Inverse Trig Functions

You’ve used the $\sin^{-1}$ button for years to find angles. This page looks at what that button really is: a function, $\arcsin x$, with its own domain, range and graph. Because sine repeats forever, building an inverse means making a choice, and understanding that choice explains some surprising results, like why $\arcsin(\sin x)$ isn’t always $x$. All angles on this page are in radians, as on AA papers.

Key ideas

Why we need to restrict the domain

An inverse function only exists if the original function is one-to-one (it passes the horizontal line test). Sine fails badly: $\sin x = \dfrac{1}{2}$ for $x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{13\pi}{6}, \ldots$ and infinitely many more.

The fix is to restrict the domain to one piece of the graph that is one-to-one and still takes every possible output. The standard choices are:

Function	Restricted to	Takes every value in
$\sin x$	$-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}$	$-1 \le y \le 1$
$\cos x$	$0 \le x \le \pi$	$-1 \le y \le 1$
$\tan x$	$-\dfrac{\pi}{2} \lt x \lt \dfrac{\pi}{2}$	all real $y$

The three inverse functions

Swapping the domain and range of each restricted function gives the inverse:

Function	Domain	Range
$y = \arcsin x$	$\{x \in \mathbb{R} \mid -1 \le x \le 1\}$	$\{y \in \mathbb{R} \mid -\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}\}$
$y = \arccos x$	$\{x \in \mathbb{R} \mid -1 \le x \le 1\}$	$\{y \in \mathbb{R} \mid 0 \le y \le \pi\}$
$y = \arctan x$	$x \in \mathbb{R}$	$\{y \in \mathbb{R} \mid -\dfrac{\pi}{2} \lt y \lt \dfrac{\pi}{2}\}$

So $\arcsin x$ means “the angle between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ whose sine is $x$”. The notation $\sin^{-1} x$ means the same thing. It is not $\dfrac{1}{\sin x}$; that’s $\csc x$ (see reciprocal trig functions).

Graphs

Each graph is the reflection of the restricted trig graph in the line $y = x$.

$y = \arcsin x$, $y = \arccos x$ and $y = \arctan x$ (blue), each the reflection in $y = x$ of the restricted trig graph (dashed).

• $y = \arcsin x$ is increasing, from $\left(-1, -\dfrac{\pi}{2}\right)$ to $\left(1, \dfrac{\pi}{2}\right)$. It’s an odd function: $\arcsin(-x) = -\arcsin x$.
• $y = \arccos x$ is decreasing, from $(-1, \pi)$ to $(1, 0)$, passing through $\left(0, \dfrac{\pi}{2}\right)$. It is not odd: $\arccos(-x) = \pi - \arccos x$.
• $y = \arctan x$ is increasing for all $x$, with horizontal asymptotes $y = -\dfrac{\pi}{2}$ and $y = \dfrac{\pi}{2}$ (the vertical asymptotes of $\tan x$, reflected). It’s odd: $\arctan(-x) = -\arctan x$.

Evaluating exact values

To find $\arcsin a$, $\arccos a$ or $\arctan a$: find a special angle with the right ratio, and then make sure it lies in the range of the inverse function. Negative inputs are where students slip:

• $\arcsin$ and $\arctan$ of a negative number give a negative angle (fourth quadrant).
• $\arccos$ of a negative number gives an obtuse angle between $\dfrac{\pi}{2}$ and $\pi$ (second quadrant).

Compositions

Because they’re inverses,

$$\sin(\arcsin x) = x \text{ for } -1 \le x \le 1 \qquad\text{but}\qquad \arcsin(\sin x) = x \text{ only for } -\frac{\pi}{2} \le x \le \frac{\pi}{2}$$

and similarly for cosine (with $0 \le x \le \pi$) and tangent (with $-\dfrac{\pi}{2} \lt x \lt \dfrac{\pi}{2}$). Outside those intervals, $\arcsin(\sin x)$ gives the angle in the range that has the same sine as $x$:

$y = \arcsin(\sin x)$ zigzags between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$. It equals $x$ only on the middle piece.

For a mixed composition like $\sin(\arccos x)$, let $\theta = \arccos x$, draw a right triangle (or use $\sin^2\theta + \cos^2\theta = 1$), and use the range of $\theta$ to decide the sign.

Looking ahead: in calculus you’ll differentiate these functions; see derivatives of inverse trig functions.

