Inverse trigonometric functions

AS · P1 · 13 min

If $\sin\theta = \tfrac{1}{2}$ , what is $\theta$ ? There are infinitely many answers ( $30^\circ$ , $150^\circ$ , $390^\circ$ , and so on), so "undoing" sine needs a rule for choosing one. The notations $\sin^{-1}x$ , $\cos^{-1}x$ and $\tan^{-1}x$ give that rule: each picks a single answer, called the principal value. Paper 1 tests the ranges of principal values, exact principal values, and inverses of functions such as $f(x) = 3 - 2\sin x$ on a restricted domain, a favourite way to combine trigonometry with the functions section.

Why sine needs a restricted domain

A function has an inverse only if it is one-one: each output comes from exactly one input. On all real numbers, $\sin x$ is many-one. Every value between $-1$ and $1$ is taken infinitely often.

The fix is to keep just one piece of the graph that takes every value from $-1$ to $1$ exactly once. For sine, the natural piece is $-\tfrac{\pi}{2} \le x \le \tfrac{\pi}{2}$ (that is, $-90^\circ$ to $90^\circ$ ), where the curve rises steadily from $-1$ to $1$ . On that domain sine is one-one, and its inverse is $\sin^{-1}$ .

y = sin x y = 0.5 x = -pi/2 x = pi/2

The line $y = \tfrac{1}{2}$ meets $y = \sin x$ many times, but only once between $x = -\tfrac{\pi}{2}$ and $x = \tfrac{\pi}{2}$ , at $x = \tfrac{\pi}{6}$ . That one is $\sin^{-1}\tfrac{1}{2}$ .

Cosine is restricted to $0 \le x \le \pi$ , where it falls steadily from $1$ to $-1$ . Tangent is restricted to $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ , between two asymptotes, where it rises through every real number.

Principal values

Definition

$\sin^{-1}x$ , $\cos^{-1}x$ and $\tan^{-1}x$ denote the principal values of the inverse trigonometric relations: the unique angle in the stated range whose sine, cosine or tangent is $x$ .

Key result

	Defined for	Principal value lies in (radians)	In degrees
$\sin^{-1}x$	$-1 \le x \le 1$	$-\tfrac{\pi}{2} \le \sin^{-1}x \le \tfrac{\pi}{2}$	$-90^\circ$ to $90^\circ$
$\cos^{-1}x$	$-1 \le x \le 1$	$0 \le \cos^{-1}x \le \pi$	$0^\circ$ to $180^\circ$
$\tan^{-1}x$	all real $x$	$-\tfrac{\pi}{2} < \tan^{-1}x < \tfrac{\pi}{2}$	$-90^\circ$ to $90^\circ$ (not inclusive)

These are exactly the answers your calculator gives. A quick way to remember the ranges:

$\sin^{-1}$ and $\tan^{-1}$ give angles in the first or fourth quadrant (right half of the unit circle), so a negative input gives a negative angle.
$\cos^{-1}$ gives angles in the first or second quadrant (top half), so a negative input gives an obtuse angle, never a negative one.

Watch out

$\sin^{-1}x$ is not $\dfrac{1}{\sin x}$ . The $-1$ means "inverse function", as in $f^{-1}$ . The reciprocal $\dfrac{1}{\sin x}$ is written $(\sin x)^{-1}$ or $\operatorname{cosec}x$ , which is Paper 3 content. Note the different convention for powers: $\sin^2 x$ does mean $(\sin x)^2$ .

Doing and undoing

Because $\sin^{-1}$ undoes sine on the restricted domain:

$\sin\left(\sin^{-1}x\right) = x$ for every $x$ from $-1$ to $1$ ;
$\sin^{-1}(\sin x) = x$ only when $x$ is in the principal range $-\tfrac{\pi}{2} \le x \le \tfrac{\pi}{2}$ .

Outside the principal range, $\sin^{-1}(\sin x)$ returns the angle in the principal range with the same sine. For example, $\sin\tfrac{5\pi}{6} = \tfrac{1}{2}$ , so $\sin^{-1}\left(\sin\tfrac{5\pi}{6}\right) = \tfrac{\pi}{6}$ , not $\tfrac{5\pi}{6}$ . The same applies to cosine and tangent with their own ranges.

