression and slope.

Learning outcomes (4)
  • Find the domain and range of the principal trigonometric functions and the inverse trigonometric functions.
  • Draw the graphs of the inverse trigonometric functions of cosine, sine, tangent, secant, cosecant and cotangent (the book says 'within the domain from to ': the trigonometric curves are drawn on and their principal parts are reflected in ).
  • State, prove and apply the addition and subtraction formulae of inverse trigonometric functions.
  • Apply inverse trigonometric functions to real-life word problems, such as mechanical engineering, architecture (height of a building), angles of elevation and depression, and the angles of bridges in scale models.

8.1The inverse trigonometric functions

You already know how to go from an angle to a ratio, for example . An inverse trigonometric function goes the other way. It starts from the value of a ratio and gives back the angle. If the cosine of an angle is , the inverse cosine function gives the angle , that is radians.

There is one inverse function for each of the six trigonometric functions. Each has two common names:

Inverse functionWritten asThe output is
inverse cosine or the angle whose cosine is
inverse sine or the angle whose sine is
inverse tangent or the angle whose tangent is
inverse secant or the angle whose secant is
inverse cosecant or the angle whose cosecant is
inverse cotangent or the angle whose cotangent is

Why the domain must be restricted

A function has an inverse only if it is one-to-one: each output comes from exactly one input. The trigonometric functions are periodic, so they repeat their values again and again. For example, has infinitely many solutions:

Solutions of
−1.5
−1
−0.5
0.5
1
1.5
O
The line meets infinitely often, but it meets the principal part () only once, at .

The horizontal line cuts the cosine curve many times, so cosine fails the horizontal line test and has no inverse on the whole real line. To fix this we keep only one piece of the curve on which every value from to occurs exactly once. The function on this piece is called a principal function, and its inverse is the inverse trigonometric function. For cosine the piece used is , so the only answer to there is .

Restricted (principal) domain
The interval to which a trigonometric function is limited so that it becomes one-to-one, for example for cosine.
Principal function
The trigonometric function defined only on its restricted domain, written with a capital letter: , , , , , .
Principal value
The value of an inverse trigonometric function: the one angle in the fixed range that has the given ratio. For example, the principal value of is .

8.1.1 to 8.1.3Principal cosine function and inverse cosine functionOften examined

Principal cosine function (8.1.1)

We need an interval on which has only one solution for every in . On the cosine curve falls steadily from (at ) to (at ), so it takes every value in exactly once. Other intervals, such as , would also work, but by convention we use .

The cosine function defined on is called the principal cosine function. For each the equation has a unique solution. We write it as , read as ' is the real number in whose cosine is '.

Inverse cosine function (8.1.2)

Here is a real number (a value of cosine) and is an angle in radians. Three values follow straight from the definition:

ValueReason
and
Example 1
Find the principal value of .
  1. Let the required angle be . By the definition, its cosine is and it must lie in the range .
  2. Cosine is negative, so lies in Quadrant II, between and .
  3. The reference angle is because . In Quadrant II the angle is minus the reference angle.
  4. Check: and lies in .

Domain and range of the inverse cosine function (8.1.3)

An inverse function swaps inputs and outputs, so the domain and the range of the principal function swap:

FunctionDomainRange

The domain of an inverse trigonometric function is a set of real numbers (values of a ratio). Its range is a set of angles, measured in radians or degrees.

Graph by reflection

The graph of an inverse function is the reflection of the original graph in the line , because every point becomes . So we draw on and reflect it in .

and its reflection
O
Each point of the principal cosine curve becomes on the inverse cosine curve: and .

8.1.4 to 8.1.6Principal sine function and inverse sine functionOften examined

Principal sine function (8.1.4)

On the sine curve rises steadily from to , so it takes every value in exactly once. On this interval, has a unique solution for each .

The sine function defined on is called the principal sine function. The unique solution of is written : ' is the real number in whose sine is '.

Example 2
Find the principal value of .
  1. Let the required angle be . Its sine is and it must lie in the range of .
  2. Sine is negative, so lies between and (Quadrant IV, measured clockwise).
  3. The reference angle is because . In Quadrant IV the principal angle is the negative of the reference angle.
  4. Check: .
Check point (page 233)
Find the principal value of (i) (ii) .
  1. (i) with . The value is positive, so is in Quadrant I, and .
  2. (ii) with . The value is positive, so is in Quadrant I, and .

Inverse sine function (8.1.5)

ValueReason

Domain and range of the inverse sine function (8.1.6)

FunctionDomainRange

As for cosine, the graph of is the reflection of the principal sine curve in the line .

and its reflection
O
The principal sine curve runs from to ; its mirror image in runs from to .

8.1.7 to 8.1.9Principal tangent function and inverse tangent functionOften examined

Principal tangent function (8.1.7)

The tangent function is defined for every real x except the odd multiples of \frac{\pi}{2}, where \cos x=0. So its domain is \mathbb{R}-\left\{(2n+1)\frac{\pi}{2}:n\in\mathbb{Z}\right\} and its range is all of \mathbb{R}. Because \tan x repeats every \pi, the equation \tan x=k has infinitely many solutions for every real k.

y=\tan x from -2\pi to 2\pi
\tan x=1 has a solution in every branch, but only one in the principal branch \left(-\frac{\pi}{2},\frac{\pi}{2}\right), namely \frac{\pi}{4}.

Between -\frac{\pi}{2} and \frac{\pi}{2} the tangent curve rises from -\infty to +\infty and takes every real value exactly once. The end points are left out (an open interval) because \tan x is not defined there.

\operatorname{Tan} x=\tan x,\qquad x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right),\qquad \operatorname{Tan} x\in(-\infty,\infty)

This is the principal tangent function. The unique solution of \operatorname{Tan} x=k is written x=\operatorname{Tan}^{-1} k: 'x is the real number in \left(-\frac{\pi}{2},\frac{\pi}{2}\right) whose tangent is k'.

Example 3
Find the principal value of \operatorname{Tan}^{-1}(\sqrt3).
  1. Let the required angle be y.
    y=\operatorname{Tan}^{-1}(\sqrt3)\iff\tan y=\sqrt3,\quad y\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)
  2. Tangent is positive, so y lies in Quadrant I. Since \tan\frac{\pi}{3}=\sqrt3:
    y=\dfrac{\pi}{3}
\operatorname{Tan}^{-1}(\sqrt3)=\dfrac{\pi}{3}\ \ (=60^\circ)

Inverse tangent function (8.1.8)

y=\operatorname{Tan}^{-1} x\iff x=\operatorname{Tan} y,\qquad x\in(-\infty,\infty),\quad y\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)
ValueReason
\operatorname{Tan}^{-1}(1)=\dfrac{\pi}{4}\tan\dfrac{\pi}{4}=1
\operatorname{Tan}^{-1}(0)=0\tan 0=0
\operatorname{Tan}^{-1} x\to\dfrac{\pi}{2} as x\to+\infty\tan y\to+\infty as y\to\dfrac{\pi}{2} from below
\operatorname{Tan}^{-1} x\to-\dfrac{\pi}{2} as x\to-\infty\tan y\to-\infty as y\to-\dfrac{\pi}{2} from above
Check point (page 235)
Find the principal value of \operatorname{Tan}^{-1}\left(-\dfrac{1}{\sqrt3}\right).
  1. \tan y=-\frac{1}{\sqrt3} with y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). The value is negative, so y is between -\frac{\pi}{2} and 0.
  2. The reference angle is \frac{\pi}{6} because \tan\frac{\pi}{6}=\frac{1}{\sqrt3}, so the principal angle is its negative.
    y=-\dfrac{\pi}{6}
\operatorname{Tan}^{-1}\left(-\dfrac{1}{\sqrt3}\right)=-\dfrac{\pi}{6}

Domain and range of the inverse tangent function (8.1.9)

FunctionDomainRange
y=\operatorname{Tan} x\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)(-\infty,\infty)
y=\operatorname{Tan}^{-1} x(-\infty,\infty)\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)
y=\operatorname{Tan} x and its reflection y=\operatorname{Tan}^{-1} x
The vertical asymptotes x=\pm\frac{\pi}{2} of \operatorname{Tan} x become the horizontal asymptotes y=\pm\frac{\pi}{2} of \operatorname{Tan}^{-1} x.

8.1.10 to 8.1.12Principal cotangent function and inverse cotangent functionOften examined

Principal cotangent function (8.1.10)

Like tangent, \cot x repeats every \pi, so \cot x=k has infinitely many solutions for each real k. On the open interval (0,\pi) the cotangent falls steadily from +\infty to -\infty and takes each real value exactly once. The end points 0 and \pi are left out because \cot x=\frac{\cos x}{\sin x} is not defined where \sin x=0.

Principal cotangent function, 0<x<\pi
On (0,\pi) the cotangent falls from +\infty to -\infty, taking every real value once. The lines x=0 and x=\pi are asymptotes.
\operatorname{Cot} x=\cot x,\qquad x\in(0,\pi),\qquad \operatorname{Cot} x\in(-\infty,\infty)

This is the principal cotangent function. The unique solution of \operatorname{Cot} x=k is written x=\operatorname{Cot}^{-1} k: 'x is the real number in (0,\pi) whose cotangent is k'.

Example 4 (corrected)
Find the principal value of \operatorname{Cot}^{-1}(-\sqrt3).
  1. Let the required angle be y. Its range is (0,\pi), not \left(-\frac{\pi}{2},\frac{\pi}{2}\right).
    y=\operatorname{Cot}^{-1}(-\sqrt3)\iff\cot y=-\sqrt3,\quad y\in(0,\pi)
  2. Cotangent is negative, so y lies in Quadrant II, between \frac{\pi}{2} and \pi.
  3. The reference angle is \frac{\pi}{6} because \cot\frac{\pi}{6}=\sqrt3. In Quadrant II the angle is \pi minus the reference angle.
    y=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}
  4. Check: \cot\frac{5\pi}{6}=\dfrac{\cos\frac{5\pi}{6}}{\sin\frac{5\pi}{6}}=\dfrac{-\frac{\sqrt3}{2}}{\frac12}=-\sqrt3, and \frac{5\pi}{6}\in(0,\pi).
\operatorname{Cot}^{-1}(-\sqrt3)=\dfrac{5\pi}{6}\ \ (=150^\circ)
Check point (page 236)
Find the principal value of \operatorname{Cot}^{-1}(1).
  1. \cot y=1 with y\in(0,\pi). The value is positive, so y is in Quadrant I, and \cot\frac{\pi}{4}=1.
    \operatorname{Cot}^{-1}(1)=\dfrac{\pi}{4}
\operatorname{Cot}^{-1}(1)=\dfrac{\pi}{4}

Inverse cotangent function (8.1.11)

y=\operatorname{Cot}^{-1} x\iff x=\operatorname{Cot} y,\qquad x\in(-\infty,\infty),\quad y\in(0,\pi)
ValueReason
\operatorname{Cot}^{-1}(0)=\dfrac{\pi}{2}\cot\dfrac{\pi}{2}=0
\operatorname{Cot}^{-1} x\to0 as x\to+\infty\cot y\to+\infty as y\to0 from above
\operatorname{Cot}^{-1} x\to\pi as x\to-\infty\cot y\to-\infty as y\to\pi from below

Domain and range of the inverse cotangent function (8.1.12)

FunctionDomainRange
y=\operatorname{Cot} x(0,\pi)(-\infty,\infty)
y=\operatorname{Cot}^{-1} x(-\infty,\infty)(0,\pi)

(The book's heading line here says 'inverse tangent function y=\operatorname{Cot}^{-1} x'; it means the inverse cotangent function.)

y=\operatorname{Cot} x and its reflection y=\operatorname{Cot}^{-1} x
The asymptotes x=0 and x=\pi of \operatorname{Cot} x become the asymptotes y=0 and y=\pi of \operatorname{Cot}^{-1} x.

8.1.13 to 8.1.15Principal secant function and inverse secant functionOften examined

Principal secant function (8.1.13)

\sec x=\dfrac{1}{\cos x} is defined wherever \cos x\ne0, that is for x\ne(2n+1)\frac{\pi}{2}, n\in\mathbb{Z}. Since |\cos x|\le1, its values satisfy |\sec x|\ge1, so the range of secant is \mathbb{R}-(-1,1). The equation \sec x=k has no solution when -1<k<1 and infinitely many solutions for every other k; for example \sec x=1 at x=0,\pm2\pi,\pm4\pi,\dots

y=\sec x from -2\pi to 2\pi
\sec x never takes values strictly between -1 and 1. The principal part is 0\le x\le\pi, x\ne\frac{\pi}{2}: it runs from (0,1) up to the asymptote, then from below up to (\pi,-1).

Cosine is one-to-one on [0,\pi], so its reciprocal is one-to-one there too. We must only remove x=\frac{\pi}{2}, where \cos x=0. Other choices such as (\pi,2\pi) are possible, but by convention we take [0,\pi] without \frac{\pi}{2}.

\operatorname{Sec} x=\sec x,\qquad x\in\left[0,\tfrac{\pi}{2}\right)\cup\left(\tfrac{\pi}{2},\pi\right],\qquad \operatorname{Sec} x\in(-\infty,-1]\cup[1,\infty)

This is the principal secant function. The unique solution of \operatorname{Sec} x=k, for k\in\mathbb{R}-(-1,1), is written x=\operatorname{Sec}^{-1} k: 'x is the real number in \left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right] whose secant is k'.

Example 5
Find the principal value of \operatorname{Sec}^{-1}(-2).
  1. Let the required angle be y.
    y=\operatorname{Sec}^{-1}(-2)\iff\sec y=-2,\quad y\in[0,\pi],\ y\ne\dfrac{\pi}{2}
  2. Change to cosine, because \sec y=\frac{1}{\cos y}.
    \cos y=-\dfrac12
  3. Cosine (and so secant) is negative, so y is in Quadrant II. The reference angle is \frac{\pi}{3}.
    y=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}
\operatorname{Sec}^{-1}(-2)=\dfrac{2\pi}{3}\ \ (=120^\circ)
Check point (page 238)
Find the principal value of \operatorname{Sec}^{-1}\left(\dfrac{2}{\sqrt3}\right).
  1. \sec y=\frac{2}{\sqrt3} means \cos y=\frac{\sqrt3}{2}, with y\in[0,\pi], y\ne\frac{\pi}{2}. Positive, so Quadrant I.
    \operatorname{Sec}^{-1}\left(\dfrac{2}{\sqrt3}\right)=\operatorname{Cos}^{-1}\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\pi}{6}
\operatorname{Sec}^{-1}\left(\dfrac{2}{\sqrt3}\right)=\dfrac{\pi}{6}

Inverse secant function (8.1.14)

\begin{aligned}&y=\operatorname{Sec}^{-1} x\iff x=\operatorname{Sec} y\\ &x\in(-\infty,-1]\cup[1,\infty)\\ &y\in[0,\pi],\ y\ne\dfrac{\pi}{2}\end{aligned}
ValueReason
\operatorname{Sec}^{-1}(1)=0\sec 0=1
\operatorname{Sec}^{-1}(-1)=\pi\sec\pi=-1
\operatorname{Sec}^{-1} x\to\dfrac{\pi}{2} as x\to\pm\infty\sec y\to\pm\infty as y\to\dfrac{\pi}{2}

Domain and range of the inverse secant function (8.1.15)

FunctionDomainRange
y=\operatorname{Sec} x[0,\pi], x\ne\dfrac{\pi}{2}\mathbb{R}-(-1,1)
y=\operatorname{Sec}^{-1} x\mathbb{R}-(-1,1)[0,\pi], y\ne\dfrac{\pi}{2}

(The book writes the range of \operatorname{Sec}^{-1} x as 'x\in[0,\pi],\ x\ne\frac{\pi}{2}'; the variable should be y, the output.)

y=\operatorname{Sec} x and its reflection y=\operatorname{Sec}^{-1} x
Reflection in y=x turns the vertical asymptote x=\frac{\pi}{2} into the horizontal asymptote y=\frac{\pi}{2}; (0,1)\to(1,0) and (\pi,-1)\to(-1,\pi).

8.1.16 to 8.1.18Principal cosecant function and inverse cosecant functionOften examined

Principal cosecant function (8.1.16)

\csc x=\dfrac{1}{\sin x} is defined wherever \sin x\ne0, that is for x\ne n\pi. Like secant, its values satisfy |\csc x|\ge1, so \csc x=k has many solutions for each k\in\mathbb{R}-(-1,1) and none for -1<k<1.

y=\csc x from -2\pi to 2\pi
The principal part is -\frac{\pi}{2}\le x\le\frac{\pi}{2}, x\ne0: from \left(-\frac{\pi}{2},-1\right) down to the asymptote x=0, and from the asymptote down to \left(\frac{\pi}{2},1\right).