Worked examples

Example 1: Exact values

Find the exact value of each.

• (a) $\arcsin\left(-\dfrac{1}{2}\right)$
• (b) $\arccos\left(-\dfrac{\sqrt{2}}{2}\right)$
• (c) $\arctan\sqrt{3}$
• (d) $\arccos(-1)$

Solution. (a) $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, and the range of arcsin allows negative angles, so $\arcsin\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$.

(b) $\cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$. For a negative cosine, arccos gives a second-quadrant angle: $\arccos\left(-\dfrac{\sqrt{2}}{2}\right) = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$. (Not $-\dfrac{\pi}{4}$, which isn’t in the range $0 \le y \le \pi$.)

(c) $\tan\dfrac{\pi}{3} = \sqrt{3}$, and $\dfrac{\pi}{3}$ is in the range, so $\arctan\sqrt{3} = \dfrac{\pi}{3}$.

(d) $\cos\pi = -1$, so $\arccos(-1) = \pi$.

Example 2: When arcsin(sin x) isn’t x

Find the exact value of each.

• (a) $\arcsin\left(\sin\dfrac{5\pi}{6}\right)$
• (b) $\arccos\left(\cos\left(-\dfrac{\pi}{3}\right)\right)$
• (c) $\arctan\left(\tan\dfrac{3\pi}{4}\right)$

Solution. Work from the inside out.

(a) $\sin\dfrac{5\pi}{6} = \dfrac{1}{2}$, and $\arcsin\dfrac{1}{2} = \dfrac{\pi}{6}$. So the answer is $\dfrac{\pi}{6}$, not $\dfrac{5\pi}{6}$, because $\dfrac{5\pi}{6}$ is outside the range of arcsin. (This is the marked point on the zigzag graph.)

(b) $\cos\left(-\dfrac{\pi}{3}\right) = \dfrac{1}{2}$, and $\arccos\dfrac{1}{2} = \dfrac{\pi}{3}$.

(c) $\tan\dfrac{3\pi}{4} = -1$, and $\arctan(-1) = -\dfrac{\pi}{4}$.

Example 3: Compositions with a triangle

Find the exact value of (a) $\sin\left(\arccos\dfrac{3}{5}\right)$ and (b) $\tan\left(\arcsin\left(-\dfrac{5}{13}\right)\right)$. Then (c) write $\cos(\arctan x)$ in terms of $x$.

Solution. (a) Let $\theta = \arccos\dfrac{3}{5}$, so $\cos\theta = \dfrac{3}{5}$ and $0 \le \theta \le \pi$. In a right triangle with adjacent side $3$ and hypotenuse $5$, the opposite side is $4$. Sine is positive for $0 \le \theta \le \pi$, so

$$\sin\left(\arccos\frac{3}{5}\right) = \frac{4}{5}$$

(b) Let $\theta = \arcsin\left(-\dfrac{5}{13}\right)$, so $\sin\theta = -\dfrac{5}{13}$ and $\theta$ is in the fourth quadrant ($-\dfrac{\pi}{2} \lt \theta \lt 0$). Then $\cos\theta = \sqrt{1 - \dfrac{25}{169}} = \dfrac{12}{13}$ (positive in the fourth quadrant), and

$$\tan\theta = \frac{-5/13}{12/13} = -\frac{5}{12}$$

(c) Let $\theta = \arctan x$, so $\tan\theta = x = \dfrac{x}{1}$ with $-\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2}$. A right triangle with opposite side $x$ and adjacent side $1$ has hypotenuse $\sqrt{1 + x^2}$. Cosine is positive on this interval, so

$$\cos(\arctan x) = \frac{1}{\sqrt{1 + x^2}}$$

Example 4: A transformed inverse function

Let $f(x) = 2\arcsin(x - 1)$.

• (a) Find the domain and range of $f$.
• (b) Find $f(1.5)$ exactly.
• (c) Solve $f(x) = \dfrac{\pi}{2}$.

Solution. (a) The input of arcsin must be between $-1$ and $1$:

$$-1 \le x - 1 \le 1 \quad\Rightarrow\quad 0 \le x \le 2$$

The arcsin part is between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$, so doubling it gives values between $-\pi$ and $\pi$. Domain $\{x \in \mathbb{R} \mid 0 \le x \le 2\}$, range $\{y \in \mathbb{R} \mid -\pi \le y \le \pi\}$.

