Inverse trigonometric functions
If ${\sin x=\dfrac{1}{2}}$, what is ${x}$? Going backward, from the value of a function to its input, is the job of an inverse function. This page defines inverses for sine, cosine, and tangent.
You met these on a calculator in Solving right triangles. Inverse functions in general are explained in Inverse functions in College Algebra.
The problem: sine is not one-to-one
A function has an inverse only if it is one-to-one: each output comes from exactly one input. On a graph, every horizontal line must cross the graph at most once.
Sine fails this test badly. The value ${\dfrac{1}{2}}$ comes from ${\dfrac{\pi}{6}}$, ${\dfrac{5\pi}{6}}$, ${-\dfrac{7\pi}{6}}$, and infinitely many other inputs:
The fix is to use only part of the graph. On the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, sine rises steadily from ${-1}$ to ${1}$. It takes every value between ${-1}$ and ${1}$ exactly once. So this piece has an inverse.
The inverse sine function
${\sin^{-1}x=y}$ means ${\sin y=x}$ and ${-\dfrac{\pi}{2}\le y\le\dfrac{\pi}{2}}$
In words: ${\sin^{-1}x}$ is the angle between ${-\dfrac{\pi}{2}}$ and ${\dfrac{\pi}{2}}$ whose sine is ${x}$. It is also called the arcsine, written ${\arcsin x}$. Its domain is ${[-1,1]}$, because sine only takes values in that interval.
Watch out: ${\sin^{-1}x}$ does not mean ${\dfrac{1}{\sin x}}$. That is ${(\sin x)^{-1}}$, which is ${\csc x}$.
As with every inverse function, the graph of ${y=\sin^{-1}x}$ is the mirror image of the restricted sine graph across the line ${y=x}$:
The inverse cosine and inverse tangent
Cosine is restricted to a different interval, ${[0,\pi]}$, where it falls steadily from ${1}$ to ${-1}$. Tangent is restricted to $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, between two of its asymptotes.
${\cos^{-1}x=y}$ means ${\cos y=x}$ and ${0\le y\le\pi}$
${\tan^{-1}x=y}$ means ${\tan y=x}$ and ${-\dfrac{\pi}{2}<y<\dfrac{\pi}{2}}$
They are also called the arccosine and the arctangent. Here are their domains and ranges, and their graphs:
| Function | Domain | Range |
|---|---|---|
| ${\sin^{-1}x}$ | ${[-1,1]}$ | $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ |
| ${\cos^{-1}x}$ | ${[-1,1]}$ | ${[0,\pi]}$ |
| ${\tan^{-1}x}$ | all reals | $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ |
Since tangent takes every real value, ${\tan^{-1}x}$ is defined for every ${x}$. Its graph levels off toward the lines ${y=\pm\dfrac{\pi}{2}}$, which are horizontal asymptotes.
Exact values
To find an inverse value, ask: which angle in the allowed range has this sine (or cosine, or tangent)?
Example 1: Find the exact value.
(a) ${\sin^{-1}\dfrac{1}{2}}$ (b) $\sin^{-1}\left(-\dfrac{\sqrt{2}}{2}\right)$
(c) ${\cos^{-1}\left(-\dfrac{1}{2}\right)}$ (d) ${\tan^{-1}(-1)}$
Solution:
(a) The angle in $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ whose sine is ${\dfrac{1}{2}}$ is ${\dfrac{\pi}{6}}$. So
${\sin^{-1}\dfrac{1}{2}=\dfrac{\pi}{6}}$
(b) Sine is negative, so the angle is in ${\left[-\dfrac{\pi}{2},0\right)}$. Since ${\sin\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}}$, the answer is ${-\dfrac{\pi}{4}}$.
(c) The angle must be in ${[0,\pi]}$. Cosine is negative, so it is in quadrant II, with reference angle ${\dfrac{\pi}{3}}$. The answer is
${\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}}$
(d) The angle in $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ whose tangent is ${-1}$ is ${-\dfrac{\pi}{4}}$.
Example 2: Use a calculator to find ${\sin^{-1}(0.6)}$ in radians and in degrees.
Solution:
In radian mode, ${\sin^{-1}(0.6)\approx 0.6435}$. In degree mode, ${\sin^{-1}(0.6)\approx 36.87^\circ}$.
Cancellation properties
A function and its inverse undo each other, but only on the right intervals:
${\sin(\sin^{-1}x)=x}$ for ${-1\le x\le 1}$
${\sin^{-1}(\sin x)=x}$ for ${-\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}}$
Cosine and tangent work the same way, each with its own interval.
Example 3: Find the exact value.
(a) ${\sin\left(\sin^{-1}\dfrac{2}{3}\right)}$ (b) $\sin^{-1}\left(\sin\dfrac{3\pi}{4}\right)$
Solution:
(a) ${\dfrac{2}{3}}$ is between ${-1}$ and ${1}$, so the functions cancel. The answer is ${\dfrac{2}{3}}$.
(b) Careful: ${\dfrac{3\pi}{4}}$ is not in $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, so they do not cancel. Work from the inside out:
${\sin\dfrac{3\pi}{4}=\dfrac{\sqrt{2}}{2}}$
Then
$\sin^{-1}\dfrac{\sqrt{2}}{2}=\dfrac{\pi}{4}$
The answer is ${\dfrac{\pi}{4}}$, not ${\dfrac{3\pi}{4}}$.
Compositions with a triangle
To find something like ${\cos\left(\sin^{-1}\dfrac{3}{5}\right)}$, give the inner angle a name and draw a right triangle.
Example 4: Find the exact value of ${\cos\left(\sin^{-1}\dfrac{3}{5}\right)}$.
Solution:
Let ${\theta=\sin^{-1}\dfrac{3}{5}}$. Then ${\sin\theta=\dfrac{3}{5}}$, and ${\theta}$ is in quadrant I because ${\dfrac{3}{5}>0}$. Draw a right triangle with opposite side ${3}$ and hypotenuse ${5}$. The third side is ${\sqrt{25-9}=4}$:
So $\cos\left(\sin^{-1}\dfrac{3}{5}\right)=\cos\theta=\dfrac{4}{5}$.
Example 5: Write ${\tan(\cos^{-1}x)}$ as an algebraic expression in ${x}$, for ${0<x\le 1}$.
Solution:
Let ${\theta=\cos^{-1}x}$, so ${\cos\theta=x=\dfrac{x}{1}}$. Draw a right triangle with adjacent side ${x}$ and hypotenuse ${1}$. The opposite side is ${\sqrt{1-x^2}}$. So
$\begin{align*}\tan(\cos^{-1}x)&=\dfrac{\text{opp}}{\text{adj}}\\&=\dfrac{\sqrt{1-x^2}}{x}\end{align*}$
Summary
- ${\sin^{-1}x}$ is the angle in $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ whose sine is ${x}$. Its domain is ${[-1,1]}$.
- ${\cos^{-1}x}$ is the angle in ${[0,\pi]}$ whose cosine is ${x}$.
- ${\tan^{-1}x}$ is the angle in $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ whose tangent is ${x}$. It is defined for all ${x}$.
- ${\sin(\sin^{-1}x)=x}$ always, but ${\sin^{-1}(\sin x)=x}$ only when ${x}$ is in $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.
- For compositions such as ${\cos(\sin^{-1}x)}$, name the angle and draw a right triangle.