Sine is one-to-one on \left[-\frac{\pi}{2},\frac{\pi}{2}\right], so its reciprocal is one-to-one there, except at x=0 where \sin x=0. The end points \pm\frac{\pi}{2} are kept because \csc\left(\pm\frac{\pi}{2}\right)=\pm1.

\operatorname{Csc} x=\csc x,\qquad x\in\left[-\tfrac{\pi}{2},0\right)\cup\left(0,\tfrac{\pi}{2}\right],\qquad \operatorname{Csc} x\in(-\infty,-1]\cup[1,\infty)

This is the principal cosecant function. The unique solution of \operatorname{Csc} x=k is written x=\operatorname{Csc}^{-1} k: 'x is the real number in \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right] whose cosecant is k'.

Example 6
Find the principal value of \operatorname{Csc}^{-1}(2).
  1. Let the required angle be y.
    y=\operatorname{Csc}^{-1}(2)\iff\csc y=2,\quad y\in\left[-\tfrac{\pi}{2},0\right)\cup\left(0,\tfrac{\pi}{2}\right]
  2. Change to sine, because \csc y=\frac{1}{\sin y}.
    \sin y=\dfrac12
  3. Positive, so y is in Quadrant I, and \sin\frac{\pi}{6}=\frac12.
    y=\dfrac{\pi}{6}
\operatorname{Csc}^{-1}(2)=\dfrac{\pi}{6}\ \ (=30^\circ)
Check point (page 239)
Find the principal value of \operatorname{Csc}^{-1}(-\sqrt2).
  1. \csc y=-\sqrt2 means \sin y=-\frac{1}{\sqrt2}, with y\in\left[-\frac{\pi}{2},0\right) because the value is negative. The reference angle is \frac{\pi}{4}.
    \operatorname{Csc}^{-1}(-\sqrt2)=-\dfrac{\pi}{4}
\operatorname{Csc}^{-1}(-\sqrt2)=-\dfrac{\pi}{4}

Inverse cosecant function (8.1.17)

y=\operatorname{Csc}^{-1} x\iff x=\operatorname{Csc} y,\qquad x\in(-\infty,-1]\cup[1,\infty),\quad y\in\left[-\tfrac{\pi}{2},0\right)\cup\left(0,\tfrac{\pi}{2}\right]
ValueReason
\operatorname{Csc}^{-1}(-1)=-\dfrac{\pi}{2}\csc\left(-\dfrac{\pi}{2}\right)=-1
\operatorname{Csc}^{-1}(\sqrt2)=\dfrac{\pi}{4}\csc\dfrac{\pi}{4}=\sqrt2
\operatorname{Csc}^{-1} x\to0 as x\to\pm\infty\csc y\to\pm\infty as y\to0

Domain and range of the inverse cosecant function (8.1.18)

FunctionDomainRange
y=\operatorname{Csc} x\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right], x\ne0\mathbb{R}-(-1,1)
y=\operatorname{Csc}^{-1} x\mathbb{R}-(-1,1)\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right], y\ne0
y=\operatorname{Csc} x and its reflection y=\operatorname{Csc}^{-1} x
The vertical asymptote x=0 of \operatorname{Csc} x becomes the horizontal asymptote y=0 (the x-axis) of \operatorname{Csc}^{-1} x.

8.1Finding principal values: summary and methodOften examined

Every question of the type 'find the principal value' is solved the same way. First learn the six ranges; they decide where the answer must lie.

Ranges of the six inverse trigonometric functions.
FunctionDomainRange (principal values)Input >0 givesInput <0 gives
\operatorname{Sin}^{-1} x[-1,1]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]Quadrant IQuadrant IV (negative angle)
\operatorname{Cos}^{-1} x[-1,1][0,\pi]Quadrant IQuadrant II
\operatorname{Tan}^{-1} x\mathbb{R}\left(-\frac{\pi}{2},\frac{\pi}{2}\right)Quadrant IQuadrant IV (negative angle)
\operatorname{Cot}^{-1} x\mathbb{R}(0,\pi)Quadrant IQuadrant II
\operatorname{Sec}^{-1} x|x|\ge1[0,\pi]-\left\{\frac{\pi}{2}\right\}Quadrant IQuadrant II
\operatorname{Csc}^{-1} x|x|\ge1\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}Quadrant IQuadrant IV (negative angle)
Method for any principal value
  1. Write y=F^{-1}(k) as F(y)=k and write down the range in which y must lie.
  2. Ignore the sign of k and find the acute reference angle \alpha with F(\alpha)=|k| from the table of standard values.
  3. Place the answer in the range. If k\ge0, then y=\alpha. If k<0:
    y=-\alpha\ \text{ for }\operatorname{Sin}^{-1},\ \operatorname{Tan}^{-1},\ \operatorname{Csc}^{-1};\qquad y=\pi-\alpha\ \text{ for }\operatorname{Cos}^{-1},\ \operatorname{Cot}^{-1},\ \operatorname{Sec}^{-1}
  4. Check that y is in the range and that F(y)=k.
Standard values used for exact principal values.
\theta0\frac{\pi}{6}\ (30^\circ)\frac{\pi}{4}\ (45^\circ)\frac{\pi}{3}\ (60^\circ)\frac{\pi}{2}\ (90^\circ)
\sin\theta0\frac12\frac{1}{\sqrt2}\frac{\sqrt3}{2}1
\cos\theta1\frac{\sqrt3}{2}\frac{1}{\sqrt2}\frac120
\tan\theta0\frac{1}{\sqrt3}1\sqrt3not defined
\cot\thetanot defined\sqrt31\frac{1}{\sqrt3}0
\sec\theta1\frac{2}{\sqrt3}\sqrt22not defined
\csc\thetanot defined2\sqrt2\frac{2}{\sqrt3}1

Properties I to VIProperties of inverse trigonometric functionsOften examined

These six groups of properties follow directly from the definitions and ranges. They are used to simplify expressions, to change one inverse function into another, and to prove identities. In each property, notice the condition on x: it is part of the property.

Property I: inverse of a function (angle in, angle out)

\begin{aligned}\operatorname{Cos}^{-1}(\operatorname{Cos} x)&=x,\ x\in[0,\pi]\\ \operatorname{Sin}^{-1}(\operatorname{Sin} x)&=x,\ x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\\ \operatorname{Tan}^{-1}(\operatorname{Tan} x)&=x,\ x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\\ \operatorname{Cot}^{-1}(\operatorname{Cot} x)&=x,\ x\in(0,\pi)\\ \operatorname{Sec}^{-1}(\operatorname{Sec} x)&=x,\ x\in\left[0,\tfrac{\pi}{2}\right)\cup\left(\tfrac{\pi}{2},\pi\right]\\ \operatorname{Csc}^{-1}(\operatorname{Csc} x)&=x,\ x\in\left[-\tfrac{\pi}{2},0\right)\cup\left(0,\tfrac{\pi}{2}\right]\end{aligned}

These hold only when x is already in the principal interval. Outside it, the answer is the angle in the principal interval that has the same trigonometric value.

Property I in a picture: y=\operatorname{Sin}^{-1}(\sin x)
The curve agrees with y=x only between the dotted lines x=\pm\frac{\pi}{2}. Outside that strip it zigzags, for example \operatorname{Sin}^{-1}\left(\sin\frac{5\pi}{6}\right)=\frac{\pi}{6}.
Example (Property I outside the interval)
Evaluate \operatorname{Sin}^{-1}\left(\sin\dfrac{5\pi}{6}\right) and \operatorname{Cos}^{-1}\left(\cos\dfrac{4\pi}{3}\right).
  1. \frac{5\pi}{6} is not in \left[-\frac{\pi}{2},\frac{\pi}{2}\right], so Property I does not apply directly. Evaluate the inside first.
    \sin\dfrac{5\pi}{6}=\sin\left(\pi-\dfrac{\pi}{6}\right)=\dfrac12\;\Rightarrow\;\operatorname{Sin}^{-1}\left(\dfrac12\right)=\dfrac{\pi}{6}
  2. \frac{4\pi}{3} is not in [0,\pi]. Evaluate the inside first.
    \cos\dfrac{4\pi}{3}=\cos\left(\pi+\dfrac{\pi}{3}\right)=-\dfrac12\;\Rightarrow\;\operatorname{Cos}^{-1}\left(-\dfrac12\right)=\dfrac{2\pi}{3}
\operatorname{Sin}^{-1}\left(\sin\dfrac{5\pi}{6}\right)=\dfrac{\pi}{6},\qquad \operatorname{Cos}^{-1}\left(\cos\dfrac{4\pi}{3}\right)=\dfrac{2\pi}{3}

Property II: function of the inverse (number in, number out)

\begin{aligned}&\left.\begin{aligned}\cos(\operatorname{Cos}^{-1} x)&=x\\ \sin(\operatorname{Sin}^{-1} x)&=x\end{aligned}\right\}\ x\in[-1,1]\\[6pt] &\left.\begin{aligned}\tan(\operatorname{Tan}^{-1} x)&=x\\ \cot(\operatorname{Cot}^{-1} x)&=x\end{aligned}\right\}\ x\in(-\infty,\infty)\\[6pt] &\left.\begin{aligned}\sec(\operatorname{Sec}^{-1} x)&=x\\ \csc(\operatorname{Csc}^{-1} x)&=x\end{aligned}\right\}\ x\in(-\infty,-1]\cup[1,\infty)\end{aligned}

This always works for every x in the domain, because \operatorname{Cos}^{-1} x is, by definition, an angle whose cosine is x. For example \cos\left(\operatorname{Cos}^{-1} 0.3\right)=0.3, while \cos\left(\operatorname{Cos}^{-1} 2\right) has no meaning because 2 is outside the domain.

Property III: negative arguments

\begin{aligned}&\left.\begin{aligned}\operatorname{Sin}^{-1}(-x)&=-\operatorname{Sin}^{-1} x\\ \operatorname{Cos}^{-1}(-x)&=\pi-\operatorname{Cos}^{-1} x\end{aligned}\right\}\ x\in[-1,1]\\[6pt] &\left.\begin{aligned}\operatorname{Tan}^{-1}(-x)&=-\operatorname{Tan}^{-1} x\\ \operatorname{Cot}^{-1}(-x)&=\pi-\operatorname{Cot}^{-1} x\end{aligned}\right\}\ x\in\mathbb{R}\\[6pt] &\left.\begin{aligned}\operatorname{Csc}^{-1}(-x)&=-\operatorname{Csc}^{-1} x\\ \operatorname{Sec}^{-1}(-x)&=\pi-\operatorname{Sec}^{-1} x\end{aligned}\right\}\ |x|\ge1\end{aligned}

Derivation \operatorname{Cos}^{-1}(-x)=\pi-\operatorname{Cos}^{-1} x

Aim. Show that \operatorname{Cos}^{-1}(-x)=\pi-\operatorname{Cos}^{-1} x for -1\le x\le1.

  1. Name the angle. Let \theta=\operatorname{Cos}^{-1} x. By the definition, \cos\theta=x and \theta lies in the range.
    \cos\theta=x,\qquad 0\le\theta\le\pi
  2. Use the identity \cos(\pi-\theta)=-\cos\theta to produce -x.
    \cos(\pi-\theta)=-\cos\theta=-x
  3. Check that \pi-\theta is a principal value of \operatorname{Cos}^{-1}. From 0\le\theta\le\pi we get 0\le\pi-\theta\le\pi, which is the range of \operatorname{Cos}^{-1}.
    0\le\pi-\theta\le\pi
  4. So \pi-\theta is the angle in [0,\pi] whose cosine is -x.
    \operatorname{Cos}^{-1}(-x)=\pi-\theta=\pi-\operatorname{Cos}^{-1} x
Result\operatorname{Cos}^{-1}(-x)=\pi-\operatorname{Cos}^{-1} x,\qquad -1\le x\le1

Changing the sign of the input reflects the angle about \frac{\pi}{2}. For example \operatorname{Cos}^{-1}\left(-\frac12\right)=\pi-\frac{\pi}{3}=\frac{2\pi}{3}. The same proof with \cot(\pi-\theta)=-\cot\theta or \sec(\pi-\theta)=-\sec\theta gives the results for \operatorname{Cot}^{-1} and \operatorname{Sec}^{-1}.

Derivation \operatorname{Sin}^{-1}(-x)=-\operatorname{Sin}^{-1} x

Aim. Show that the inverse sine is an odd function.

  1. Let \theta=\operatorname{Sin}^{-1} x, so \sin\theta=x and \theta is in the principal range.
    \sin\theta=x,\qquad -\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}
  2. Sine is odd, so \sin(-\theta)=-\sin\theta=-x.
    \sin(-\theta)=-x
  3. Multiplying the inequality by -1 shows that -\theta is also in \left[-\frac{\pi}{2},\frac{\pi}{2}\right], so it is the principal value for -x.
    \operatorname{Sin}^{-1}(-x)=-\theta=-\operatorname{Sin}^{-1} x
Result\operatorname{Sin}^{-1}(-x)=-\operatorname{Sin}^{-1} x,\qquad -1\le x\le1

The graph of y=\operatorname{Sin}^{-1} x is symmetric about the origin. The same steps with \tan(-\theta)=-\tan\theta and \csc(-\theta)=-\csc\theta prove the results for \operatorname{Tan}^{-1} and \operatorname{Csc}^{-1}.

Property IV: reciprocal arguments

\begin{aligned}\operatorname{Sin}^{-1}\left(\dfrac1x\right)&=\operatorname{Csc}^{-1} x\\ \operatorname{Cos}^{-1}\left(\dfrac1x\right)&=\operatorname{Sec}^{-1} x\\ x&\in(-\infty,-1]\cup[1,\infty)\end{aligned}
\operatorname{Tan}^{-1}\left(\dfrac1x\right)=\begin{cases}\operatorname{Cot}^{-1} x, & x>0\\ -\pi+\operatorname{Cot}^{-1} x, & x<0\end{cases}

The first two hold for every x with |x|\ge1, because \operatorname{Sin} and \operatorname{Csc} use the same interval (without 0), and \operatorname{Cos} and \operatorname{Sec} use the same interval (without \frac{\pi}{2}). Tangent and cotangent use different intervals, which is why the third one splits into two cases.

Derivation \operatorname{Tan}^{-1}\left(\dfrac1x\right) in terms of \operatorname{Cot}^{-1} x

Aim. Show that \operatorname{Tan}^{-1}\frac1x=\operatorname{Cot}^{-1} x for x>0 and \operatorname{Tan}^{-1}\frac1x=\operatorname{Cot}^{-1} x-\pi for x<0.

  1. Let \theta=\operatorname{Cot}^{-1} x. Then \cot\theta=x with 0<\theta<\pi, and \theta\ne\frac{\pi}{2} because x\ne0. So \tan\theta=\frac{1}{\cot\theta}=\frac1x.
    \tan\theta=\dfrac1x,\qquad 0<\theta<\pi
  2. **Case x>0.** Then \cot\theta>0, so 0<\theta<\frac{\pi}{2}. This is inside the range of \operatorname{Tan}^{-1}, so \theta is the principal value.
    \operatorname{Tan}^{-1}\dfrac1x=\theta=\operatorname{Cot}^{-1} x
  3. **Case x<0.** Now \frac{\pi}{2}<\theta<\pi, which is outside \left(-\frac{\pi}{2},\frac{\pi}{2}\right). Tangent has period \pi, so \tan(\theta-\pi)=\tan\theta=\frac1x, and \theta-\pi lies in \left(-\frac{\pi}{2},0\right).
    \operatorname{Tan}^{-1}\dfrac1x=\theta-\pi=\operatorname{Cot}^{-1} x-\pi
Result\operatorname{Tan}^{-1}\dfrac1x=\begin{cases}\operatorname{Cot}^{-1} x, & x>0\\ \operatorname{Cot}^{-1} x-\pi, & x<0\end{cases}

For negative x the two answers differ by \pi. Example: \operatorname{Cot}^{-1}(-1)=\frac{3\pi}{4} but \operatorname{Tan}^{-1}(-1)=-\frac{\pi}{4}=\frac{3\pi}{4}-\pi.

Property V: complementary pairs

\begin{aligned}\operatorname{Sin}^{-1} x+\operatorname{Cos}^{-1} x&=\dfrac{\pi}{2},\ \ x\in[-1,1]\\[6pt] \operatorname{Tan}^{-1} x+\operatorname{Cot}^{-1} x&=\dfrac{\pi}{2},\ \ x\in\mathbb{R}\\[6pt] \operatorname{Sec}^{-1} x+\operatorname{Csc}^{-1} x&=\dfrac{\pi}{2},\ \ |x|\ge1\end{aligned}

(The book prints the last one as \operatorname{Sec}^{-1}(-x)+\operatorname{Csc}^{-1}(-x)=\frac{\pi}{2}. Since -x runs over the same set |x|\ge1, it is the same statement.)