(b) $f(1.5) = 2\arcsin(0.5) = 2 \times \dfrac{\pi}{6} = \dfrac{\pi}{3}$.

(c)

$$2\arcsin(x - 1) = \frac{\pi}{2} \quad\Rightarrow\quad \arcsin(x - 1) = \frac{\pi}{4} \quad\Rightarrow\quad x - 1 = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} \quad\Rightarrow\quad x = 1 + \frac{\sqrt{2}}{2}$$

Check: $1 + \dfrac{\sqrt{2}}{2} \approx 1.71$ is in the domain. ✓

Common mistakes

Reading $\sin^{-1} x$ as $\dfrac{1}{\sin x}$. The $-1$ means inverse. $\sin^{-1}(0.5) = \dfrac{\pi}{6} \approx 0.524$, but $\dfrac{1}{\sin 0.5} \approx 2.09$.

Giving an angle outside the range. $\arccos\left(-\dfrac{1}{2}\right)$ is $\dfrac{2\pi}{3}$, not $-\dfrac{\pi}{3}$ or $\dfrac{4\pi}{3}$. Each inverse function gives exactly one answer, and it must lie in that function’s range.

Assuming $\arcsin(\sin x) = x$ always. It’s only true for $-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}$. Evaluate the inside first, then take the inverse (Example 2).

Getting the sign wrong in compositions. In $\sin(\arccos x)$, the angle is between $0$ and $\pi$, where sine is never negative, so the answer is $+\sqrt{1 - x^2}$. In $\cos(\arcsin x)$, the angle is between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$, where cosine is never negative. Use the range to decide.

Forgetting the domain restriction on arcsin and arccos. $\arcsin 2$ and $\arccos(-1.5)$ don’t exist: no angle has a sine of $2$. Equations like $\arccos x = \dfrac{4\pi}{3}$ have no solution, because $\dfrac{4\pi}{3}$ is outside the range.

Calculator in degree mode. On AA papers, answers are expected in radians unless stated. $\arctan 1$ should be $\dfrac{\pi}{4} \approx 0.785$, not $45$.

Practice

1. (Warm-up) Find the exact value of each.

• (a) $\arcsin\dfrac{\sqrt{3}}{2}$
• (b) $\arccos\dfrac{1}{2}$
• (c) $\arctan(-1)$
• (d) $\arccos 0$

Solution

(a) $\dfrac{\pi}{3}$

(b) $\dfrac{\pi}{3}$

(c) $-\dfrac{\pi}{4}$ (arctan gives a negative angle for a negative input)

(d) $\dfrac{\pi}{2}$

2. (Warm-up) State the domain and range of $y = \arccos x$, and explain why $\arccos 2$ is undefined.

Solution

Domain $\{x \in \mathbb{R} \mid -1 \le x \le 1\}$, range $\{y \in \mathbb{R} \mid 0 \le y \le \pi\}$.

$\arccos 2$ would be an angle whose cosine is $2$, but cosine is always between $-1$ and $1$. So $2$ is not in the domain.

3. (Core) Find the exact value of each.

• (a) $\arccos\left(\cos\dfrac{4\pi}{3}\right)$
• (b) $\arcsin\left(\sin\dfrac{3\pi}{4}\right)$
• (c) $\tan(\arctan 7)$

Solution

(a) $\cos\dfrac{4\pi}{3} = -\dfrac{1}{2}$, and $\arccos\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3}$.

(b) $\sin\dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}$, and $\arcsin\dfrac{\sqrt{2}}{2} = \dfrac{\pi}{4}$.

(c) $7$. Tangent undoes arctan for every real input.

4. (Core) Find the exact value of (a) $\cos\left(\arcsin\dfrac{2}{3}\right)$ and (b) $\tan\left(\arccos\left(-\dfrac{1}{4}\right)\right)$.

Solution

(a) Let $\theta = \arcsin\dfrac{2}{3}$, so $\sin\theta = \dfrac{2}{3}$ and $-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$, where cosine is not negative:

$$\cos\theta = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$$

(b) Let $\theta = \arccos\left(-\dfrac{1}{4}\right)$, so $\cos\theta = -\dfrac{1}{4}$ and $\theta$ is in the second quadrant, where sine is positive:

$$\sin\theta = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4} \qquad \tan\theta = \frac{\sqrt{15}/4}{-1/4} = -\sqrt{15}$$

5. (Core) Show that $\sin(\arccos x) = \sqrt{1 - x^2}$ for $-1 \le x \le 1$. Explain why there is no $\pm$ sign.

