Trigonometric functions are not one-to-one because their values repeat periodically and that the horizontal lines $y=c$ intersect the graphs in an infinite number of points, if at all, as we at once see from Figure 1 (recall the horizontal line test in Section One-to-One Functions). Therefore, they cannot have inverses unless we restrict their domains to intervals on which they are one-to-one.

 (a) Graph of $y=\sin x$ (b) Graph of $y=\tan x$
Figure 1: Trigonometric functions are not one-to-one as they do not pass the horizontal line test.

Table of Contents

Inverse of sine

If we look at the graph of $y=\sin x$ or if we consider the unit circle, we realize that the sine function on the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ is increasing from $-1$ to $1$. So by restricting its domain to this interval, we make it a one-to-one function whose domain is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and its range is $[-1,1]$. The inverse of the sine function, denoted by “$\sin^{-1}x$” or  “$\arcsin x$” , is a one-to-one function whose domain is $[-1,1]$ and its range is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. The graph of $y=\arcsin x$ is obtained by reflecting the graph of $y=\sin x$ (restricted to the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$) in the line $y=x$ (see Figure 2).

•  The two symbols “$\sin^{-1}x$” and “$\arcsin x$” are equivalent and can be used interchangeably. The first one is read “the inverse sine of $x$” and the second “the arc sine of $x$.”

• Again note that $\sin^{-1}x\neq\frac{1}{\sin x}$.

$y=\arcsin x$   means  $y$  is a number in the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ for which $\sin y=x$
•  The graph of $y=\arcsin x$ is symmetric about the origin, which shows $y=\arcsin x$ is an odd function. To prove it algebraically, we need to show $\arcsin(-x)=-\arcsin x$. Let
$y=\arcsin(-x).$ We know it means
\begin{align*}
\sin y & =-x\\
\Rightarrow-\sin y & =x\\
\Rightarrow\sin(-y) & =x & \text{the sine is an odd function}\\
\Rightarrow-y & =\arcsin x\\
\Rightarrow y & =-\arcsin x\\
\Rightarrow\arcsin(-x) & =-\arcsin x
\end{align*}

 Figure 2: The graph of $y=\arcsin x$ is obtained by reflecting the graph of $y=\sin x$ restricted to the interval $[-\pi/2,\pi/2]$ in the line $y=x$.

Inverse of cosine

The cosine and tangent functions can be inverted in a similar fashion. By considering the unit circle or looking at the graph of $y=\cos x$, we realize that $y=\cos x$ is not one-to-one on $[-\pi/2,\pi/2]$. So we had to choose a different interval for the cosine function. If we restrict the domain of the cosine function to the interval $[0,\pi]$, we can make it one-to-one, so that it has an inverse function denoted by $\cos^{-1}x$ or $\arccos x$. The graph of $y=\arccos x$ is shown in Figure 3.

• The domain of $y=\arccos x$ is $[-1,1]$ and its range is $[0,\pi]$.

$y=\arccos x$ means $y$ is a number in the interval $[0,\pi]$ for which $\cos y=x$

• The graph of $y=\arccos x$ is neither symmetric about the $y$-axis nor is symmetric about the origin, which means that $y=\arccos x$ is neither odd nor even.

 Figure 3: The graph of $y=\arccos x$ is obtained by reflecting the graph of $y=\cos x$ restricted to the interval $[0,\pi]$ in the line $y=x$.

Inverse of Tangent

For the tangent function, we choose the open interval $(-\frac{\pi}{2},\frac{\pi}{2})$ to perform the inversion. The resulting function is denoted by “$\tan^{-1}x$” or “$\arctan x$.”

• The domain of $y=\arctan x$ is $(-\infty,\infty)$ and its range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

$y=\arctan x$ means $y$  is a number in the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ for which $\tan y=x$.

The graph of $y=\arctan x$ is shown in Figure 4. This figure shows that the inverse tangent function is an odd function.

 Figure 4: The graph of $y=\arctan x$ is obtained by reflecting the graph of $y=\tan x$ restricted to the open interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ in the line $y=x$.
• Recall that if $f$ and $g$ are inverse functions of each other then
$f(g(x))=x,\qquad g(f(x))=x$ for every $x$ in the domain of the inside function, which are $g$ and $f$, respectively (Theorem 1 in Section on Inverse Functions). The following table summarizes some properties of the inverse trigonometric functions. Note that here we deal with the restricted domains  of the trigonometric functions; otherwise, their inverses do not exist.