Derivation \operatorname{Sin}^{-1} x+\operatorname{Cos}^{-1} x=\dfrac{\pi}{2}

Aim. Prove that \operatorname{Sin}^{-1} x+\operatorname{Cos}^{-1} x=\dfrac{\pi}{2} for -1\le x\le1.

  1. Let \theta=\operatorname{Sin}^{-1} x. By the definition:
    \sin\theta=x,\qquad -\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}
  2. The complementary-angle identity turns sine into cosine.
    \cos\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta=x
  3. Check the range. Multiply the inequality for \theta by -1 and add \frac{\pi}{2}:
    -\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}\;\Rightarrow\;0\le\dfrac{\pi}{2}-\theta\le\pi
  4. So \frac{\pi}{2}-\theta is the angle in [0,\pi] whose cosine is x, that is, the principal value of \operatorname{Cos}^{-1} x.
    \operatorname{Cos}^{-1} x=\dfrac{\pi}{2}-\theta=\dfrac{\pi}{2}-\operatorname{Sin}^{-1} x
  5. Rearrange.
    \operatorname{Sin}^{-1} x+\operatorname{Cos}^{-1} x=\dfrac{\pi}{2}
Result\operatorname{Sin}^{-1} x+\operatorname{Cos}^{-1} x=\dfrac{\pi}{2},\qquad -1\le x\le1

The two angles are complementary, like the two acute angles of a right triangle. For example \operatorname{Sin}^{-1}(0.3)+\operatorname{Cos}^{-1}(0.3)=\frac{\pi}{2} without any calculation, and \operatorname{Cos}^{-1} x=\frac{\pi}{2}-\operatorname{Sin}^{-1} x lets you change one function into the other.

Derivation \operatorname{Tan}^{-1} x+\operatorname{Cot}^{-1} x=\dfrac{\pi}{2}

Aim. Prove that \operatorname{Tan}^{-1} x+\operatorname{Cot}^{-1} x=\dfrac{\pi}{2} for every real x.

  1. Let \theta=\operatorname{Tan}^{-1} x, so \tan\theta=x with -\frac{\pi}{2}<\theta<\frac{\pi}{2}.
  2. Use \cot\left(\frac{\pi}{2}-\theta\right)=\tan\theta.
    \cot\left(\dfrac{\pi}{2}-\theta\right)=x
  3. From -\frac{\pi}{2}<\theta<\frac{\pi}{2} we get 0<\frac{\pi}{2}-\theta<\pi, the range of \operatorname{Cot}^{-1}. Hence \operatorname{Cot}^{-1} x=\frac{\pi}{2}-\theta.
    \operatorname{Tan}^{-1} x+\operatorname{Cot}^{-1} x=\theta+\left(\dfrac{\pi}{2}-\theta\right)=\dfrac{\pi}{2}
Result\operatorname{Tan}^{-1} x+\operatorname{Cot}^{-1} x=\dfrac{\pi}{2},\qquad x\in\mathbb{R}

The same idea with \csc\left(\frac{\pi}{2}-\theta\right)=\sec\theta proves \operatorname{Sec}^{-1} x+\operatorname{Csc}^{-1} x=\frac{\pi}{2}.

Property VI: changing one inverse function into another

To express \operatorname{Sin}^{-1} x through the other functions, draw a right triangle with an acute angle \theta, opposite side x and hypotenuse 1. Pythagoras gives the adjacent side \sqrt{1-x^2}, and all six ratios of \theta can be read off.

Derivation Property VI for \operatorname{Sin}^{-1} x

Aim. Write \operatorname{Sin}^{-1} x in terms of the other five inverse functions, for 0<x<1.

For \theta=\operatorname{Sin}^{-1} x: opposite x, hypotenuse 1, adjacent \sqrt{1-x^2}.
  1. Let \theta=\operatorname{Sin}^{-1} x. For 0<x<1 the angle is acute, so it fits in a right triangle with opposite side x and hypotenuse 1.
    \sin\theta=\dfrac{x}{1},\qquad 0<\theta<\dfrac{\pi}{2}
  2. Pythagoras gives the third side.
    \text{adjacent}=\sqrt{1^2-x^2}=\sqrt{1-x^2}
  3. Read the other ratios from the triangle.
    \cos\theta=\sqrt{1-x^2},\ \ \tan\theta=\dfrac{x}{\sqrt{1-x^2}},\ \ \cot\theta=\dfrac{\sqrt{1-x^2}}{x},\ \ \sec\theta=\dfrac{1}{\sqrt{1-x^2}},\ \ \csc\theta=\dfrac1x
  4. \theta is acute, so it lies in the principal range of every inverse function. Taking the inverse of each ratio gives \theta back.
Result\operatorname{Sin}^{-1} x=\operatorname{Cos}^{-1}\sqrt{1-x^2}=\operatorname{Tan}^{-1}\dfrac{x}{\sqrt{1-x^2}}=\operatorname{Cot}^{-1}\dfrac{\sqrt{1-x^2}}{x}=\operatorname{Sec}^{-1}\dfrac{1}{\sqrt{1-x^2}}=\operatorname{Csc}^{-1}\dfrac1x

One right triangle converts any inverse function into any other. The same method gives the chains for \operatorname{Cos}^{-1} x and \operatorname{Tan}^{-1} x below.

\operatorname{Cos}^{-1} x=\operatorname{Sin}^{-1}\sqrt{1-x^2}=\operatorname{Tan}^{-1}\dfrac{\sqrt{1-x^2}}{x}=\operatorname{Cot}^{-1}\dfrac{x}{\sqrt{1-x^2}}=\operatorname{Sec}^{-1}\dfrac1x=\operatorname{Csc}^{-1}\dfrac{1}{\sqrt{1-x^2}}

for 0<x<1 (triangle: adjacent x, hypotenuse 1)

\operatorname{Tan}^{-1} x=\operatorname{Sin}^{-1}\dfrac{x}{\sqrt{1+x^2}}=\operatorname{Cos}^{-1}\dfrac{1}{\sqrt{1+x^2}}=\operatorname{Cot}^{-1}\dfrac1x=\operatorname{Sec}^{-1}\sqrt{1+x^2}=\operatorname{Csc}^{-1}\dfrac{\sqrt{1+x^2}}{x}

for x>0 (triangle: opposite x, adjacent 1, hypotenuse \sqrt{1+x^2})

Example (Property VI)
Express \operatorname{Tan}^{-1}\dfrac34 in terms of \operatorname{Sin}^{-1} and of \operatorname{Cos}^{-1}.
  1. Let \theta=\operatorname{Tan}^{-1}\frac34. It is acute; draw a triangle with opposite 3 and adjacent 4.
    \text{hypotenuse}=\sqrt{3^2+4^2}=5
  2. Read \sin\theta and \cos\theta and invert (the angle is acute, so it is the principal value).
    \operatorname{Tan}^{-1}\dfrac34=\operatorname{Sin}^{-1}\dfrac35=\operatorname{Cos}^{-1}\dfrac45
\operatorname{Tan}^{-1}\dfrac34=\operatorname{Sin}^{-1}\dfrac35=\operatorname{Cos}^{-1}\dfrac45\ (\approx0.6435)

8.2Graphs of inverse trigonometric functions

A graph shows at a glance what the domain and range only list: where the function starts and ends, whether it rises or falls, its symmetry, and its asymptotes. Every graph in this section is the reflection of the principal trigonometric function in the line y=x.

  1. Draw the axes. The x-axis carries the input (a real number); the y-axis carries the output angle in radians, marked in steps of \frac{\pi}{6} or \frac{\pi}{4}.
  2. Make a table of values from the standard angles, for example \operatorname{Cos}^{-1}\frac12=\frac{\pi}{3}.
  3. Plot the points, join them with a smooth curve, mark included end points with filled dots, and draw asymptotes as dashed lines.
Domain, range and key points (corrected version of the book's table on pages 243 and 244).
FunctionDomainRangeKey points
y=\operatorname{Cos}^{-1} x[-1,1][0,\pi](-1,\pi), \left(0,\frac{\pi}{2}\right), (1,0)
y=\operatorname{Sin}^{-1} x[-1,1]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\left(-1,-\frac{\pi}{2}\right), (0,0), \left(1,\frac{\pi}{2}\right)
y=\operatorname{Tan}^{-1} x\mathbb{R}\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\left(-1,-\frac{\pi}{4}\right), (0,0), \left(1,\frac{\pi}{4}\right)
y=\operatorname{Cot}^{-1} x\mathbb{R}(0,\pi)\left(-1,\frac{3\pi}{4}\right), \left(0,\frac{\pi}{2}\right), \left(1,\frac{\pi}{4}\right)
y=\operatorname{Sec}^{-1} x(-\infty,-1]\cup[1,\infty)[0,\pi]-\left\{\frac{\pi}{2}\right\}(-2,\frac{2\pi}{3}), (-1,\pi), (1,0), (2,\frac{\pi}{3})
y=\operatorname{Csc}^{-1} x(-\infty,-1]\cup[1,\infty)\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}(-2,-\frac{\pi}{6}), (-1,-\frac{\pi}{2}), (1,\frac{\pi}{2}), (2,\frac{\pi}{6})
Shape of each graph. All six functions are non-periodic and one-to-one.
FunctionSymmetryIncreasing or decreasingAsymptotes
\operatorname{Cos}^{-1} xneither even nor odd; symmetric about the point \left(0,\frac{\pi}{2}\right)strictly decreasingnone
\operatorname{Sin}^{-1} xodd (symmetric about the origin)strictly increasingnone
\operatorname{Tan}^{-1} xoddstrictly increasinghorizontal: y=\pm\frac{\pi}{2}
\operatorname{Cot}^{-1} xneither; symmetric about \left(0,\frac{\pi}{2}\right)strictly decreasinghorizontal: y=0 and y=\pi
\operatorname{Sec}^{-1} xneither; symmetric about \left(0,\frac{\pi}{2}\right)increasing on each of its two intervalshorizontal: y=\frac{\pi}{2}
\operatorname{Csc}^{-1} xodddecreasing on each of its two intervalshorizontal: y=0

Shifting, stretching and reflecting

  • Adding a constant outside moves the graph up or down: y=\operatorname{Sin}^{-1} x+c. Adding inside moves it left or right: y=\operatorname{Sin}^{-1}(x-c) is shifted c units right.
  • Multiplying outside stretches the graph vertically and multiplies the range: y=2\operatorname{Sin}^{-1} x has range [-\pi,\pi]. Multiplying x inside compresses it horizontally and divides the domain: y=\operatorname{Sin}^{-1}(2x) has domain \left[-\frac12,\frac12\right].
  • A negative factor reflects the graph: y=-\operatorname{Sin}^{-1} x in the x-axis, y=\operatorname{Sin}^{-1}(-x) in the y-axis.
Example (transformed graph)
Find the domain and range of y=2\operatorname{Sin}^{-1}(x-1) and sketch it.
  1. Domain. The input of \operatorname{Sin}^{-1} must be in [-1,1].
    -1\le x-1\le1\;\Rightarrow\;0\le x\le2
  2. Range. \operatorname{Sin}^{-1}(x-1) takes every value in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]; doubling gives:
    -\pi\le 2\operatorname{Sin}^{-1}(x-1)\le\pi
  3. Graph. It is y=\operatorname{Sin}^{-1} x moved 1 unit right and stretched vertically by 2. End points: (0,-\pi) and (2,\pi); centre (1,0).
Shift and stretch of y=\operatorname{Sin}^{-1} x
Replacing x by x-1 moves the curve 1 unit right; multiplying by 2 doubles every height, so the range becomes [-\pi,\pi].
D=[0,2],\qquad R=[-\pi,\pi]

8.2.1Graph of y = Cos⁻¹xOften examined

The graph of y=\operatorname{Cos}^{-1} x shows the angle y whose cosine is x. The domain is x\in[-1,1] and the range is y\in[0,\pi], so every output is an angle from 0 to \pi radians.

Construction steps
  1. Draw the axes. Mark the x-axis from -1 to 1 and the y-axis from 0 to \pi in steps of \frac{\pi}{6}, using a suitable scale.
  2. Plot the points of the table.
  3. Join them with a smooth curve and mark both end points with filled dots.
Table of values for y=\operatorname{Cos}^{-1} x.
x1\frac{\sqrt3}{2}\approx0.87\frac{1}{\sqrt2}\approx0.710.50-0.5-0.71-0.87-1
y=\operatorname{Cos}^{-1} x0\frac{\pi}{6}\frac{\pi}{4}\frac{\pi}{3}\frac{\pi}{2}\frac{2\pi}{3}\frac{3\pi}{4}\frac{5\pi}{6}\pi
degrees0^\circ30^\circ45^\circ60^\circ90^\circ120^\circ135^\circ150^\circ180^\circ
Graph of y=\operatorname{Cos}^{-1} x
Domain [-1,1], range [0,\pi]. The curve falls from (-1,\pi) through \left(0,\frac{\pi}{2}\right) to (1,0).
Properties
  • Domain and range: domain [-1,1], range [0,\pi].
  • Key points: (1,0), \left(0,\frac{\pi}{2}\right), (-1,\pi). The y-intercept is \frac{\pi}{2} and the x-intercept is 1.
  • Appearance: a smooth curve that starts at (-1,\pi), passes through \left(0,\frac{\pi}{2}\right) and ends at (1,0).
  • Monotonicity: strictly decreasing; as x increases, the angle decreases.
  • Symmetry: neither even nor odd. Because \operatorname{Cos}^{-1}(-x)=\pi-\operatorname{Cos}^{-1} x, the point (-x,\pi-y) is on the graph whenever (x,y) is, so the graph is symmetric about the point \left(0,\frac{\pi}{2}\right).
  • Reflection: it is the reflection of y=\operatorname{Cos} x (0\le x\le\pi) in the line y=x; the coordinates of every point are swapped.
  • Inverse relations: \cos(\operatorname{Cos}^{-1} x)=x for x\in[-1,1] and \operatorname{Cos}^{-1}(\cos x)=x for x\in[0,\pi].

Derivation \operatorname{Cos}^{-1} x in terms of \operatorname{Sin}^{-1}\sqrt{1-x^2}

Aim. Show that \operatorname{Cos}^{-1} x=\operatorname{Sin}^{-1}\sqrt{1-x^2} for 0\le x\le1 and \operatorname{Cos}^{-1} x=\pi-\operatorname{Sin}^{-1}\sqrt{1-x^2} for -1\le x<0.

  1. Let \theta=\operatorname{Cos}^{-1} x, so \cos\theta=x with 0\le\theta\le\pi. On this interval \sin\theta\ge0, so the positive root is correct.
    \sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-x^2}
  2. **Case 0\le x\le1.** Then 0\le\theta\le\frac{\pi}{2}, which is inside the range of \operatorname{Sin}^{-1}, so \theta itself is the principal value.
    \operatorname{Sin}^{-1}\sqrt{1-x^2}=\theta=\operatorname{Cos}^{-1} x
  3. **Case -1\le x<0.** Now \frac{\pi}{2}<\theta\le\pi, outside the range of \operatorname{Sin}^{-1}. But \sin(\pi-\theta)=\sin\theta=\sqrt{1-x^2} and 0\le\pi-\theta<\frac{\pi}{2}.
    \operatorname{Sin}^{-1}\sqrt{1-x^2}=\pi-\theta\;\Rightarrow\;\operatorname{Cos}^{-1} x=\pi-\operatorname{Sin}^{-1}\sqrt{1-x^2}
Result\operatorname{Cos}^{-1} x=\begin{cases}\operatorname{Sin}^{-1}\sqrt{1-x^2}, & 0\le x\le1\\ \pi-\operatorname{Sin}^{-1}\sqrt{1-x^2}, & -1\le x<0\end{cases}

The formula changes at x=0 because \operatorname{Cos}^{-1} gives obtuse angles for negative inputs while \operatorname{Sin}^{-1} never does.

8.2.2Graph of y = Sin⁻¹xOften examined

The graph of y=\operatorname{Sin}^{-1} x shows the angle y whose sine is x. The domain is x\in[-1,1] and the range is y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].

