## Section2.12Inverse Trigonometric Functions

One very useful application of implicit differentiation is to find the derivatives of inverse functions. We have already used this approach to find the derivative of the inverse of the exponential function — the logarithm.

We are now going to consider the problem of finding the derivatives of the inverses of trigonometric functions. Now is a very good time to go back and reread Section 0.6 on inverse functions — especially Definition 0.6.4. Most importantly, given a function $f(x)\text{,}$ its inverse function $f^{-1}(x)$ only exists, with domain $D\text{,}$ when $f(x)$ passes the “horizontal line test”, which says that for each $Y$ in $D$ the horizontal line $y=Y$ intersects the graph $y=f(x)$ exactly once. (That is, $f(x)$ is a one-to-one function.)

Let us start by playing with the sine function and determine how to restrict the domain of $\sin x$ so that its inverse function exists.

Let $y=f(x)=\sin(x)\text{.}$ We would like to find the inverse function which takes $y$ and returns to us a unique $x$-value so that $\sin(x)=y\text{.}$

• For each real number $Y\text{,}$ the number of $x$-values that obey $\sin(x)=Y\text{,}$ is exactly the number of times the horizontal straight line $y=Y$ intersects the graph of $\sin(x)\text{.}$
• When $-1\le Y\le 1\text{,}$ the horizontal line intersects the graph infinitely many times. This is illustrated in the figure above by the line $y=0.3\text{.}$
• On the other hand, when $Y \lt -1$ or $Y \gt 1\text{,}$ the line $y=Y$ never intersects the graph of $\sin(x)\text{.}$ This is illustrated in the figure above by the line $y=-1.2\text{.}$

This is exactly the horizontal line test and it shows that the sine function is not one-to-one.

Now consider the function

\begin{align*} y &= \sin(x) & \text{with domain } -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \end{align*}

This function has the same formula but the domain has been restricted so that, as we'll now show, the horizontal line test is satisfied.

As we saw above when $|Y| \gt 1$ no $x$ obeys $\sin(x)=Y$ and, for each $-1\le Y\le 1\text{,}$ the line $y=Y$ (illustrated in the figure above with $y=0.3$) crosses the curve $y=\sin(x)$ infinitely many times, so that there are infinitely many $x$'s that obey $f(x)=\sin x=Y\text{.}$ However exactly one of those crossings (the dot in the figure) has $-\frac{\pi}{2}\le x \le\frac{\pi}{2}\text{.}$

That is, for each $-1\le Y \le 1\text{,}$ there is exactly one $x\text{,}$ call it $X\text{,}$ that obeys both

\begin{align*} \sin X &= Y &\text{and} && -\frac{\pi}{2}\le X \le \frac{\pi}{2} \end{align*}

That unique value, $X\text{,}$ is typically denoted $\arcsin(Y)\text{.}$ That is

\begin{align*} \sin( \arcsin(Y) ) &= Y & \text{and} && -\frac{\pi}{2}\le \arcsin(Y) \le\frac{\pi}{2} \end{align*}

Renaming $Y\rightarrow x\text{,}$ the inverse function $\arcsin(x)$ is defined for all $-1 \le x \le 1$ and is determined by the equation

Note that many texts will use $\sin^{-1}(x)$ to denote arcsine, however we will use $\arcsin(x)$ since we feel that it is clearer  1 The main reason being that people frequently confuse $\sin^{-1} (x)$ with $(\sin(x))^{-1} = \frac{1}{\sin x}\text{.}$ We feel that prepending the prefix “arc” less likely to lead to such confusion. The notations $\textrm{asin}(x)$ and $\textrm{Arcsin}(x)$ are also used.; the reader should recognise both.

Since

and $-\frac{\pi}{2}\le \frac{\pi}{6},\frac{\pi}{2}\le \frac{\pi}{2}\text{,}$ we have

\begin{equation*} \arcsin 1= \frac{\pi}{2}\qquad \arcsin \frac{1}{2}= \frac{\pi}{6} \end{equation*}

Even though

\begin{equation*} \sin(2\pi)=0 \end{equation*}

it is not true that $\arcsin 0 =2\pi\text{,}$ and it is not true that $\arcsin\big(\sin(2\pi)\big) =2\pi\text{,}$ because $2\pi$ is not between $-\frac{\pi}{2}$ and $\frac{\pi}{2}\text{.}$ More generally

\begin{align*} \arcsin\big(\sin(x)\big) &=\text{ the unique angle } \theta \text{ between } -\frac{\pi}{2} \text{ and } \frac{\pi}{2} \text{ obeying } \sin \theta =\sin x\\ &= x\quad\text{if and only if $-\frac{\pi}{2}\le x\le \frac{\pi}{2}$} \end{align*}

So, for example, $\arcsin\big(\sin\big(\frac{11\pi}{16}\big)\big)$ cannot be $\frac{11\pi}{16}$ because $\frac{11\pi}{16}$ is bigger than $\frac{\pi}{2}\text{.}$ So how do we find the correct answer? Start by sketching the graph of $\sin(x)\text{.}$

It looks like the graph of $\sin x$ is symmetric about $x=\frac{\pi}{2}\text{.}$ The mathematical way to say that “the graph of $\sin x$ is symmetric about $x=\frac{\pi}{2}$” is “$\sin(\frac{\pi}{2}-\theta)= \sin(\frac{\pi}{2}+\theta)$” for all $\theta\text{.}$ That is indeed true  2 Indeed both are equal to $\cos \theta\text{.}$ You can see this by playing with the trig identities in Appendix A.8. .

Now $\frac{11\pi}{16}=\frac{\pi}{2} +\frac{3\pi}{16}$ so

\begin{equation*} \sin\Big(\frac{11\pi}{16}\Big) =\sin\Big(\frac{\pi}{2}+\frac{3\pi}{16}\Big) =\sin\Big(\frac{\pi}{2}-\frac{3\pi}{16}\Big) =\sin\Big(\frac{5\pi}{16}\Big) \end{equation*}

and, since $\frac{5\pi}{16}$ is indeed between $-\frac{\pi}{2}$ and $\frac{\pi}{2}\text{,}$

\begin{equation*} \arcsin\Big(\sin\Big(\frac{11\pi}{16}\Big)\Big) =\frac{5\pi}{16}\qquad\Big(\text{and not $\frac{11\pi}{16}$}\Big). \end{equation*}