A trigonometric function of an inverse

To find something like $\tan\left(\cos^{-1}\tfrac{1}{3}\right)$ exactly, let $\theta = \cos^{-1}\tfrac{1}{3}$ . Then $\cos\theta = \tfrac{1}{3}$ and $\theta$ is in the principal range, which fixes its quadrant. Draw the triangle, as in Exact values and angles of any size, and read off the value with the correct sign.

Inverses of trigonometric functions

A function such as $f(x) = 3 - 2\sin x$ is one-one only on a suitable domain. Once it is, its inverse is found in the usual way: write $y = f(x)$ , rearrange to make $x$ the subject, and use $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ at the last step. The domain of $f^{-1}$ is the range of $f$ .

Inverse of a trigonometric function

Check $f$ is one-one on its domain: the inside of the trigonometric function must stay within one principal range (or one monotonic stretch of the graph).
Find the range of $f$ , using $-1 \le \sin \le 1$ and the given domain.
Write $y = f(x)$ and isolate the trigonometric function, for example $\sin x = \dfrac{3 - y}{2}$ .
Apply the inverse: $x = \sin^{-1}\left(\dfrac{3 - y}{2}\right)$ .
Swap letters: $f^{-1}(x) = \sin^{-1}\left(\dfrac{3 - x}{2}\right)$ , with domain equal to the range of $f$ .

For "find the largest value of $k$ for which $f$ is one-one on $0 \le x \le k$ ", find where the graph first turns back. For $\cos x$ on $0 \le x \le k$ , that is $k = \pi$ ; for $\cos\tfrac{x}{2}$ it is where $\tfrac{x}{2} = \pi$ , so $k = 2\pi$ .

Tip

The syllabus says the graphs of the inverse trigonometric functions are not required. They are the reflections of the restricted graphs in $y = x$ , like every inverse (see One-one and inverse functions). Knowing this helps you check ranges, but you will not be asked to sketch them.

y = sin(x) y = arcsin(x) y = x

The part of $y = \sin x$ between $x = -\tfrac{\pi}{2}$ and $x = \tfrac{\pi}{2}$ and the curve $y = \sin^{-1}x$ are reflections of each other in $y = x$ . The inverse curve stops at $x = \pm 1$ , because $\sin^{-1}x$ is only defined for $-1 \le x \le 1$ .

Worked examples

Exact principal values

Find the exact values of (a) $\sin^{-1}\left(-\tfrac{1}{2}\right)$ , (b) $\cos^{-1}\left(-\tfrac{1}{2}\right)$ , (c) $\tan^{-1}\left(-\sqrt{3}\right)$ , (d) $\cos^{-1}\left(-\tfrac{1}{\sqrt{2}}\right)$ , giving your answers in radians.

Solution

(a) $\sin\tfrac{\pi}{6} = \tfrac{1}{2}$ , and $\sin^{-1}$ of a negative number is negative: $\sin^{-1}\left(-\tfrac{1}{2}\right) = -\tfrac{\pi}{6}$ .

(b) $\cos\tfrac{\pi}{3} = \tfrac{1}{2}$ . The answer must lie in $[0, \pi]$ with negative cosine, so it is in the second quadrant: $\pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}$ .

(d) $\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}$ , second quadrant: $\pi - \tfrac{\pi}{4} = \tfrac{3\pi}{4}$ .

Undoing outside the principal range

Find the exact values of (a) $\sin^{-1}\left(\sin\tfrac{5\pi}{6}\right)$ and (b) $\cos^{-1}\left(\cos\tfrac{4\pi}{3}\right)$ .

Solution

(a) $\sin\tfrac{5\pi}{6} = \tfrac{1}{2}$ , and $\sin^{-1}\tfrac{1}{2} = \tfrac{\pi}{6}$ . (Not $\tfrac{5\pi}{6}$ , which is outside $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ .)

(b) $\cos\tfrac{4\pi}{3} = -\tfrac{1}{2}$ , and $\cos^{-1}\left(-\tfrac{1}{2}\right) = \tfrac{2\pi}{3}$ . (Not $\tfrac{4\pi}{3}$ , which is outside $[0, \pi]$ .)

A trigonometric function of an inverse

Find the exact values of (a) $\tan\left(\cos^{-1}\tfrac{1}{3}\right)$ and (b) $\sin\left(\tan^{-1}\left(-\tfrac{3}{4}\right)\right)$ .