Construction steps
  1. Draw the axes. Mark the x-axis from -1 to 1 and the y-axis from -\frac{\pi}{2} to \frac{\pi}{2} in steps of \frac{\pi}{6}.
  2. Plot the points of the table and join them with a smooth curve; mark the end points \left(\pm1,\pm\frac{\pi}{2}\right) with filled dots.
Table of values for y=\operatorname{Sin}^{-1} x (0.87\approx\frac{\sqrt3}{2}, 0.71\approx\frac{1}{\sqrt2}; the book rounds 0.87 to 0.9).
x-1-0.87-0.71-0.500.50.710.871
y=\operatorname{Sin}^{-1} x-\frac{\pi}{2}-\frac{\pi}{3}-\frac{\pi}{4}-\frac{\pi}{6}0\frac{\pi}{6}\frac{\pi}{4}\frac{\pi}{3}\frac{\pi}{2}
degrees-90^\circ-60^\circ-45^\circ-30^\circ0^\circ30^\circ45^\circ60^\circ90^\circ
Graph of y=\operatorname{Sin}^{-1} x
Domain [-1,1], range \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. The curve rises from \left(-1,-\frac{\pi}{2}\right) through the origin to \left(1,\frac{\pi}{2}\right).
Properties
  • Domain and range: domain [-1,1], range \left[-\frac{\pi}{2},\frac{\pi}{2}\right].
  • Key points: \left(-1,-\frac{\pi}{2}\right), (0,0), \left(1,\frac{\pi}{2}\right). The only intercept is the origin.
  • Monotonicity: strictly increasing.
  • Odd function: \operatorname{Sin}^{-1}(-x)=-\operatorname{Sin}^{-1} x, so the graph is symmetric about the origin.
  • Reflection: it is the reflection of y=\operatorname{Sin} x \left(-\frac{\pi}{2}\le x\le\frac{\pi}{2}\right) in the line y=x.
  • Inverse relations: \sin(\operatorname{Sin}^{-1} x)=x for x\in[-1,1] and \operatorname{Sin}^{-1}(\sin x)=x for x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].
  • Identities: \operatorname{Sin}^{-1} x=\operatorname{Cos}^{-1}\sqrt{1-x^2} for 0\le x\le1, and \operatorname{Sin}^{-1} x=-\operatorname{Cos}^{-1}\sqrt{1-x^2} for -1\le x<0 (the minus sign is needed because \operatorname{Cos}^{-1} is never negative).

8.2.3Graph of y = Tan⁻¹x

The graph of y=\operatorname{Tan}^{-1} x shows the angle y whose tangent is x. The domain is all real numbers and the range is the open interval \left(-\frac{\pi}{2},\frac{\pi}{2}\right).

Construction steps
  1. Draw the axes. The x-axis must be long, because the domain is all of \mathbb{R}; mark the y-axis from -\frac{\pi}{2} to \frac{\pi}{2}.
  2. Draw the horizontal asymptotes y=\pm\frac{\pi}{2} as dashed lines.
  3. Plot the points of the table and join them smoothly, making the curve flatten towards the asymptotes at both ends.
Table of values for y=\operatorname{Tan}^{-1} x (the book rounds 1.73 to 1.7 and 0.58 to 0.6).
x\to-\infty-\sqrt3\approx-1.73-1-\frac{1}{\sqrt3}\approx-0.5800.5811.73\to+\infty
y=\operatorname{Tan}^{-1} x\to-\frac{\pi}{2}-\frac{\pi}{3}-\frac{\pi}{4}-\frac{\pi}{6}0\frac{\pi}{6}\frac{\pi}{4}\frac{\pi}{3}\to\frac{\pi}{2}
Graph of y=\operatorname{Tan}^{-1} x
Domain \mathbb{R}, range \left(-\frac{\pi}{2},\frac{\pi}{2}\right). The curve approaches the asymptotes y=\pm\frac{\pi}{2} but never reaches them.
Properties
  • Domain and range: domain (-\infty,\infty), range \left(-\frac{\pi}{2},\frac{\pi}{2}\right).
  • Key points: \left(-1,-\frac{\pi}{4}\right), (0,0), \left(1,\frac{\pi}{4}\right). The book also lists '\left(\pm\infty,\pm\frac{\pi}{2}\right)': these are the directions the curve approaches, not points on it.
  • Asymptotes: y=-\frac{\pi}{2} as x\to-\infty and y=\frac{\pi}{2} as x\to+\infty.
  • Shape: an S-shaped curve through the origin, strictly increasing.
  • Odd function: \operatorname{Tan}^{-1}(-x)=-\operatorname{Tan}^{-1} x; the graph is symmetric about the origin.
  • Reflection: the reflection of y=\operatorname{Tan} x \left(-\frac{\pi}{2}<x<\frac{\pi}{2}\right) in y=x.
  • Inverse relations: \tan(\operatorname{Tan}^{-1} x)=x for all x, and \operatorname{Tan}^{-1}(\tan x)=x for x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).

Derivation \operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1}\dfrac1x=\pm\dfrac{\pi}{2}

Aim. Show that \operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1}\frac1x=\frac{\pi}{2} for x>0 and -\frac{\pi}{2} for x<0.

  1. From Property IV:
    \operatorname{Tan}^{-1}\dfrac1x=\begin{cases}\operatorname{Cot}^{-1} x, & x>0\\ \operatorname{Cot}^{-1} x-\pi, & x<0\end{cases}
  2. Add \operatorname{Tan}^{-1} x and use Property V, \operatorname{Tan}^{-1} x+\operatorname{Cot}^{-1} x=\frac{\pi}{2}.
    x>0:\ \ \operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1}\dfrac1x=\operatorname{Tan}^{-1} x+\operatorname{Cot}^{-1} x=\dfrac{\pi}{2}
  3. For negative x the extra -\pi remains.
    x<0:\ \ \operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1}\dfrac1x=\dfrac{\pi}{2}-\pi=-\dfrac{\pi}{2}
Result\operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1}\dfrac1x=\begin{cases}\dfrac{\pi}{2}, & x>0\\ -\dfrac{\pi}{2}, & x<0\end{cases}

Example: \operatorname{Tan}^{-1} 2+\operatorname{Tan}^{-1}\frac12=\frac{\pi}{2}, and \operatorname{Tan}^{-1}(-\sqrt3)+\operatorname{Tan}^{-1}\left(-\frac{1}{\sqrt3}\right)=-\frac{\pi}{3}-\frac{\pi}{6}=-\frac{\pi}{2}.

8.2.4Graph of y = Cot⁻¹x

The graph of y=\operatorname{Cot}^{-1} x shows the angle y in (0,\pi) whose cotangent is x. The domain is all real numbers and the range is (0,\pi).

Construction steps
  1. Draw the axes; mark the y-axis from 0 to \pi in steps of \frac{\pi}{6}.
  2. Draw the asymptotes y=0 (the x-axis) and y=\pi.
  3. Plot the points of the table and join them smoothly.
Table of values for y=\operatorname{Cot}^{-1} x (1.73\approx\sqrt3, 0.58\approx\frac{1}{\sqrt3}).
x\to-\infty-1.73-1-0.5800.5811.73\to+\infty
y=\operatorname{Cot}^{-1} x\to\pi\frac{5\pi}{6}\frac{3\pi}{4}\frac{2\pi}{3}\frac{\pi}{2}\frac{\pi}{3}\frac{\pi}{4}\frac{\pi}{6}\to0
Graph of y=\operatorname{Cot}^{-1} x
Domain \mathbb{R}, range (0,\pi). The curve falls from near y=\pi (far left) through \left(0,\frac{\pi}{2}\right) towards y=0 (the x-axis, far right).
Properties
  • Domain and range: domain (-\infty,\infty), range (0,\pi).
  • Key points: \left(-1,\frac{3\pi}{4}\right), \left(0,\frac{\pi}{2}\right), \left(1,\frac{\pi}{4}\right). There is no x-intercept, because \operatorname{Cot}^{-1} x is never 0.
  • Asymptotes: y=\pi as x\to-\infty and y=0 as x\to+\infty.
  • Monotonicity: strictly decreasing (an S-shaped curve going down).
  • Symmetry: neither even nor odd. \operatorname{Cot}^{-1}(-x)=\pi-\operatorname{Cot}^{-1} x means the graph is symmetric about the point \left(0,\frac{\pi}{2}\right).
  • Reflection: the reflection of y=\operatorname{Cot} x (0<x<\pi) in y=x.
  • Inverse relations: \cot(\operatorname{Cot}^{-1} x)=x for all x, and \operatorname{Cot}^{-1}(\cot x)=x for x\in(0,\pi).
  • Identity: \operatorname{Cot}^{-1} x=\operatorname{Tan}^{-1}\frac1x for x>0, and \operatorname{Cot}^{-1} x=\pi+\operatorname{Tan}^{-1}\frac1x for x<0.

8.2.5Graph of y = Sec⁻¹x

The graph of y=\operatorname{Sec}^{-1} x shows the angle y in [0,\pi], y\ne\frac{\pi}{2}, whose secant is x. The domain is (-\infty,-1]\cup[1,\infty), so the graph has two separate branches, and the range is \left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right].

Construction steps
  1. Draw the axes; mark the y-axis from 0 to \pi and draw the asymptote y=\frac{\pi}{2} as a dashed line.
  2. Leave the strip -1<x<1 empty: no point of the graph lies there.
  3. Plot the points of the table; join each branch smoothly, with filled dots at (1,0) and (-1,\pi).
Table of values for y=\operatorname{Sec}^{-1} x (the book rounds 1.15 to 1.2 and 1.41 to 1.4).
x1\frac{2}{\sqrt3}\approx1.15\sqrt2\approx1.412\to\pm\infty-2-1.41-1.15-1
y=\operatorname{Sec}^{-1} x0\frac{\pi}{6}\frac{\pi}{4}\frac{\pi}{3}\to\frac{\pi}{2}\frac{2\pi}{3}\frac{3\pi}{4}\frac{5\pi}{6}\pi
Graph of y=\operatorname{Sec}^{-1} x
Two branches: from (1,0) rising towards y=\frac{\pi}{2}, and from (-1,\pi) falling towards y=\frac{\pi}{2} as x\to-\infty. Nothing is drawn for -1<x<1.
Properties
  • Domain and range: domain (-\infty,-1]\cup[1,\infty), range \left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right].
  • Key points: (-2,\frac{2\pi}{3}), (-1,\pi), (1,0), (2,\frac{\pi}{3}). The only x-intercept is (1,0); there is no y-intercept because 0 is not in the domain.
  • Asymptote: the horizontal line y=\frac{\pi}{2}, approached as x\to\pm\infty.
  • Monotonicity: increasing on each branch: from 0 up towards \frac{\pi}{2} on [1,\infty), and from about \frac{\pi}{2} up to \pi on (-\infty,-1]. It is not increasing across the gap, because \operatorname{Sec}^{-1}(-1)=\pi is larger than \operatorname{Sec}^{-1}(1)=0.
  • Symmetry: neither even nor odd; \operatorname{Sec}^{-1}(-x)=\pi-\operatorname{Sec}^{-1} x gives symmetry about the point \left(0,\frac{\pi}{2}\right).
  • Reflection: the reflection of y=\operatorname{Sec} x (0\le x\le\pi, x\ne\frac{\pi}{2}) in y=x.
  • Inverse relations: \sec(\operatorname{Sec}^{-1} x)=x for |x|\ge1, and \operatorname{Sec}^{-1}(\sec x)=x for x\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right].
  • Identity: \operatorname{Sec}^{-1} x=\operatorname{Cos}^{-1}\frac1x for |x|\ge1.

8.2.6Graph of y = Csc⁻¹x

The graph of y=\operatorname{Csc}^{-1} x shows the angle y in \left[-\frac{\pi}{2},\frac{\pi}{2}\right], y\ne0, whose cosecant is x. The domain is (-\infty,-1]\cup[1,\infty) and the range is \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right], so the outputs are angles between -\frac{\pi}{2} and \frac{\pi}{2}.

Construction steps
  1. Draw the axes; mark the y-axis from -\frac{\pi}{2} to \frac{\pi}{2} in steps of \frac{\pi}{6}.
  2. Leave the strip -1<x<1 empty. The x-axis (y=0) is the asymptote.
  3. Plot the points of the table and join each branch smoothly, with filled dots at \left(\pm1,\pm\frac{\pi}{2}\right).
Table of values for y=\operatorname{Csc}^{-1} x (1.15\approx\frac{2}{\sqrt3}, 1.41\approx\sqrt2).
x11.151.412\to+\infty\to-\infty-2-1.41-1.15-1
y=\operatorname{Csc}^{-1} x\frac{\pi}{2}\frac{\pi}{3}\frac{\pi}{4}\frac{\pi}{6}\to0^{+}\to0^{-}-\frac{\pi}{6}-\frac{\pi}{4}-\frac{\pi}{3}-\frac{\pi}{2}
Graph of y=\operatorname{Csc}^{-1} x
Two branches that approach the x-axis (y=0): from \left(1,\frac{\pi}{2}\right) down towards 0, and from \left(-1,-\frac{\pi}{2}\right) up towards 0 as x\to-\infty.
Properties
  • Domain and range: domain (-\infty,-1]\cup[1,\infty), range \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right].
  • Key points: \left(-2,-\frac{\pi}{6}\right), \left(-1,-\frac{\pi}{2}\right), \left(1,\frac{\pi}{2}\right), \left(2,\frac{\pi}{6}\right). There are no intercepts.
  • Asymptote: the horizontal line y=0, approached as x\to\pm\infty.
  • Monotonicity: decreasing on each branch: from \frac{\pi}{2} down towards 0 on [1,\infty), and from just below 0 down to -\frac{\pi}{2} on (-\infty,-1].
  • Odd function: \operatorname{Csc}^{-1}(-x)=-\operatorname{Csc}^{-1} x; the graph is symmetric about the origin.
  • Reflection: the reflection of y=\operatorname{Csc} x \left(-\frac{\pi}{2}\le x\le\frac{\pi}{2},\ x\ne0\right) in y=x.
  • Inverse relations: \csc(\operatorname{Csc}^{-1} x)=x for |x|\ge1, and \operatorname{Csc}^{-1}(\csc x)=x for x\in\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right].
  • Identity: \operatorname{Csc}^{-1} x=\operatorname{Sin}^{-1}\frac1x for |x|\ge1.

Addition and subtraction formulaeOften examined

Two inverse trigonometric functions can often be combined into one. The idea is simple: name the two angles, use the compound-angle formula for \tan(\alpha\pm\beta), \sin(\alpha\pm\beta) or \cos(\alpha\pm\beta), and then check that the combined angle lies in the principal range.

Inverse tangent

Derivation \operatorname{Tan}^{-1} A+\operatorname{Tan}^{-1} B=\operatorname{Tan}^{-1}\dfrac{A+B}{1-AB}

Aim. Prove the addition formula for inverse tangents when AB<1.

Assumptions.
  • A and B are real numbers with AB<1.
  1. Name the angles. Let \alpha=\operatorname{Tan}^{-1} A and \beta=\operatorname{Tan}^{-1} B.
    \tan\alpha=A,\quad \tan\beta=B,\qquad -\dfrac{\pi}{2}<\alpha,\ \beta<\dfrac{\pi}{2}
  2. Use the compound-angle formula for tangent and substitute.
    \tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\dfrac{A+B}{1-AB}
  3. Check the range. \cos\alpha>0 and \cos\beta>0, and \cos(\alpha+\beta)=\cos\alpha\cos\beta\,(1-\tan\alpha\tan\beta)=\cos\alpha\cos\beta\,(1-AB). When AB<1 this is positive, so \alpha+\beta lies in \left(-\frac{\pi}{2},\frac{\pi}{2}\right).
    AB<1\;\Rightarrow\;-\dfrac{\pi}{2}<\alpha+\beta<\dfrac{\pi}{2}
  4. So \alpha+\beta is the principal value of the inverse tangent of \frac{A+B}{1-AB}.
    \operatorname{Tan}^{-1} A+\operatorname{Tan}^{-1} B=\operatorname{Tan}^{-1}\dfrac{A+B}{1-AB}
Result\operatorname{Tan}^{-1} A+\operatorname{Tan}^{-1} B=\operatorname{Tan}^{-1}\dfrac{A+B}{1-AB},\qquad AB<1

The sum of two angles is found from their tangents without finding either angle. If A,B>0 and AB>1, the sum is more than \frac{\pi}{2} and you must add \pi: \operatorname{Tan}^{-1} A+\operatorname{Tan}^{-1} B=\pi+\operatorname{Tan}^{-1}\frac{A+B}{1-AB}.

\operatorname{Tan}^{-1} A-\operatorname{Tan}^{-1} B=\operatorname{Tan}^{-1}\dfrac{A-B}{1+AB},\qquad AB>-1

Replace B by -B in the addition formula and use \operatorname{Tan}^{-1}(-B)=-\operatorname{Tan}^{-1} B.

2\operatorname{Tan}^{-1} A=\operatorname{Tan}^{-1}\dfrac{2A}{1-A^2},\qquad -1<A<1

Put B=A in the addition formula (the condition AB<1 becomes A^2<1).

Inverse sine and inverse cosine

Derivation \operatorname{Sin}^{-1} A\pm\operatorname{Sin}^{-1} B

Aim. Prove that \operatorname{Sin}^{-1} A\pm\operatorname{Sin}^{-1} B=\operatorname{Sin}^{-1}\left(A\sqrt{1-B^2}\pm B\sqrt{1-A^2}\right).