Solution

Let $\theta = \arccos x$, so $\cos\theta = x$ and $0 \le \theta \le \pi$. From $\sin^2\theta + \cos^2\theta = 1$:

$$\sin^2\theta = 1 - x^2 \quad\Rightarrow\quad \sin\theta = \pm\sqrt{1 - x^2}$$

For $0 \le \theta \le \pi$, $\sin\theta \ge 0$, so we take the positive root: $\sin(\arccos x) = \sqrt{1 - x^2}$.

6. (Core) Let $g(x) = 3\arctan(2x)$.

• (a) State the range of $g$ and the equations of its horizontal asymptotes.
• (b) Find $g\left(\dfrac{1}{2}\right)$ exactly.
• (c) Solve $g(x) = \pi$.

Solution

(a) $\arctan(2x)$ takes every value strictly between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$, so $g$ has range $\{y \in \mathbb{R} \mid -\dfrac{3\pi}{2} \lt y \lt \dfrac{3\pi}{2}\}$, with horizontal asymptotes $y = \dfrac{3\pi}{2}$ and $y = -\dfrac{3\pi}{2}$.

(b) $g\left(\dfrac{1}{2}\right) = 3\arctan 1 = 3 \times \dfrac{\pi}{4} = \dfrac{3\pi}{4}$.

(c)

$$3\arctan(2x) = \pi \quad\Rightarrow\quad \arctan(2x) = \frac{\pi}{3} \quad\Rightarrow\quad 2x = \tan\frac{\pi}{3} = \sqrt{3} \quad\Rightarrow\quad x = \frac{\sqrt{3}}{2}$$

7. (Core) Solve each equation, or explain why there is no solution.

• (a) $\arcsin(2x - 1) = \dfrac{\pi}{6}$
• (b) $\arccos x = \dfrac{2\pi}{3}$
• (c) $\arccos x = \dfrac{4\pi}{3}$

Solution

(a) $2x - 1 = \sin\dfrac{\pi}{6} = \dfrac{1}{2}$, so $2x = \dfrac{3}{2}$ and $x = \dfrac{3}{4}$.

(b) $x = \cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$.

(c) No solution: the range of arccos is $0 \le y \le \pi$, and $\dfrac{4\pi}{3} \gt \pi$. (Taking the cosine of both sides would give $x = -\dfrac{1}{2}$, but $\arccos\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3}$, not $\dfrac{4\pi}{3}$.)

8. (Challenge) Prove that $\arcsin x + \arccos x = \dfrac{\pi}{2}$ for all $x$ with $-1 \le x \le 1$.

Solution

Let $\theta = \arcsin x$. Then $\sin\theta = x$ and $-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$.

Consider the angle $\dfrac{\pi}{2} - \theta$:

• $\cos\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta = x$, and
• from $-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$ we get $0 \le \dfrac{\pi}{2} - \theta \le \pi$, which is exactly the range of arccos.

So $\dfrac{\pi}{2} - \theta$ is the angle between $0$ and $\pi$ whose cosine is $x$, which means $\arccos x = \dfrac{\pi}{2} - \theta$. Therefore

$$\arcsin x + \arccos x = \theta + \left(\frac{\pi}{2} - \theta\right) = \frac{\pi}{2}$$

9. (Challenge) Show that $\arctan\dfrac{1}{2} + \arctan\dfrac{1}{3} = \dfrac{\pi}{4}$. (You’ll need the compound angle formula for $\tan(A + B)$.)

Solution

Let $A = \arctan\dfrac{1}{2}$ and $B = \arctan\dfrac{1}{3}$, so $\tan A = \dfrac{1}{2}$ and $\tan B = \dfrac{1}{3}$.

$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1$$

Both $\dfrac{1}{2}$ and $\dfrac{1}{3}$ are between $0$ and $1$, so $A$ and $B$ are each between $0$ and $\dfrac{\pi}{4}$, and $A + B$ is between $0$ and $\dfrac{\pi}{2}$. The only angle in that interval with tangent $1$ is $\dfrac{\pi}{4}$, so $A + B = \dfrac{\pi}{4}$.