Solution

(a) Let $\theta = \cos^{-1}\tfrac{1}{3}$ , so $\cos\theta = \tfrac{1}{3}$ with $0 \le \theta \le \pi$ . Since $\cos\theta > 0$ , $\theta$ is acute. Triangle: adjacent $1$ , hypotenuse $3$ , opposite $\sqrt{8} = 2\sqrt{2}$ . So $\tan\theta = 2\sqrt{2}$ .

(b) Let $\phi = \tan^{-1}\left(-\tfrac{3}{4}\right)$ , so $\tan\phi = -\tfrac{3}{4}$ with $-\tfrac{\pi}{2} < \phi < 0$ (fourth quadrant, sine negative). Triangle $3$ , $4$ , $5$ gives $\sin\phi = -\tfrac{3}{5}$ .

Inverse of a sine function

The function $f$ is defined by $f(x) = 3 - 2\sin x$ for $-\tfrac{\pi}{2} \le x \le \tfrac{\pi}{2}$ .

(a) Explain why $f$ has an inverse, and state the range of $f$ .

(b) Find an expression for $f^{-1}(x)$ and state its domain.

Solution

(a) On $-\tfrac{\pi}{2} \le x \le \tfrac{\pi}{2}$ , $\sin x$ increases from $-1$ to $1$ , so $f(x) = 3 - 2\sin x$ decreases from $5$ to $1$ . A decreasing function is one-one, so $f$ has an inverse. Range: $1 \le f(x) \le 5$ .

(b)

y = 3 - 2\sin x \quad\Rightarrow\quad \sin x = \frac{3 - y}{2} \quad\Rightarrow\quad x = \sin^{-1}\left(\frac{3 - y}{2}\right)

So $f^{-1}(x) = \sin^{-1}\left(\dfrac{3 - x}{2}\right)$ , with domain $1 \le x \le 5$ .

The largest domain and an inverse cosine

The function $f$ is defined by $f(x) = 2 - 3\cos x$ for $0 \le x \le p$ .

(a) State the largest value of $p$ for which $f$ has an inverse.

(b) For this value of $p$ , find $f^{-1}(x)$ and state its domain.

Solution

(a) $\cos x$ decreases from $1$ to $-1$ on $0 \le x \le \pi$ and then increases again. So $f$ is one-one up to $p = \pi$ , and no further.

(b) On $0 \le x \le \pi$ , $f$ increases from $2 - 3 = -1$ to $2 + 3 = 5$ , so its range is $-1 \le f(x) \le 5$ .

y = 2 - 3\cos x \quad\Rightarrow\quad \cos x = \frac{2 - y}{3} \quad\Rightarrow\quad f^{-1}(x) = \cos^{-1}\left(\frac{2 - x}{3}\right)

Domain of $f^{-1}$ : $-1 \le x \le 5$ .

Solving, one-one and the inverse together

The function $f$ is defined by $f(x) = 5 + 3\cos\left(\tfrac{1}{2}x\right)$ for $0 \le x \le 2\pi$ .

(a) Solve the equation $f(x) = 7$ , giving your answer correct to 3 significant figures.

(b) Explain why $f$ has an inverse.

Solution

(a) $3\cos\left(\tfrac{1}{2}x\right) = 2$ , so $\cos\left(\tfrac{1}{2}x\right) = \tfrac{2}{3}$ . Since $0 \le \tfrac{1}{2}x \le \pi$ , there is only one solution, the principal value:

\tfrac{1}{2}x = \cos^{-1}\tfrac{2}{3} = 0.8411 \quad\Rightarrow\quad x = 1.68

(b) As $x$ goes from $0$ to $2\pi$ , $\tfrac{1}{2}x$ goes from $0$ to $\pi$ , where cosine is decreasing. So $f$ is decreasing, hence one-one, hence it has an inverse.

y = 5 + 3\cos\left(\tfrac{1}{2}x\right) \quad\Rightarrow\quad \cos\left(\tfrac{1}{2}x\right) = \frac{y - 5}{3} \quad\Rightarrow\quad x = 2\cos^{-1}\left(\frac{y - 5}{3}\right)

So $f^{-1}(x) = 2\cos^{-1}\left(\dfrac{x - 5}{3}\right)$ for $2 \le x \le 8$ .

Watch out

Negative inverse cosines. $\cos^{-1}\left(-\tfrac{1}{2}\right)$ is $\tfrac{2\pi}{3}$ , never $-\tfrac{\pi}{3}$ . The principal range of $\cos^{-1}$ is $0$ to $\pi$ .