  1. Let \alpha=\operatorname{Sin}^{-1} A and \beta=\operatorname{Sin}^{-1} B, so \sin\alpha=A, \sin\beta=B with \alpha,\beta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].
  2. On this interval cosine is never negative, so take the positive roots.
    \cos\alpha=\sqrt{1-A^2},\qquad \cos\beta=\sqrt{1-B^2}
  3. Use \sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta.
    \sin(\alpha\pm\beta)=A\sqrt{1-B^2}\pm B\sqrt{1-A^2}
  4. If \alpha\pm\beta lies in \left[-\frac{\pi}{2},\frac{\pi}{2}\right], it is the principal value of the inverse sine of that number.
    \alpha\pm\beta=\operatorname{Sin}^{-1}\left(A\sqrt{1-B^2}\pm B\sqrt{1-A^2}\right)
Result\operatorname{Sin}^{-1} A\pm\operatorname{Sin}^{-1} B=\operatorname{Sin}^{-1}\left(A\sqrt{1-B^2}\pm B\sqrt{1-A^2}\right)

Valid when the combined angle stays in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. For 0\le A,B\le1 this is always true for the difference, and true for the sum when A^2+B^2\le1.

Derivation \operatorname{Cos}^{-1} A\pm\operatorname{Cos}^{-1} B

Aim. Prove that \operatorname{Cos}^{-1} A\pm\operatorname{Cos}^{-1} B=\operatorname{Cos}^{-1}\left(AB\mp\sqrt{1-A^2}\sqrt{1-B^2}\right).

  1. Let \alpha=\operatorname{Cos}^{-1} A and \beta=\operatorname{Cos}^{-1} B, so \cos\alpha=A, \cos\beta=B with \alpha,\beta\in[0,\pi].
  2. On [0,\pi] sine is never negative.
    \sin\alpha=\sqrt{1-A^2},\qquad \sin\beta=\sqrt{1-B^2}
  3. Use \cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta.
    \cos(\alpha\pm\beta)=AB\mp\sqrt{1-A^2}\sqrt{1-B^2}
  4. If \alpha\pm\beta lies in [0,\pi] it is the principal value.
    \alpha\pm\beta=\operatorname{Cos}^{-1}\left(AB\mp\sqrt{1-A^2}\sqrt{1-B^2}\right)
Result\operatorname{Cos}^{-1} A\pm\operatorname{Cos}^{-1} B=\operatorname{Cos}^{-1}\left(AB\mp\sqrt{1-A^2}\sqrt{1-B^2}\right)

The sum formula needs \alpha+\beta\le\pi, which holds when A+B\ge0. The difference formula needs \alpha\ge\beta, which holds when A\le B.

Worked examples

Example (tangent sum)
Show that \operatorname{Tan}^{-1}\dfrac12+\operatorname{Tan}^{-1}\dfrac13=\dfrac{\pi}{4}.
  1. Here A=\frac12, B=\frac13 and AB=\frac16<1, so the addition formula applies.
    \operatorname{Tan}^{-1}\dfrac12+\operatorname{Tan}^{-1}\dfrac13=\operatorname{Tan}^{-1}\dfrac{\frac12+\frac13}{1-\frac12\cdot\frac13}
  2. Simplify the fraction.
    \dfrac{\frac56}{\frac56}=1\;\Rightarrow\;\operatorname{Tan}^{-1} 1=\dfrac{\pi}{4}
\operatorname{Tan}^{-1}\dfrac12+\operatorname{Tan}^{-1}\dfrac13=\dfrac{\pi}{4}
Example (sine sum)
Show that \operatorname{Sin}^{-1}\dfrac35+\operatorname{Sin}^{-1}\dfrac{8}{17}=\operatorname{Sin}^{-1}\dfrac{77}{85}.
  1. Both numbers are positive and \left(\frac35\right)^2+\left(\frac{8}{17}\right)^2=0.36+0.22<1, so the sum formula applies. Find the square roots first.
    \sqrt{1-\tfrac{9}{25}}=\dfrac45,\qquad \sqrt{1-\tfrac{64}{289}}=\dfrac{15}{17}
  2. Substitute into A\sqrt{1-B^2}+B\sqrt{1-A^2}.
    \dfrac35\cdot\dfrac{15}{17}+\dfrac{8}{17}\cdot\dfrac45=\dfrac{45}{85}+\dfrac{32}{85}=\dfrac{77}{85}
\operatorname{Sin}^{-1}\dfrac35+\operatorname{Sin}^{-1}\dfrac{8}{17}=\operatorname{Sin}^{-1}\dfrac{77}{85}
Example (cosine sum)
Show that \operatorname{Cos}^{-1}\dfrac45+\operatorname{Cos}^{-1}\dfrac{12}{13}=\operatorname{Cos}^{-1}\dfrac{33}{65}.
  1. A+B>0, so the sum formula applies. The square roots are \sqrt{1-\frac{16}{25}}=\frac35 and \sqrt{1-\frac{144}{169}}=\frac{5}{13}.
  2. Substitute into AB-\sqrt{1-A^2}\sqrt{1-B^2}.
    \dfrac45\cdot\dfrac{12}{13}-\dfrac35\cdot\dfrac{5}{13}=\dfrac{48}{65}-\dfrac{15}{65}=\dfrac{33}{65}
\operatorname{Cos}^{-1}\dfrac45+\operatorname{Cos}^{-1}\dfrac{12}{13}=\operatorname{Cos}^{-1}\dfrac{33}{65}
Example (when AB > 1)
Show that \operatorname{Tan}^{-1}1+\operatorname{Tan}^{-1}2+\operatorname{Tan}^{-1}3=\pi.
  1. For \operatorname{Tan}^{-1}2+\operatorname{Tan}^{-1}3, AB=6>1 with both positive, so add \pi.
    \operatorname{Tan}^{-1}2+\operatorname{Tan}^{-1}3=\pi+\operatorname{Tan}^{-1}\dfrac{2+3}{1-6}=\pi+\operatorname{Tan}^{-1}(-1)=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}
  2. Add \operatorname{Tan}^{-1}1=\frac{\pi}{4}.
    \dfrac{\pi}{4}+\dfrac{3\pi}{4}=\pi
\operatorname{Tan}^{-1}1+\operatorname{Tan}^{-1}2+\operatorname{Tan}^{-1}3=\pi

Applications: angles of elevation, depression and slopesOften examined

In practical problems we usually know lengths and want an angle. The inverse trigonometric functions give that angle directly from a ratio of two sides of a right triangle:

\begin{aligned}\theta&=\operatorname{Tan}^{-1}\left(\dfrac{\text{opposite}}{\text{adjacent}}\right)\\[4pt] \theta&=\operatorname{Sin}^{-1}\left(\dfrac{\text{opposite}}{\text{hypotenuse}}\right)\\[4pt] \theta&=\operatorname{Cos}^{-1}\left(\dfrac{\text{adjacent}}{\text{hypotenuse}}\right)\end{aligned}

For an acute angle the ratio is positive, and the principal value is exactly the acute angle we want. This is what the \sin^{-1}, \cos^{-1} and \tan^{-1} keys of a calculator give.

Line of sight
The straight line from the observer's eye to the object being looked at.
Angle of elevation
The angle between the horizontal and the line of sight when the object is above the observer.
Angle of depression
The angle between the horizontal and the line of sight when the object is below the observer.
Angle of inclination (of a ramp or road)
The angle a slope makes with the horizontal: \theta=\operatorname{Tan}^{-1}\left(\dfrac{\text{rise}}{\text{run}}\right).
Gradient
The ratio \dfrac{\text{rise}}{\text{run}}=\tan\theta, often written as '1 in 12' and so on.
Both angles are measured from the horizontal line through the observer's eye O, never from the vertical.
Example 1 (ramp)
A wheelchair ramp rises 1\,\mathrm{m} over a horizontal distance of 12\,\mathrm{m}. Find its angle of inclination and its length.
Rise, run and length of a ramp; \tan\theta=\frac{\text{rise}}{\text{run}}.
  1. The rise is opposite \theta and the run is adjacent, so use the inverse tangent.
    \theta=\operatorname{Tan}^{-1}\left(\dfrac{1}{12}\right)\approx4.76^\circ
  2. The length is the hypotenuse (Pythagoras).
    L=\sqrt{1^2+12^2}=\sqrt{145}\approx12.04\,\mathrm{m}
\theta\approx4.76^\circ,\qquad L\approx12.04\,\mathrm{m}
Example 2 (height of a building)
From a point 50\,\mathrm{m} from the foot of a building, the angle of elevation of its top is 35^\circ. Find the height. Then, from the height and a 60\,\mathrm{m} shadow cast by the same building, find the angle of elevation of the sun.
  1. Height: the height is opposite the angle and 50\,\mathrm{m} is adjacent.
    h=50\tan35^\circ\approx35.0\,\mathrm{m}
  2. Sun's elevation: now the sides are known and the angle is wanted, so use the inverse tangent.
    \theta=\operatorname{Tan}^{-1}\left(\dfrac{35.0}{60}\right)\approx30.3^\circ
h\approx35.0\,\mathrm{m},\qquad \theta\approx30.3^\circ

Derivation Height from two angles of elevation

Aim. A tower of height h is seen at an angle of elevation \alpha from a point C. After walking a distance d straight towards it to D, the angle is \beta (\beta>\alpha). Find h.

Tower BT seen from C (angle \alpha) and from D (angle \beta), with CD=d and DB=x.
  1. Let x=DB, the distance from D to the base B. Then CB=x+d.
  2. Right triangle DBT:
    \tan\beta=\dfrac{h}{x}\;\Rightarrow\;x=\dfrac{h}{\tan\beta}
  3. Right triangle CBT:
    \tan\alpha=\dfrac{h}{x+d}\;\Rightarrow\;x+d=\dfrac{h}{\tan\alpha}
  4. Subtract the two expressions to remove the unknown x.
    d=\dfrac{h}{\tan\alpha}-\dfrac{h}{\tan\beta}=h\,\dfrac{\tan\beta-\tan\alpha}{\tan\alpha\tan\beta}
  5. Solve for h.
    h=\dfrac{d\tan\alpha\tan\beta}{\tan\beta-\tan\alpha}
Resulth=\dfrac{d\tan\alpha\tan\beta}{\tan\beta-\tan\alpha},\qquad x=\dfrac{h}{\tan\beta}

Two sightings from a measured distance apart fix the height, even when the foot of the tower cannot be reached. With \alpha=30^\circ, \beta=60^\circ, d=40\,\mathrm{m} it gives h=20\sqrt3\approx34.6\,\mathrm{m} (FBISE 2026, Set B).

Example 3 (angle of depression)
From the top of a cliff 80\,\mathrm{m} high, a boat is seen 150\,\mathrm{m} from the foot of the cliff. Find the angle of depression of the boat.
  1. The angle of depression at the top equals the angle of elevation of the top seen from the boat (alternate angles). In that right triangle the cliff is opposite and 150\,\mathrm{m} is adjacent.
    \theta=\operatorname{Tan}^{-1}\left(\dfrac{80}{150}\right)\approx28.07^\circ
\theta\approx28.1^\circ
Example 4 (ladder, mechanical set-up)
A 5\,\mathrm{m} ladder leans against a vertical wall with its foot 1.5\,\mathrm{m} from the wall. Find the angle it makes with the ground and the height it reaches.
  1. The distance from the wall is adjacent to the angle and the ladder is the hypotenuse, so use the inverse cosine.
    \theta=\operatorname{Cos}^{-1}\left(\dfrac{1.5}{5}\right)=\operatorname{Cos}^{-1}(0.3)\approx72.54^\circ
  2. Height by Pythagoras.
    h=\sqrt{5^2-1.5^2}=\sqrt{22.75}\approx4.77\,\mathrm{m}
\theta\approx72.5^\circ,\qquad h\approx4.77\,\mathrm{m}
Example 5 (scale model of a bridge)
In a scale model of a cable-stayed bridge, a cable runs from the top of a tower 12\,\mathrm{cm} high to a point on the deck 30\,\mathrm{cm} from the foot of the tower. Find the angle between the cable and the deck. Does it change in the real bridge, built at scale 100:1?
  1. The tower is opposite the angle and the deck distance is adjacent.
    \theta=\operatorname{Tan}^{-1}\left(\dfrac{12}{30}\right)=\operatorname{Tan}^{-1}(0.4)\approx21.80^\circ
  2. In the real bridge both lengths are multiplied by 100: 12\,\mathrm{m} and 30\,\mathrm{m}. The ratio \frac{1200}{3000}=0.4 is unchanged, so the angle is the same.
\theta\approx21.8^\circ\ \text{in the model and in the real bridge}

Formula sheet

Inverse cosine
y=\operatorname{Cos}^{-1} x\iff\operatorname{Cos} y=x,\quad x\in[-1,1],\ y\in[0,\pi]
x: a real number (value of cosine); y: the principal angle
Units: y in radians (or degrees)
Inverse sine
y=\operatorname{Sin}^{-1} x\iff\operatorname{Sin} y=x,\quad x\in[-1,1],\ y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]
x: value of sine; y: the principal angle
Unit: y in radians
Inverse tangent
y=\operatorname{Tan}^{-1} x\iff\operatorname{Tan} y=x,\quad x\in\mathbb{R},\ y\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)
x: any real number; y: the principal angle
Unit: y in radians
Inverse cotangent
y=\operatorname{Cot}^{-1} x\iff\operatorname{Cot} y=x,\quad x\in\mathbb{R},\ y\in(0,\pi)
x: any real number; y: the principal angle
Unit: y in radians
Inverse secant
y=\operatorname{Sec}^{-1} x\iff\operatorname{Sec} y=x,\quad |x|\ge1,\ y\in[0,\pi]-\left\{\tfrac{\pi}{2}\right\}
x: value of secant, x\le-1 or x\ge1
Unit: y in radians
Inverse cosecant
y=\operatorname{Csc}^{-1} x\iff\operatorname{Csc} y=x,\quad |x|\ge1,\ y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]-\{0\}
x: value of cosecant, x\le-1 or x\ge1
Unit: y in radians
Property I (angle in, angle out)
\operatorname{Sin}^{-1}(\sin x)=x\ \ \left(x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\right),\qquad \operatorname{Cos}^{-1}(\cos x)=x\ \ (x\in[0,\pi])
true only when x is in the principal interval; similarly for the other four
Unit: radians
Property II (number in, number out)
\begin{aligned}&\left.\begin{aligned}\sin(\operatorname{Sin}^{-1} x)&=x\\ \cos(\operatorname{Cos}^{-1} x)&=x\end{aligned}\right\}\ (|x|\le1)\\[4pt] &\tan(\operatorname{Tan}^{-1} x)=x\ \ (x\in\mathbb{R})\end{aligned}
true for every x in the domain
Unit: none
Negative arguments (odd functions)
\begin{aligned}\operatorname{Sin}^{-1}(-x)&=-\operatorname{Sin}^{-1} x\\ \operatorname{Tan}^{-1}(-x)&=-\operatorname{Tan}^{-1} x\\ \operatorname{Csc}^{-1}(-x)&=-\operatorname{Csc}^{-1} x\end{aligned}
x in the domain of each function
Unit: radians
Negative arguments (range 0 to π)
\begin{aligned}\operatorname{Cos}^{-1}(-x)&=\pi-\operatorname{Cos}^{-1} x\\ \operatorname{Cot}^{-1}(-x)&=\pi-\operatorname{Cot}^{-1} x\\ \operatorname{Sec}^{-1}(-x)&=\pi-\operatorname{Sec}^{-1} x\end{aligned}
x in the domain of each function
Unit: radians
Reciprocal arguments
\operatorname{Sin}^{-1}\dfrac1x=\operatorname{Csc}^{-1} x,\qquad \operatorname{Cos}^{-1}\dfrac1x=\operatorname{Sec}^{-1} x\qquad(|x|\ge1)
x\le-1 or x\ge1
Unit: radians
Tan and Cot of reciprocals
\operatorname{Tan}^{-1}\dfrac1x=\begin{cases}\operatorname{Cot}^{-1} x, & x>0\\ \operatorname{Cot}^{-1} x-\pi, & x<0\end{cases}
x\ne0
Unit: radians
Complementary pairs
\operatorname{Sin}^{-1} x+\operatorname{Cos}^{-1} x=\dfrac{\pi}{2},\quad \operatorname{Tan}^{-1} x+\operatorname{Cot}^{-1} x=\dfrac{\pi}{2},\quad \operatorname{Sec}^{-1} x+\operatorname{Csc}^{-1} x=\dfrac{\pi}{2}
|x|\le1, x\in\mathbb{R}, |x|\ge1 respectively
Unit: radians
Tan of a number and its reciprocal
\operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1}\dfrac1x=\begin{cases}\dfrac{\pi}{2}, & x>0\\ -\dfrac{\pi}{2}, & x<0\end{cases}
x\ne0
Unit: radians
Property VI for sine
\operatorname{Sin}^{-1} x=\operatorname{Cos}^{-1}\sqrt{1-x^2}=\operatorname{Tan}^{-1}\dfrac{x}{\sqrt{1-x^2}}=\operatorname{Csc}^{-1}\dfrac1x
0<x<1 (acute angle; right triangle with opposite x, hypotenuse 1)
Unit: radians
Property VI for cosine
\operatorname{Cos}^{-1} x=\operatorname{Sin}^{-1}\sqrt{1-x^2}=\operatorname{Tan}^{-1}\dfrac{\sqrt{1-x^2}}{x}=\operatorname{Sec}^{-1}\dfrac1x
0<x<1
Unit: radians
Property VI for tangent
\operatorname{Tan}^{-1} x=\operatorname{Sin}^{-1}\dfrac{x}{\sqrt{1+x^2}}=\operatorname{Cos}^{-1}\dfrac{1}{\sqrt{1+x^2}}=\operatorname{Cot}^{-1}\dfrac1x
x>0
Unit: radians
Cosine through sine for negative x
\begin{aligned}\operatorname{Cos}^{-1} x&=\pi-\operatorname{Sin}^{-1}\sqrt{1-x^2}\ \ (-1\le x<0)\\ \operatorname{Sin}^{-1} x&=-\operatorname{Cos}^{-1}\sqrt{1-x^2}\ \ (-1\le x<0)\end{aligned}
use these instead of Property VI when x is negative
Unit: radians
Domain of a transformed inverse
y=\operatorname{Sin}^{-1}(ax+b):\ -1\le ax+b\le1;\qquad y=\operatorname{Csc}^{-1}(ax+b):\ |ax+b|\ge1
put the whole inside expression into the domain condition
Unit: none
Inverse tangent addition
\operatorname{Tan}^{-1} A+\operatorname{Tan}^{-1} B=\operatorname{Tan}^{-1}\dfrac{A+B}{1-AB}\quad(AB<1)
if A,B>0 and AB>1, add \pi
Unit: radians
Inverse tangent subtraction
\operatorname{Tan}^{-1} A-\operatorname{Tan}^{-1} B=\operatorname{Tan}^{-1}\dfrac{A-B}{1+AB}\quad(AB>-1)
A, B real
Unit: radians
Double inverse tangent
2\operatorname{Tan}^{-1} A=\operatorname{Tan}^{-1}\dfrac{2A}{1-A^2}\quad(|A|<1)
A real, -1<A<1
Unit: radians
Inverse sine sum and difference
\operatorname{Sin}^{-1} A\pm\operatorname{Sin}^{-1} B=\operatorname{Sin}^{-1}\left(A\sqrt{1-B^2}\pm B\sqrt{1-A^2}\right)
valid when the combined angle is in \left[-\frac{\pi}{2},\frac{\pi}{2}\right] (for A,B\ge0: sum needs A^2+B^2\le1)
Unit: radians
Inverse cosine sum and difference
\operatorname{Cos}^{-1} A\pm\operatorname{Cos}^{-1} B=\operatorname{Cos}^{-1}\left(AB\mp\sqrt{1-A^2}\sqrt{1-B^2}\right)
sum: A+B\ge0; difference: A\le B
Unit: radians
Angle from two sides
\theta=\operatorname{Tan}^{-1}\dfrac{\text{opposite}}{\text{adjacent}}=\operatorname{Sin}^{-1}\dfrac{\text{opposite}}{\text{hypotenuse}}=\operatorname{Cos}^{-1}\dfrac{\text{adjacent}}{\text{hypotenuse}}
\theta: an acute angle of a right triangle
Units: degrees or radians
Angle of inclination of a slope
\theta=\operatorname{Tan}^{-1}\dfrac{\text{rise}}{\text{run}},\qquad L=\sqrt{\text{rise}^2+\text{run}^2}
rise: vertical gain; run: horizontal distance; L: length along the slope
Units: \theta in degrees; lengths in \mathrm{m}
Height from two angles of elevation
h=\dfrac{d\tan\alpha\tan\beta}{\tan\beta-\tan\alpha}
\alpha, \beta: angles of elevation from two points d apart on a line towards the object (\beta>\alpha)
h and d in the same unit, for example \mathrm{m}

Key terms

Inverse trigonometric function
A function that gives the angle whose trigonometric ratio is a given number, for example \operatorname{Sin}^{-1} x is the angle whose sine is x.
One-to-one function
A function in which different inputs always give different outputs. Only one-to-one functions have inverses.
Horizontal line test
A function is one-to-one if no horizontal line cuts its graph more than once. The full trigonometric functions fail this test.
Restricted (principal) domain
The interval on which a trigonometric function is kept so that it becomes one-to-one: [0,\pi] for cosine, \left[-\frac{\pi}{2},\frac{\pi}{2}\right] for sine, and so on.
Principal function
A trigonometric function restricted to its principal domain, written with a capital letter such as \operatorname{Cos} x or \operatorname{Tan} x.
Principal value
The single value of an inverse trigonometric function, lying in its fixed range; for example the principal value of \operatorname{Cos}^{-1}\left(-\frac12\right) is \frac{2\pi}{3}.
Principal cosine function
\operatorname{Cos} x=\cos x for 0\le x\le\pi; its inverse is \operatorname{Cos}^{-1} x.
Principal sine function
\operatorname{Sin} x=\sin x for -\frac{\pi}{2}\le x\le\frac{\pi}{2}; its inverse is \operatorname{Sin}^{-1} x.
Principal tangent function
\operatorname{Tan} x=\tan x for -\frac{\pi}{2}<x<\frac{\pi}{2}; its inverse is \operatorname{Tan}^{-1} x.
Principal cotangent function
\operatorname{Cot} x=\cot x for 0<x<\pi; its inverse is \operatorname{Cot}^{-1} x.
Principal secant function
\operatorname{Sec} x=\sec x for 0\le x\le\pi, x\ne\frac{\pi}{2}; its inverse is \operatorname{Sec}^{-1} x.
Principal cosecant function
\operatorname{Csc} x=\csc x for -\frac{\pi}{2}\le x\le\frac{\pi}{2}, x\ne0; its inverse is \operatorname{Csc}^{-1} x.
Reference angle
The acute angle that has the same trigonometric ratio as |k|; used to place a principal value in the correct quadrant.
Reflection in y = x
The graph of an inverse function is the mirror image of the original graph in the line y=x: each point (a,b) becomes (b,a).
Horizontal asymptote
A horizontal line that a graph approaches but never reaches, for example y=\pm\frac{\pi}{2} for y=\operatorname{Tan}^{-1} x.
Odd function
A function with f(-x)=-f(x); its graph is symmetric about the origin. \operatorname{Sin}^{-1}, \operatorname{Tan}^{-1} and \operatorname{Csc}^{-1} are odd.
Angle of elevation
The angle between the horizontal and the line of sight to an object above the observer.
Angle of depression
The angle between the horizontal and the line of sight to an object below the observer.
Angle of inclination
The angle a slope (ramp, road, roof) makes with the horizontal, \theta=\operatorname{Tan}^{-1}\frac{\text{rise}}{\text{run}}.

Exercise 8.1p. 241–242

Principal values come from the six ranges and the standard angles (Q1, Q2). For compositions, evaluate the inner part first or draw a right triangle for the inner angle (Q3, Q4). Answers in Q4 are in radians, correct to three decimal places.