Cancelling blindly. $\sin^{-1}(\sin x) = x$ only for $x$ in $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ . Check the range before cancelling.

Inputs outside $[-1, 1]$ . $\sin^{-1}2$ and $\cos^{-1}(-1.5)$ do not exist. If a rearrangement gives $\sin x = 2$ , there is no solution.

Forgetting the domain of $f^{-1}$ . It is the range of $f$ , and it is usually worth a mark.

Degree and radian mode. If the domain is in radians, the inverse gives radians; set the calculator accordingly.

Exam tip

Ranges. You may be asked to state the range of principal values directly; give it with the correct inequality signs ( $\tan^{-1}$ has strict inequalities).
One-one explanations. "Explain why $f$ has an inverse" needs a reason: " $f$ is decreasing on the given domain, so it is one-one". Mention the domain.
Exact answers. When the input is $\pm\tfrac{1}{2}$ , $\pm\tfrac{1}{\sqrt{2}}$ , $\pm\tfrac{\sqrt{3}}{2}$ , $\pm 1$ , $\pm\sqrt{3}$ or $\pm\tfrac{1}{\sqrt{3}}$ , the answer is expected exactly, as a multiple of $\pi$ .
Inverse function notation. Write $f^{-1}(x) = \ldots$ in terms of $x$ , not $y$ . Brackets matter: $\sin^{-1}\left(\dfrac{3 - x}{2}\right)$ , not $\dfrac{\sin^{-1}(3 - x)}{2}$ .
Equations. The principal value is only the first solution of a trigonometric equation. Finding the others is covered in Solving trigonometric equations.

Summary

Sine, cosine and tangent are made one-one by restricting their domains; the inverses give principal values.
$-\tfrac{\pi}{2} \le \sin^{-1}x \le \tfrac{\pi}{2}$ , $0 \le \cos^{-1}x \le \pi$ , $-\tfrac{\pi}{2} < \tan^{-1}x < \tfrac{\pi}{2}$ .
$\sin^{-1}x$ and $\cos^{-1}x$ need $-1 \le x \le 1$ ; $\tan^{-1}x$ accepts any $x$ .
$\sin^{-1}x$ means the inverse function, not $\dfrac{1}{\sin x}$ .
$\sin(\sin^{-1}x) = x$ always; $\sin^{-1}(\sin x) = x$ only in the principal range.
For $\tan(\cos^{-1}k)$ and similar, draw the triangle and use the principal range to fix the sign.
Inverse of a trigonometric function: check one-one, find the range, rearrange, apply the inverse, state the domain.

Practice questions

Question

Find the exact values, in radians, of $\sin^{-1}\tfrac{\sqrt{3}}{2}$ , $\cos^{-1}0$ , $\tan^{-1}(-1)$ and $\cos^{-1}\left(-\tfrac{\sqrt{3}}{2}\right)$ .
Find, in degrees, $\sin^{-1}\left(-\tfrac{\sqrt{2}}{2}\right)$ , $\cos^{-1}(-1)$ and $\tan^{-1}\tfrac{1}{\sqrt{3}}$ .
Find the exact values of $\sin^{-1}\left(\sin\tfrac{2\pi}{3}\right)$ , $\cos^{-1}\left(\cos\left(-\tfrac{\pi}{4}\right)\right)$ and $\tan^{-1}\left(\tan\tfrac{3\pi}{4}\right)$ .
Find the exact values of $\cos\left(\sin^{-1}\tfrac{3}{5}\right)$ and $\tan\left(\sin^{-1}\left(-\tfrac{5}{13}\right)\right)$ .
Explain why $\cos^{-1}2$ is undefined, and why $\sin^{-1}(\sin\pi) \neq \pi$ .
Verify that $\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}$ when $x = \tfrac{1}{2}$ and when $x = -\tfrac{1}{2}$ .
The function $f$ is defined by $f(x) = 1 + 2\sin 2x$ for $-\tfrac{\pi}{4} \le x \le \tfrac{\pi}{4}$ . State the range of $f$ and find $f^{-1}(x)$ .
The function $g$ is defined by $g(x) = 2\cos x + 3$ for $0 \le x \le k$ . (a) State the largest value of $k$ for which $g$ has an inverse. (b) For this value of $k$ , find $g^{-1}(x)$ , state its domain, and find the exact value of $g^{-1}(4)$ .
The function $f$ is defined by $f(x) = 3 - 2\tan x$ for $-\tfrac{\pi}{4} \le x \le \tfrac{\pi}{4}$ . (a) Find the range of $f$ . (b) Find $f^{-1}(x)$ . (c) Find the exact value of $f^{-1}\left(3 - \tfrac{2}{\sqrt{3}}\right)$ . (d) Explain why the equation $f(x) = 3 + 2\sqrt{3}$ has no solution.
Solve the equation $\sin^{-1}x = \cos^{-1}x$ .

Answers

$\tfrac{\pi}{3}$ ; $\tfrac{\pi}{2}$ ; $-\tfrac{\pi}{4}$ ; $\pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$ .
$-45^\circ$ ; $180^\circ$ ; $30^\circ$ .
$\sin\tfrac{2\pi}{3} = \tfrac{\sqrt{3}}{2}$ , so the answer is $\tfrac{\pi}{3}$ . $\cos\left(-\tfrac{\pi}{4}\right) = \tfrac{1}{\sqrt{2}}$ , so the answer is $\tfrac{\pi}{4}$ . $\tan\tfrac{3\pi}{4} = -1$ , so the answer is $-\tfrac{\pi}{4}$ .
$\theta = \sin^{-1}\tfrac{3}{5}$ is acute, triangle $3$ , $4$ , $5$ : $\cos\theta = \tfrac{4}{5}$ . $\phi = \sin^{-1}\left(-\tfrac{5}{13}\right)$ is in $\left(-\tfrac{\pi}{2}, 0\right)$ , triangle $5$ , $12$ , $13$ , tangent negative: $\tan\phi = -\tfrac{5}{12}$ .
Cosine only takes values from $-1$ to $1$ , so no angle has cosine $2$ . $\sin\pi = 0$ and $\sin^{-1}0 = 0$ , because the principal value must lie in $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ , and $\pi$ does not.
$x = \tfrac{1}{2}$ : $\tfrac{\pi}{6} + \tfrac{\pi}{3} = \tfrac{\pi}{2}$ . $x = -\tfrac{1}{2}$ : $-\tfrac{\pi}{6} + \tfrac{2\pi}{3} = \tfrac{\pi}{2}$ .
$2x$ runs from $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ , so $\sin 2x$ runs from $-1$ to $1$ and the range is $-1 \le f(x) \le 3$ . $y = 1 + 2\sin 2x$ gives $\sin 2x = \tfrac{y - 1}{2}$ , so $f^{-1}(x) = \tfrac{1}{2}\sin^{-1}\left(\dfrac{x - 1}{2}\right)$ for $-1 \le x \le 3$ .
(a) $k = \pi$ . (b) The range of $g$ is $1 \le g(x) \le 5$ . $\cos x = \tfrac{y - 3}{2}$ , so $g^{-1}(x) = \cos^{-1}\left(\dfrac{x - 3}{2}\right)$ for $1 \le x \le 5$ . $g^{-1}(4) = \cos^{-1}\tfrac{1}{2} = \tfrac{\pi}{3}$ .
(a) $\tan x$ runs from $-1$ to $1$ , so $f(x)$ runs from $5$ down to $1$ : $1 \le f(x) \le 5$ . (b) $\tan x = \tfrac{3 - y}{2}$ , so $f^{-1}(x) = \tan^{-1}\left(\dfrac{3 - x}{2}\right)$ for $1 \le x \le 5$ . (c) $\dfrac{3 - \left(3 - \frac{2}{\sqrt{3}}\right)}{2} = \dfrac{1}{\sqrt{3}}$ , and $\tan^{-1}\tfrac{1}{\sqrt{3}} = \tfrac{\pi}{6}$ . (d) $3 + 2\sqrt{3} \approx 6.46$ is outside the range $1 \le f(x) \le 5$ . (Equivalently, it needs $\tan x = -\sqrt{3}$ , i.e. $x = -\tfrac{\pi}{3}$ , which is not in the domain.)
Let $\theta = \sin^{-1}x = \cos^{-1}x$ . Then $\theta$ lies in both $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ and $[0, \pi]$ , so $0 \le \theta \le \tfrac{\pi}{2}$ , and $\sin\theta = \cos\theta = x$ . In this interval that happens only at $\theta = \tfrac{\pi}{4}$ , so $x = \tfrac{\sqrt{2}}{2}$ .

Why sine needs a restricted domain

Principal values

Doing and undoing

A trigonometric function of an inverse

Inverses of trigonometric functions

Worked examples

Practice questions

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