1
Find the principal values of each of the following without using a calculator.
Solution
(i) \operatorname{Cos}^{-1}\left(\dfrac{\sqrt3}{2}\right)
  1. Let y=\operatorname{Cos}^{-1}\frac{\sqrt3}{2}. Then \cos y=\frac{\sqrt3}{2} with y\in[0,\pi].
  2. The value is positive, so y is in Quadrant I. Since \cos\frac{\pi}{6}=\frac{\sqrt3}{2}:
    y=\dfrac{\pi}{6}
\operatorname{Cos}^{-1}\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\pi}{6}
(ii) \operatorname{Sin}^{-1}(1)
  1. Let y=\operatorname{Sin}^{-1}1. Then \sin y=1 with y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].
  2. The only angle in this interval with sine 1 is the right end point.
    y=\dfrac{\pi}{2}
\operatorname{Sin}^{-1}(1)=\dfrac{\pi}{2}
(iii) \operatorname{Tan}^{-1}(\sqrt3)
  1. Let y=\operatorname{Tan}^{-1}\sqrt3. Then \tan y=\sqrt3 with y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).
  2. Positive, so Quadrant I. Since \tan\frac{\pi}{3}=\sqrt3:
    y=\dfrac{\pi}{3}
\operatorname{Tan}^{-1}(\sqrt3)=\dfrac{\pi}{3}
(iv) \operatorname{Cot}^{-1}\left(\dfrac{\sqrt3}{3}\right)
  1. First simplify: \frac{\sqrt3}{3}=\frac{\sqrt3}{\sqrt3\cdot\sqrt3}=\frac{1}{\sqrt3}.
  2. Let y=\operatorname{Cot}^{-1}\frac{1}{\sqrt3}. Then \cot y=\frac{1}{\sqrt3} with y\in(0,\pi). Positive, so Quadrant I.
  3. Since \cot\frac{\pi}{3}=\frac{1}{\sqrt3} (equivalently \tan y=\sqrt3):
    y=\dfrac{\pi}{3}
\operatorname{Cot}^{-1}\left(\dfrac{\sqrt3}{3}\right)=\dfrac{\pi}{3}
(v) \operatorname{Sec}^{-1}\left(\dfrac{2\sqrt3}{3}\right)
  1. Simplify: \frac{2\sqrt3}{3}=\frac{2}{\sqrt3}. Let y=\operatorname{Sec}^{-1}\frac{2}{\sqrt3}, so \sec y=\frac{2}{\sqrt3} with y\in[0,\pi], y\ne\frac{\pi}{2}.
  2. Change to cosine: \cos y=\frac{\sqrt3}{2}. Positive, so Quadrant I.
    y=\dfrac{\pi}{6}
\operatorname{Sec}^{-1}\left(\dfrac{2\sqrt3}{3}\right)=\dfrac{\pi}{6}
(vi) \operatorname{Csc}^{-1}(-\sqrt2)
  1. Let y=\operatorname{Csc}^{-1}(-\sqrt2), so \csc y=-\sqrt2, that is \sin y=-\frac{1}{\sqrt2}, with y\in\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right].
  2. Negative, so y is between -\frac{\pi}{2} and 0. The reference angle is \frac{\pi}{4}.
    y=-\dfrac{\pi}{4}
\operatorname{Csc}^{-1}(-\sqrt2)=-\dfrac{\pi}{4}
(vii) \operatorname{Cos}^{-1}\left(-\dfrac12\right)
  1. Let y=\operatorname{Cos}^{-1}\left(-\frac12\right), so \cos y=-\frac12 with y\in[0,\pi].
  2. Negative, so Quadrant II. The reference angle is \frac{\pi}{3}.
    y=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}
\operatorname{Cos}^{-1}\left(-\dfrac12\right)=\dfrac{2\pi}{3}
(viii) \operatorname{Tan}^{-1}\left(\dfrac{1}{\sqrt3}\right)
  1. Let y=\operatorname{Tan}^{-1}\frac{1}{\sqrt3}, so \tan y=\frac{1}{\sqrt3} with y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).
  2. Positive, so Quadrant I. Since \tan\frac{\pi}{6}=\frac{1}{\sqrt3}:
    y=\dfrac{\pi}{6}
\operatorname{Tan}^{-1}\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{6}
(ix) \operatorname{Csc}^{-1}(-2)
  1. Let y=\operatorname{Csc}^{-1}(-2), so \csc y=-2, that is \sin y=-\frac12, with y in the range of \operatorname{Csc}^{-1}.
  2. Negative, so y is between -\frac{\pi}{2} and 0. The reference angle is \frac{\pi}{6}.
    y=-\dfrac{\pi}{6}
\operatorname{Csc}^{-1}(-2)=-\dfrac{\pi}{6}
2
Find sum of principal values of the following inverse trigonometric expressions without using calculator.
Solution
(i) \operatorname{Tan}^{-1}(1)+\operatorname{Cos}^{-1}\left(-\dfrac12\right)+\operatorname{Sin}^{-1}\left(-\dfrac12\right)
  1. Find each principal value separately.
    \operatorname{Tan}^{-1}1=\dfrac{\pi}{4},\qquad \operatorname{Cos}^{-1}\left(-\dfrac12\right)=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3},\qquad \operatorname{Sin}^{-1}\left(-\dfrac12\right)=-\dfrac{\pi}{6}
  2. Add, using the common denominator 12.
    \dfrac{\pi}{4}+\dfrac{2\pi}{3}-\dfrac{\pi}{6}=\dfrac{3\pi+8\pi-2\pi}{12}=\dfrac{9\pi}{12}=\dfrac{3\pi}{4}
\dfrac{3\pi}{4}
(ii) \operatorname{Cos}^{-1}\left(\dfrac12\right)+2\operatorname{Sin}^{-1}\left(\dfrac12\right)
  1. Principal values:
    \operatorname{Cos}^{-1}\dfrac12=\dfrac{\pi}{3},\qquad \operatorname{Sin}^{-1}\dfrac12=\dfrac{\pi}{6}
  2. Substitute.
    \dfrac{\pi}{3}+2\cdot\dfrac{\pi}{6}=\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}
\dfrac{2\pi}{3}
(iii) \operatorname{Tan}^{-1}(\sqrt3)-\operatorname{Sec}^{-1}(-2)
  1. Principal values (\sec y=-2 means \cos y=-\frac12, Quadrant II):
    \operatorname{Tan}^{-1}\sqrt3=\dfrac{\pi}{3},\qquad \operatorname{Sec}^{-1}(-2)=\dfrac{2\pi}{3}
  2. Subtract.
    \dfrac{\pi}{3}-\dfrac{2\pi}{3}=-\dfrac{\pi}{3}
-\dfrac{\pi}{3}
(iv) \operatorname{Cot}^{-1}(-\sqrt3)+\operatorname{Csc}^{-1}(-2)-\operatorname{Cos}^{-1}\left(\dfrac{\sqrt2}{2}\right)
  1. \operatorname{Cot}^{-1}(-\sqrt3): \cot y=-\sqrt3, y\in(0,\pi), Quadrant II, reference angle \frac{\pi}{6}.
    \operatorname{Cot}^{-1}(-\sqrt3)=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}
  2. \operatorname{Csc}^{-1}(-2): \sin y=-\frac12, so y=-\frac{\pi}{6}. \operatorname{Cos}^{-1}\frac{\sqrt2}{2}=\operatorname{Cos}^{-1}\frac{1}{\sqrt2}=\frac{\pi}{4}.
  3. Combine, using the common denominator 12.
    \dfrac{5\pi}{6}-\dfrac{\pi}{6}-\dfrac{\pi}{4}=\dfrac{10\pi-2\pi-3\pi}{12}=\dfrac{5\pi}{12}
\dfrac{5\pi}{12}
3
Find the exact real number value of each without using a calculator.
Solution
(i) \operatorname{Cos}^{-1}\left[\sin\left(\dfrac{\pi}{4}\right)\right]
  1. Evaluate the inside first.
    \sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}=\dfrac{1}{\sqrt2}
  2. Now find the angle in [0,\pi] with cosine \frac{1}{\sqrt2}.
    \operatorname{Cos}^{-1}\dfrac{1}{\sqrt2}=\dfrac{\pi}{4}
\dfrac{\pi}{4}
(ii) \operatorname{Sin}^{-1}\left[\cos\left(-\dfrac{2\pi}{3}\right)\right]
  1. Cosine is an even function, so the minus sign inside disappears.
    \cos\left(-\dfrac{2\pi}{3}\right)=\cos\dfrac{2\pi}{3}=-\dfrac12
  2. Now the principal value of \operatorname{Sin}^{-1}\left(-\frac12\right): negative, so between -\frac{\pi}{2} and 0.
    \operatorname{Sin}^{-1}\left(-\dfrac12\right)=-\dfrac{\pi}{6}
-\dfrac{\pi}{6}
(iii) \operatorname{Cos}^{-1}\left(\sin\dfrac{11\pi}{6}\right)
  1. \frac{11\pi}{6}=2\pi-\frac{\pi}{6} is in Quadrant IV, where sine is negative.
    \sin\dfrac{11\pi}{6}=-\sin\dfrac{\pi}{6}=-\dfrac12
  2. Principal value of \operatorname{Cos}^{-1}\left(-\frac12\right) (Quadrant II).
    \operatorname{Cos}^{-1}\left(-\dfrac12\right)=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}
\dfrac{2\pi}{3}
(iv) \operatorname{Cos}^{-1}\left(\cos\dfrac{\pi}{6}\right)
  1. \frac{\pi}{6} lies in [0,\pi], so Property I applies directly. (Check: \cos\frac{\pi}{6}=\frac{\sqrt3}{2} and \operatorname{Cos}^{-1}\frac{\sqrt3}{2}=\frac{\pi}{6}.)
    \operatorname{Cos}^{-1}\left(\cos\dfrac{\pi}{6}\right)=\dfrac{\pi}{6}
\dfrac{\pi}{6}
(v) \sin\left(\operatorname{Tan}^{-1}\dfrac34\right)
  1. Let \theta=\operatorname{Tan}^{-1}\frac34. Then \tan\theta=\frac34 and \theta is acute (positive input), so draw a right triangle with opposite 3 and adjacent 4.
  2. Hypotenuse by Pythagoras.
    \sqrt{3^2+4^2}=5
  3. Read the sine.
    \sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac35
\theta=\operatorname{Tan}^{-1}\frac34: opposite 3, adjacent 4, hypotenuse 5.
\dfrac35
(vi) \cos\left(2\operatorname{Sin}^{-1}\dfrac{\sqrt2}{2}\right)
  1. \frac{\sqrt2}{2}=\frac{1}{\sqrt2}, so the inner angle is a standard one.
    \operatorname{Sin}^{-1}\dfrac{1}{\sqrt2}=\dfrac{\pi}{4}
  2. Double it and take the cosine.
    \cos\left(2\cdot\dfrac{\pi}{4}\right)=\cos\dfrac{\pi}{2}=0
0
(vii) \sin\left[2\operatorname{Sin}^{-1}\left(\dfrac45\right)\right]
  1. Let \theta=\operatorname{Sin}^{-1}\frac45, so \sin\theta=\frac45 and \theta is acute. Then \cos\theta is positive.
    \cos\theta=\sqrt{1-\tfrac{16}{25}}=\dfrac35
  2. Use the double-angle formula \sin2\theta=2\sin\theta\cos\theta.
    \sin2\theta=2\cdot\dfrac45\cdot\dfrac35=\dfrac{24}{25}
\dfrac{24}{25}
(viii) \cos\left(\operatorname{Sin}^{-1}\dfrac{5}{13}\right)
  1. Let \theta=\operatorname{Sin}^{-1}\frac{5}{13}, so \sin\theta=\frac{5}{13} with \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. On this interval cosine is not negative.
  2. Use \cos\theta=\sqrt{1-\sin^2\theta}.
    \cos\theta=\sqrt{1-\tfrac{25}{169}}=\sqrt{\tfrac{144}{169}}=\dfrac{12}{13}
\dfrac{12}{13}
(ix) \sin\left[\operatorname{Cos}^{-1}\left(-\dfrac35\right)\right]
  1. Let \theta=\operatorname{Cos}^{-1}\left(-\frac35\right). Then \cos\theta=-\frac35 and \theta is in Quadrant II, where sine is positive.
  2. Take the positive root.
    \sin\theta=+\sqrt{1-\tfrac{9}{25}}=\dfrac45
\dfrac45
(x) \sin\left[\operatorname{Sin}^{-1}\dfrac23+\operatorname{Cos}^{-1}\dfrac12\right]
  1. Let A=\operatorname{Sin}^{-1}\frac23 and B=\operatorname{Cos}^{-1}\frac12=\frac{\pi}{3}. Both angles are acute.
    \sin A=\dfrac23,\ \ \cos A=\sqrt{1-\tfrac49}=\dfrac{\sqrt5}{3};\qquad \sin B=\dfrac{\sqrt3}{2},\ \ \cos B=\dfrac12
  2. Use \sin(A+B)=\sin A\cos B+\cos A\sin B.
    \sin(A+B)=\dfrac23\cdot\dfrac12+\dfrac{\sqrt5}{3}\cdot\dfrac{\sqrt3}{2}=\dfrac{2}{6}+\dfrac{\sqrt{15}}{6}=\dfrac{2+\sqrt{15}}{6}
\dfrac{2+\sqrt{15}}{6}\ (\approx0.979)
(xi) \cos\left(\operatorname{Sin}^{-1}\dfrac34+\operatorname{Cos}^{-1}\dfrac{5}{13}\right)
  1. Let A=\operatorname{Sin}^{-1}\frac34 and B=\operatorname{Cos}^{-1}\frac{5}{13}. Both are acute, so all their ratios are positive.
    \begin{aligned}\sin A&=\dfrac34\\[4pt] \cos A&=\sqrt{1-\tfrac{9}{16}}=\dfrac{\sqrt7}{4}\\[8pt] \cos B&=\dfrac{5}{13}\\[4pt] \sin B&=\sqrt{1-\tfrac{25}{169}}=\dfrac{12}{13}\end{aligned}
  2. Use \cos(A+B)=\cos A\cos B-\sin A\sin B.
    \cos(A+B)=\dfrac{\sqrt7}{4}\cdot\dfrac{5}{13}-\dfrac34\cdot\dfrac{12}{13}=\dfrac{5\sqrt7}{52}-\dfrac{36}{52}=\dfrac{5\sqrt7-36}{52}
\dfrac{5\sqrt7-36}{52}\ (\approx-0.438)
(xii) \cos\left[\operatorname{Sec}^{-1}(3)+\operatorname{Tan}^{-1}(2)\right]
  1. Let A=\operatorname{Sec}^{-1}3: \sec A=3, so \cos A=\frac13 and A is acute.
    \cos A=\dfrac13,\qquad \sin A=\sqrt{1-\tfrac19}=\dfrac{2\sqrt2}{3}
  2. Let B=\operatorname{Tan}^{-1}2: a right triangle with opposite 2, adjacent 1 and hypotenuse \sqrt5.
    \cos B=\dfrac{1}{\sqrt5},\qquad \sin B=\dfrac{2}{\sqrt5}
  3. Use \cos(A+B)=\cos A\cos B-\sin A\sin B.
    \cos(A+B)=\dfrac13\cdot\dfrac{1}{\sqrt5}-\dfrac{2\sqrt2}{3}\cdot\dfrac{2}{\sqrt5}=\dfrac{1-4\sqrt2}{3\sqrt5}
  4. Rationalise the denominator (multiply top and bottom by \sqrt5).
    \dfrac{1-4\sqrt2}{3\sqrt5}=\dfrac{\sqrt5-4\sqrt{10}}{15}
\dfrac{1-4\sqrt2}{3\sqrt5}=\dfrac{\sqrt5-4\sqrt{10}}{15}\ (\approx-0.694)
4
Find the unknown angles and use a calculator to evaluate the following as real numbers to three decimal places:
Solution
(i) \operatorname{Cos}^{-1}\left(\dfrac35\right)=\operatorname{Sin}^{-1}(\quad)
  1. Let \theta=\operatorname{Cos}^{-1}\frac35. It is acute: adjacent 3, hypotenuse 5, so opposite =\sqrt{25-9}=4.
  2. Read the sine and invert (acute angle, so principal value).
    \sin\theta=\dfrac45\;\Rightarrow\;\operatorname{Cos}^{-1}\dfrac35=\operatorname{Sin}^{-1}\dfrac45
  3. Calculator in radian mode.
    \operatorname{Sin}^{-1}(0.8)=0.927\ (53.130^\circ)
\operatorname{Cos}^{-1}\dfrac35=\operatorname{Sin}^{-1}\dfrac45=0.927
(ii) \operatorname{Sin}^{-1}\left(\dfrac23\right)=\operatorname{Cos}^{-1}(\quad)
  1. Let \theta=\operatorname{Sin}^{-1}\frac23 (acute): opposite 2, hypotenuse 3, adjacent \sqrt{9-4}=\sqrt5.
  2. Read the cosine.
    \cos\theta=\dfrac{\sqrt5}{3}\;\Rightarrow\;\operatorname{Sin}^{-1}\dfrac23=\operatorname{Cos}^{-1}\dfrac{\sqrt5}{3}
  3. Evaluate.
    \operatorname{Cos}^{-1}(0.7454)=0.730\ (41.810^\circ)
\operatorname{Sin}^{-1}\dfrac23=\operatorname{Cos}^{-1}\dfrac{\sqrt5}{3}=0.730
(iii) \operatorname{Sin}^{-1}\left(-\dfrac{1}{\sqrt5}\right)=-\operatorname{Cos}^{-1}(\quad)
  1. \operatorname{Sin}^{-1} is odd, so take the minus sign out.
    \operatorname{Sin}^{-1}\left(-\dfrac{1}{\sqrt5}\right)=-\operatorname{Sin}^{-1}\dfrac{1}{\sqrt5}
  2. Let \phi=\operatorname{Sin}^{-1}\frac{1}{\sqrt5} (acute): opposite 1, hypotenuse \sqrt5, adjacent 2. So \cos\phi=\frac{2}{\sqrt5}.
    \operatorname{Sin}^{-1}\left(-\dfrac{1}{\sqrt5}\right)=-\operatorname{Cos}^{-1}\dfrac{2}{\sqrt5}
  3. Evaluate.
    -\operatorname{Cos}^{-1}(0.8944)=-0.464\ (-26.565^\circ)
\operatorname{Sin}^{-1}\left(-\dfrac{1}{\sqrt5}\right)=-\operatorname{Cos}^{-1}\dfrac{2}{\sqrt5}=-0.464
(iv) \operatorname{Tan}^{-1}\left(\dfrac14\right)=\operatorname{Cos}^{-1}(\quad)
  1. Let \theta=\operatorname{Tan}^{-1}\frac14 (acute): opposite 1, adjacent 4, hypotenuse \sqrt{17}.
  2. Read the cosine.
    \cos\theta=\dfrac{4}{\sqrt{17}}\;\Rightarrow\;\operatorname{Tan}^{-1}\dfrac14=\operatorname{Cos}^{-1}\dfrac{4}{\sqrt{17}}
  3. Evaluate.
    \operatorname{Cos}^{-1}(0.9701)=0.245\ (14.036^\circ)
\operatorname{Tan}^{-1}\dfrac14=\operatorname{Cos}^{-1}\dfrac{4}{\sqrt{17}}=0.245
(v) \operatorname{Tan}^{-1}(-1.2)=-\operatorname{Cos}^{-1}(\quad)
  1. \operatorname{Tan}^{-1} is odd.
    \operatorname{Tan}^{-1}(-1.2)=-\operatorname{Tan}^{-1}(1.2)=-\operatorname{Tan}^{-1}\dfrac65
  2. Let \phi=\operatorname{Tan}^{-1}\frac65 (acute): opposite 6, adjacent 5, hypotenuse \sqrt{61}. So \cos\phi=\frac{5}{\sqrt{61}}.
    \operatorname{Tan}^{-1}(-1.2)=-\operatorname{Cos}^{-1}\dfrac{5}{\sqrt{61}}
  3. Evaluate.
    -\operatorname{Cos}^{-1}(0.6402)=-0.876\ (-50.194^\circ)
\operatorname{Tan}^{-1}(-1.2)=-\operatorname{Cos}^{-1}\dfrac{5}{\sqrt{61}}=-0.876
(vi) \operatorname{Cot}^{-1}\left(-\dfrac34\right)=-\operatorname{Sin}^{-1}(\quad)
  1. With the book's range (0,\pi), a negative input gives a Quadrant II angle.
    \operatorname{Cot}^{-1}\left(-\dfrac34\right)=\pi-\operatorname{Cot}^{-1}\dfrac34
  2. Let \phi=\operatorname{Cot}^{-1}\frac34 (acute): adjacent 3, opposite 4, hypotenuse 5. So \sin\phi=\frac45 and \phi=\operatorname{Sin}^{-1}\frac45.
    \operatorname{Cot}^{-1}\left(-\dfrac34\right)=\pi-\operatorname{Sin}^{-1}\dfrac45
  3. Evaluate.
    \pi-0.927=2.214\ (126.870^\circ)
  4. The unknown number is \frac45. If the question uses the other convention, in which the inverse cotangent of a negative number is the negative angle \operatorname{Tan}^{-1}\left(-\frac43\right), then the printed form holds exactly:
    \operatorname{Tan}^{-1}\left(-\dfrac43\right)=-\operatorname{Sin}^{-1}\dfrac45=-0.927
\operatorname{Cot}^{-1}\left(-\dfrac34\right)=\pi-\operatorname{Sin}^{-1}\dfrac45=2.214
(vii) \operatorname{Sec}^{-1}(2.041)=\operatorname{Tan}^{-1}(\quad)
  1. Let \theta=\operatorname{Sec}^{-1}2.041. The input is positive, so \theta is acute and \tan\theta>0.
  2. Use \tan^2\theta=\sec^2\theta-1.
    \tan\theta=\sqrt{2.041^2-1}=\sqrt{3.165681}=1.779
  3. So \operatorname{Sec}^{-1}(2.041)=\operatorname{Tan}^{-1}(1.779). Evaluate (for example as \operatorname{Cos}^{-1}\frac{1}{2.041}).
    \theta=1.059\ (60.663^\circ)
\operatorname{Sec}^{-1}(2.041)=\operatorname{Tan}^{-1}(1.779)=1.059
(viii) \operatorname{Sec}^{-1}(-\sqrt5)=\operatorname{Cot}^{-1}(\quad)
  1. Let \theta=\operatorname{Sec}^{-1}(-\sqrt5), so \cos\theta=-\frac{1}{\sqrt5} and \theta is in Quadrant II. There \sin\theta>0.
    \sin\theta=\sqrt{1-\tfrac15}=\dfrac{2}{\sqrt5}
  2. Then \cot\theta=\frac{\cos\theta}{\sin\theta}=-\frac12. \operatorname{Cot}^{-1} of a negative number is also in Quadrant II, so the two agree.
    \operatorname{Sec}^{-1}(-\sqrt5)=\operatorname{Cot}^{-1}\left(-\dfrac12\right)
  3. Evaluate.
    \theta=\pi-\operatorname{Tan}^{-1}2=2.034\ (116.565^\circ)
\operatorname{Sec}^{-1}(-\sqrt5)=\operatorname{Cot}^{-1}\left(-\dfrac12\right)=2.034
(ix) \operatorname{Csc}^{-1}(1.172)=\operatorname{Sin}^{-1}(\quad)
  1. Use Property IV: \operatorname{Csc}^{-1} x=\operatorname{Sin}^{-1}\frac1x for |x|\ge1.
    \operatorname{Csc}^{-1}(1.172)=\operatorname{Sin}^{-1}\dfrac{1}{1.172}=\operatorname{Sin}^{-1}(0.853)
  2. Evaluate.
    \operatorname{Sin}^{-1}(0.8532)=1.022\ (58.566^\circ)
\operatorname{Csc}^{-1}(1.172)=\operatorname{Sin}^{-1}(0.853)=1.022
(x) \operatorname{Csc}^{-1}\left(-\dfrac53\right)=\operatorname{Tan}^{-1}(\quad)
  1. Let \theta=\operatorname{Csc}^{-1}\left(-\frac53\right), so \sin\theta=-\frac35 and \theta\in\left[-\frac{\pi}{2},0\right). There \cos\theta>0.
    \cos\theta=\sqrt{1-\tfrac{9}{25}}=\dfrac45
  2. Then \tan\theta=-\frac34. \operatorname{Tan}^{-1} of a negative number is also in \left(-\frac{\pi}{2},0\right), so the two agree.
    \operatorname{Csc}^{-1}\left(-\dfrac53\right)=\operatorname{Tan}^{-1}\left(-\dfrac34\right)
  3. Evaluate.
    \operatorname{Tan}^{-1}(-0.75)=-0.644\ (-36.870^\circ)
\operatorname{Csc}^{-1}\left(-\dfrac53\right)=\operatorname{Tan}^{-1}\left(-\dfrac34\right)=-0.644

FBISE 2026 questions

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

162026 Annual I, Set B, MCQ 161 mark
The value of \sin^{-1}\left(\sin\dfrac{5\pi}{6}\right) is:
  1. A\dfrac{\pi}{6}Answer
  2. B\dfrac{5\pi}{6}
  3. C-\dfrac{\pi}{6}
  4. D\dfrac{\pi}{3}
Why

Work from the inside: \sin\dfrac{5\pi}{6}=\sin\left(\pi-\dfrac{\pi}{6}\right)=\dfrac12. The principal value of \sin^{-1}\dfrac12 is \dfrac{\pi}{6}, which lies in \left[-\frac{\pi}{2},\frac{\pi}{2}\right].

Option B is the trap: \sin^{-1}(\sin x)=x only for x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right], and \dfrac{5\pi}{6} is outside that interval. Option C has the wrong sign (\sin\frac{5\pi}{6}>0).

2 (v) OR2026 Annual I, Set B, Q.2 (v) OR4 marks
From a point on the ground, the angle of elevation to the top of a tower is 30^\circ. After walking 40 meters closer to the tower, the angle of elevation becomes 60^\circ.
Solution
(a) Find the height of the tower.
Tower BT of height h; first point C, second point D, CD=40\,\mathrm{m}, DB=x.
  1. Let h=BT be the height and x=DB the distance from the second point D to the base B. The first point C is 40\,\mathrm{m} farther, so CB=x+40.
  2. In right triangle DBT the angle at D is 60^\circ.
    \tan60^\circ=\dfrac{h}{x}\;\Rightarrow\;h=\sqrt3\,x\qquad(1)
  3. In right triangle CBT the angle at C is 30^\circ.
    \tan30^\circ=\dfrac{h}{x+40}\;\Rightarrow\;h=\dfrac{x+40}{\sqrt3}\qquad(2)
  4. Equate (1) and (2), then multiply both sides by \sqrt3.
    \sqrt3\,x=\dfrac{x+40}{\sqrt3}\;\Rightarrow\;3x=x+40\;\Rightarrow\;x=20\,\mathrm{m}
  5. Substitute in (1).
    h=20\sqrt3\approx34.64\,\mathrm{m}
h=20\sqrt3\,\mathrm{m}\approx34.6\,\mathrm{m}
(b) Find the original distance from the first point to the base of the tower.
  1. The first point C is 40\,\mathrm{m} farther from the base than D.
    CB=x+40=20+40=60\,\mathrm{m}
  2. Check with inverse tangents: the angles come back as given.
    \begin{aligned}\tan^{-1}\left(\dfrac{20\sqrt3}{60}\right)&=\tan^{-1}\left(\dfrac{1}{\sqrt3}\right)=30^\circ\\[4pt] \tan^{-1}\left(\dfrac{20\sqrt3}{20}\right)&=\tan^{-1}\sqrt3=60^\circ\end{aligned}
CB=60\,\mathrm{m}
2 (x)2026 Annual I, Set B, Q.2 (x)4 marks
Find the domain and range of f(x)=\sin^{-1}(2x-1).
Solution
  1. Domain. \sin^{-1}u is defined only for -1\le u\le1 (no angle has a sine outside this interval). Here u=2x-1.
    -1\le 2x-1\le 1
  2. Add 1 to all three parts.
    0\le 2x\le 2
  3. Divide by 2 (positive, so the signs do not change).
    0\le x\le 1\qquad\Rightarrow\qquad D_f=[0,1]
  4. Range. As x runs from 0 to 1, u=2x-1 runs through the whole interval [-1,1]. So f(x) takes every value of \sin^{-1}u, the whole principal range.
    -\dfrac{\pi}{2}\le\sin^{-1}(2x-1)\le\dfrac{\pi}{2}\qquad\Rightarrow\qquad R_f=\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]
  5. Check with the end points: f(0)=\sin^{-1}(-1)=-\frac{\pi}{2}, f\left(\frac12\right)=\sin^{-1}0=0, f(1)=\sin^{-1}1=\frac{\pi}{2}.
The graph exists only for 0\le x\le1 and covers every height from -\frac{\pi}{2} to \frac{\pi}{2}.
D_f=[0,1],\qquad R_f=\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]
4 OR2026 Annual I, Set B, Q.4 OR8 marks
Draw the graph of y=\sin^{-1}(x) for x\in[-2\pi,2\pi]. Use a proper scale. Label domain, range, intercepts, and endpoints.
Solution

The printed interval

The question asks for x\in[-2\pi,2\pi], but \sin^{-1}x is the angle whose sine is x, and the sine of every angle lies between -1 and 1. So \sin^{-1}x is defined only for -1\le x\le1; values such as \sin^{-1}2 or \sin^{-1}\pi do not exist. Since 2\pi\approx6.28, most of [-2\pi,2\pi] is outside the domain, and the graph can be drawn only on [-1,1], the part of [-2\pi,2\pi] where the function is defined.

The interval [-2\pi,2\pi] comes from the unit's learning outcome, where the sine curve is drawn from -2\pi to 2\pi and its principal part \left[-\frac{\pi}{2},\frac{\pi}{2}\right] is reflected in the line y=x (second sketch below).

Domain, range and table of values

y=\sin^{-1}x\iff \sin y=x,\qquad D=[-1,1],\qquad R=\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]
x-1-0.87-0.71-0.500.50.710.871
y-\frac{\pi}{2}-\frac{\pi}{3}-\frac{\pi}{4}-\frac{\pi}{6}0\frac{\pi}{6}\frac{\pi}{4}\frac{\pi}{3}\frac{\pi}{2}
y (decimal)-1.57-1.05-0.79-0.5200.520.791.051.57

Scale

x-axis: 1 cm represents 0.25, so the domain from -1 to 1 takes 8 cm. y-axis: 1 cm represents \frac{\pi}{12}, so each step of \frac{\pi}{6} is 2 cm and the range from -\frac{\pi}{2} to \frac{\pi}{2} takes 12 cm.

Graph

Graph of y=\sin^{-1}x
Scale: on the x-axis 1 cm =0.25; on the y-axis 1 cm =\frac{\pi}{12}. End points A and B are filled dots; the only intercept is the origin.

Labels and properties

  • Domain: [-1,1], shown by the dotted lines x=\pm1; nothing is drawn outside them.
  • Range: \left[-\frac{\pi}{2},\frac{\pi}{2}\right], shown by the dotted lines y=\pm\frac{\pi}{2}.
  • Intercepts: the graph meets both axes only at the origin O(0,0), since \sin^{-1}0=0.
  • End points: A\left(-1,-\frac{\pi}{2}\right) and B\left(1,\frac{\pi}{2}\right), both included (filled dots).
  • Shape: strictly increasing and symmetric about the origin (odd function: \sin^{-1}(-x)=-\sin^{-1}x). It is the reflection of y=\sin x, -\frac{\pi}{2}\le x\le\frac{\pi}{2}, in the line y=x.

Sketch from the sine curve on [-2π, 2π]

Where the interval [-2\pi,2\pi] comes from
Reflecting the whole sine curve from -2\pi to 2\pi in y=x gives the wavy curve x=\sin y, which a vertical line cuts many times, so it is not a function. The inverse function keeps only the solid piece from \left(-1,-\frac{\pi}{2}\right) to \left(1,\frac{\pi}{2}\right).
\begin{aligned}&D=[-1,1]\\ &R=\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\\ &\text{intercept }(0,0)\\ &\text{end points }\left(-1,-\dfrac{\pi}{2}\right),\ \left(1,\dfrac{\pi}{2}\right)\end{aligned}

Note. The printed interval [-2\pi,2\pi] is impossible for x: \sin^{-1}x exists only for -1\le x\le1. The answer explains this and draws the graph on its true domain.

172026 Annual I, Set D, MCQ 171 mark
The value of \cos^{-1}\left(\cos\dfrac{4\pi}{3}\right) is:
  1. A\dfrac{2\pi}{3}Answer
  2. B\dfrac{4\pi}{3}
  3. C\dfrac{\pi}{3}
  4. D\dfrac{5\pi}{3}
Why

\cos\dfrac{4\pi}{3}=\cos\left(\pi+\dfrac{\pi}{3}\right)=-\dfrac12, and the principal value of \cos^{-1}\left(-\dfrac12\right) is \pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}, which lies in [0,\pi].

Option B is the trap: \dfrac{4\pi}{3} is greater than \pi, so it can never be a value of \cos^{-1}. Option C, \frac{\pi}{3}, has cosine +\frac12.

2 (v) OR2026 Annual I, Set D, Q.2 (v) OR4 marks
An engineer is designing a ramp such that it rises 6\,\mathrm{m} over a horizontal distance of 18\,\mathrm{m}.
Solution
(a) Find the angle of inclination of the ramp with the horizontal using inverse trigonometric functions.
The ramp is the hypotenuse; the rise 6\,\mathrm{m} is opposite \theta and the run 18\,\mathrm{m} is adjacent.
  1. The rise is opposite the angle \theta and the horizontal distance is adjacent, so use tangent.
    \tan\theta=\dfrac{6}{18}=\dfrac13
  2. \theta is acute, so it is the principal value of the inverse tangent.
    \theta=\tan^{-1}\left(\dfrac13\right)\approx18.43^\circ\ \ (0.322\,\mathrm{rad})
\theta=\tan^{-1}\dfrac13\approx18.43^\circ
(b) Find the length of the ramp.
  1. The ramp is the hypotenuse of the right triangle (Pythagoras).
    L=\sqrt{6^2+18^2}=\sqrt{36+324}=\sqrt{360}=6\sqrt{10}
  2. Evaluate.
    L\approx18.97\,\mathrm{m}
  3. Check with the inverse sine: the angle agrees with part (a).
    \sin^{-1}\left(\dfrac{6}{6\sqrt{10}}\right)=\sin^{-1}(0.3162)\approx18.43^\circ
L=6\sqrt{10}\,\mathrm{m}\approx18.97\,\mathrm{m}

Note. Part (b) is printed 'the length of the reop'; 'reop' is a misprint for 'ramp'.

2 (x)2026 Annual I, Set D, Q.2 (x)4 marks
Find the domain and range of: f(x)=\operatorname{cosec}^{-1}(3x+2)
Solution
  1. Domain. \operatorname{cosec}^{-1}u (the book's \operatorname{Csc}^{-1} u) is defined only when |u|\ge1, that is u\le-1 or u\ge1, because the cosecant of an angle is never between -1 and 1. Here u=3x+2.
  2. Case 1:
    3x+2\ge1\;\Rightarrow\;3x\ge-1\;\Rightarrow\;x\ge-\dfrac13
  3. Case 2:
    3x+2\le-1\;\Rightarrow\;3x\le-3\;\Rightarrow\;x\le-1
  4. The domain is the union of the two cases.
    D_f=(-\infty,-1]\cup\left[-\dfrac13,\infty\right)=\mathbb{R}-\left(-1,-\dfrac13\right)
  5. Range. On this domain u=3x+2 takes every value with |u|\ge1, so f takes every value of \operatorname{Csc}^{-1} u: the whole principal range. The value 0 is excluded because \operatorname{Csc}^{-1} u=0 would need \sin0=\frac1u, that is \frac1u=0, which is impossible.
    R_f=\left[-\dfrac{\pi}{2},0\right)\cup\left(0,\dfrac{\pi}{2}\right]=\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]-\{0\}
  6. Check: f(-1)=\operatorname{Csc}^{-1}(-1)=-\frac{\pi}{2} and f\left(-\frac13\right)=\operatorname{Csc}^{-1}(1)=\frac{\pi}{2}.
No graph between x=-1 and x=-\frac13. Both branches approach the x-axis (y=0) but never reach it.
D_f=(-\infty,-1]\cup\left[-\dfrac13,\infty\right),\qquad R_f=\left[-\dfrac{\pi}{2},0\right)\cup\left(0,\dfrac{\pi}{2}\right]
4 OR2026 Annual I, Set D, Q.4 OR8 marks
Draw the graph of y=\cos^{-1}(x) for x\in[-2\pi,2\pi]. Use a proper scale. Label the domain, range, intercepts, and endpoints.
Solution

The printed interval

The question asks for x\in[-2\pi,2\pi], but \cos^{-1}x is the angle whose cosine is x, and the cosine of every angle lies between -1 and 1. So \cos^{-1}x is defined only for -1\le x\le1; for example \cos^{-1}3 does not exist. Since 2\pi\approx6.28, most of [-2\pi,2\pi] lies outside the domain, and the graph can be drawn only on [-1,1].

The interval [-2\pi,2\pi] comes from the unit's learning outcome, where the cosine curve is drawn from -2\pi to 2\pi and its principal part [0,\pi] is reflected in the line y=x (second sketch below).

Domain, range and table of values

y=\cos^{-1}x\iff \cos y=x,\qquad D=[-1,1],\qquad R=[0,\pi]
x-1-0.87-0.71-0.500.50.710.871
y\pi\frac{5\pi}{6}\frac{3\pi}{4}\frac{2\pi}{3}\frac{\pi}{2}\frac{\pi}{3}\frac{\pi}{4}\frac{\pi}{6}0
y (decimal)3.142.622.362.091.571.050.790.520

Scale

x-axis: 1 cm represents 0.25, so the domain from -1 to 1 takes 8 cm. y-axis: 1 cm represents \frac{\pi}{12}, so each step of \frac{\pi}{6} is 2 cm and the range from 0 to \pi takes 12 cm.

Graph

Graph of y=\cos^{-1}x
Scale: on the x-axis 1 cm =0.25; on the y-axis 1 cm =\frac{\pi}{12}. The y-intercept is C\left(0,\frac{\pi}{2}\right) and the x-intercept is B(1,0).

Labels and properties

  • Domain: [-1,1], shown by the dotted lines x=\pm1.
  • Range: [0,\pi]; the curve stays between the x-axis and the dotted line y=\pi.
  • Intercepts: y-intercept C\left(0,\frac{\pi}{2}\right) because \cos^{-1}0=\frac{\pi}{2}; x-intercept B(1,0) because \cos^{-1}1=0.
  • End points: A(-1,\pi) and B(1,0), both included (filled dots).
  • Shape: strictly decreasing; neither even nor odd, but symmetric about the point C\left(0,\frac{\pi}{2}\right) because \cos^{-1}(-x)=\pi-\cos^{-1}x. It is the reflection of y=\cos x, 0\le x\le\pi, in the line y=x.

Sketch from the cosine curve on [-2π, 2π]

Where the interval [-2\pi,2\pi] comes from
Reflecting the cosine curve from -2\pi to 2\pi in y=x gives the wavy curve x=\cos y, which is not a function. The inverse function keeps only the solid piece from (-1,\pi) to (1,0), the mirror image of y=\cos x on [0,\pi].
\begin{aligned}&D=[-1,1]\\ &R=[0,\pi]\\ &\text{intercepts }\left(0,\dfrac{\pi}{2}\right),\ (1,0)\\ &\text{end points }(-1,\pi),\ (1,0)\end{aligned}

Note. The printed interval [-2\pi,2\pi] is impossible for x: \cos^{-1}x exists only for -1\le x\le1. The answer explains this and draws the graph on its true domain.

Exam focus

  1. Find principal values such as \operatorname{Cos}^{-1}\left(-\frac12\right), \operatorname{Cot}^{-1}(-\sqrt3), \operatorname{Csc}^{-1}(-2), and sums of principal values. (short)
    The basic skill of the unit; Exercise 8.1 Q1 and Q2 are built on it and it appears in every MCQ on this topic.
    Go to answer
  2. MCQ: evaluate \sin^{-1}\left(\sin\frac{5\pi}{6}\right) or \cos^{-1}\left(\cos\frac{4\pi}{3}\right). (mcq)
    Asked in both 2026 sets (Set B MCQ 16, Set D MCQ 17); the trap answer is the starting angle.
    Go to answer
  3. Find the domain and range of \sin^{-1}(2x-1) or \operatorname{cosec}^{-1}(3x+2). (short)
    Section B, 4 marks, in both 2026 sets (Q.2 (x)).
    Go to answer
  4. Draw the graph of y=\sin^{-1}x or y=\cos^{-1}x with a proper scale, labelling domain, range, intercepts and end points. (long)
    Section C, 8 marks, in both 2026 sets (Q.4 OR); also know why [-2\pi,2\pi] cannot be the domain.
    Go to answer
  5. Height of a tower from two angles of elevation (30° and 60°, 40 m apart). (short)
    2026 Set B Q.2 (v) OR, 4 marks; a standard application problem.
    Go to answer
  6. Angle of inclination and length of a ramp from its rise and run. (short)
    2026 Set D Q.2 (v) OR, 4 marks; uses the inverse tangent directly.
    Go to answer
  7. Prove \operatorname{Sin}^{-1} x+\operatorname{Cos}^{-1} x=\frac{\pi}{2} and \operatorname{Cos}^{-1}(-x)=\pi-\operatorname{Cos}^{-1} x. (short)
    Short proofs that test the ranges; typical 3 to 4 mark questions.
    Go to answer
  8. Exact values such as \sin\left(\operatorname{Tan}^{-1}\frac34\right) and \cos\left(\operatorname{Sin}^{-1}\frac34+\operatorname{Cos}^{-1}\frac{5}{13}\right). (short)
    Exercise 8.1 Q3; the right-triangle method and the sign of the root decide the marks.
    Go to answer
  9. State and prove \operatorname{Tan}^{-1} A+\operatorname{Tan}^{-1} B=\operatorname{Tan}^{-1}\frac{A+B}{1-AB} and apply it, for example \operatorname{Tan}^{-1}\frac12+\operatorname{Tan}^{-1}\frac13=\frac{\pi}{4}. (long)
    Named in the unit's learning outcomes ('state, prove and apply'); the book pages for it are missing, so learn it from these notes.
    Go to answer
  10. Domain, range, symmetry and asymptotes of all six inverse functions. (short)
    First learning outcome of the unit; the book's own table has several misprints, so use the corrected table.
    Go to answer

Mark this unit as done to fill the Class 12 Mathematics ring.

Next
2026 paper questions

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