# 1.4 Inverse functions  (Page 4/11)

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## Definition

The inverse sine function, denoted ${\text{sin}}^{-1}$ or arcsin, and the inverse cosine function, denoted ${\text{cos}}^{-1}$ or arccos, are defined on the domain $D=\text{{}x|-1\le x\le 1\right\}$ as follows:

$\begin{array}{c}{\text{sin}}^{-1}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-\frac{\pi }{2}\le y\le \frac{\pi }{2};\hfill \\ {\text{cos}}^{-1}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{cos}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}0\le y\le \pi .\hfill \end{array}$

The inverse tangent function, denoted ${\text{tan}}^{-1}$ or arctan, and inverse cotangent function, denoted ${\text{cot}}^{-1}$ or arccot, are defined on the domain $D=\left\{x|-\infty as follows:

$\begin{array}{c}{\text{tan}}^{-1}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{tan}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-\frac{\pi }{2}

The inverse cosecant function, denoted ${\text{csc}}^{-1}$ or arccsc, and inverse secant function, denoted ${\text{sec}}^{-1}$ or arcsec, are defined on the domain $D=\left\{x||x|\ge 1\right\}$ as follows:

$\begin{array}{c}{\text{csc}}^{-1}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{csc}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-\frac{\pi }{2}\le y\le \frac{\pi }{2},y\ne 0;\hfill \\ {\text{sec}}^{-1}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{sec}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}0\le y\le \pi ,y\ne \pi \text{/}2.\hfill \end{array}$

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line $y=x$ ( [link] ). The graph of each of the inverse trigonometric functions is a reflection about the line y = x of the corresponding restricted trigonometric function.

Go to the following site for more comparisons of functions and their inverses.

When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate ${\text{cos}}^{-1}\left(\frac{1}{2}\right),$ we need to find an angle $\theta$ such that $\text{cos}\phantom{\rule{0.1em}{0ex}}\theta =\frac{1}{2}.$ Clearly, many angles have this property. However, given the definition of ${\text{cos}}^{-1},$ we need the angle $\theta$ that not only solves this equation, but also lies in the interval $\left[0,\pi \right].$ We conclude that ${\text{cos}}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}.$

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions $\text{sin}\left({\text{sin}}^{-1}\left(\frac{\sqrt{2}}{\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}2}\right)\right)$ and ${\text{sin}}^{-1}\left(\text{sin}\left(\pi \right)\right).$ For the first one, we simplify as follows:

$\text{sin}\left({\text{sin}}^{-1}\left(\frac{\sqrt{2}}{2}\right)\right)=\text{sin}\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}.$

For the second one, we have

${\text{sin}}^{-1}\left(\text{sin}\left(\pi \right)\right)={\text{sin}}^{-1}\left(0\right)=0.$

The inverse function is supposed to “undo” the original function, so why isn’t ${\text{sin}}^{-1}\left(\text{sin}\left(\pi \right)\right)=\pi ?$ Recalling our definition of inverse functions, a function $f$ and its inverse ${f}^{-1}$ satisfy the conditions $f\left({f}^{-1}\left(y\right)\right)=y$ for all $y$ in the domain of ${f}^{-1}$ and ${f}^{-1}\left(f\left(x\right)\right)=x$ for all $x$ in the domain of $f,$ so what happened here? The issue is that the inverse sine function, ${\text{sin}}^{-1},$ is the inverse of the restricted sine function defined on the domain $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\text{.}$ Therefore, for $x$ in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\text{,}$ it is true that ${\text{sin}}^{-1}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)=x.$ However, for values of $x$ outside this interval, the equation does not hold, even though ${\text{sin}}^{-1}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)$ is defined for all real numbers $x.$

What about $\text{sin}\left({\text{sin}}^{-1}y\right)?$ Does that have a similar issue? The answer is no . Since the domain of ${\text{sin}}^{-1}$ is the interval $\left[-1,1\right],$ we conclude that $\text{sin}\left({\text{sin}}^{-1}y\right)=y$ if $-1\le y\le 1$ and the expression is not defined for other values of $y.$ To summarize,

$\text{sin}\left({\text{sin}}^{-1}y\right)=y\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}-1\le y\le 1$

and

${\text{sin}}^{-1}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)=x\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}-\frac{\pi }{2}\le x\le \frac{\pi }{2}.$

Similarly, for the cosine function,

$\text{cos}\left({\text{cos}}^{-1}y\right)=y\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}-1\le y\le 1$

and

${\text{cos}}^{-1}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)=x\phantom{\rule{0.2em}{0ex}}\text{if}\phantom{\rule{0.2em}{0ex}}0\le x\le \pi .$

Similar properties hold for the other trigonometric functions and their inverses.

## Evaluating expressions involving inverse trigonometric functions

Evaluate each of the following expressions.

1. ${\text{sin}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
2. $\text{tan}\left({\text{tan}}^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right)$
3. ${\text{cos}}^{-1}\left(\text{cos}\left(\frac{5\pi }{4}\right)\right)$
4. ${\text{sin}}^{-1}\left(\text{cos}\left(\frac{2\pi }{3}\right)\right)$
1. Evaluating ${\text{sin}}^{-1}\left(\text{−}\sqrt{3}\text{/}2\right)$ is equivalent to finding the angle $\theta$ such that $\text{sin}\phantom{\rule{0.1em}{0ex}}\theta =\text{−}\sqrt{3}\text{/}2$ and $\text{−}\pi \text{/}2\le \theta \le \pi \text{/}2.$ The angle $\theta =\text{−}\pi \text{/}3$ satisfies these two conditions. Therefore, ${\text{sin}}^{-1}\left(\text{−}\sqrt{3}\text{/}2\right)=\text{−}\pi \text{/}3.$
2. First we use the fact that ${\text{tan}}^{-1}\left(-1\text{/}\sqrt{3}\right)=\text{−}\pi \text{/}6.$ Then $\text{tan}\left(\pi \text{/}6\right)=-1\text{/}\sqrt{3}.$ Therefore, $\text{tan}\left({\text{tan}}^{-1}\left(-1\text{/}\sqrt{3}\right)\right)=-1\text{/}\sqrt{3}.$
3. To evaluate ${\text{cos}}^{-1}\left(\text{cos}\left(5\pi \text{/}4\right)\right),$ first use the fact that $\text{cos}\left(5\pi \text{/}4\right)=\text{−}\sqrt{2}\text{/}2.$ Then we need to find the angle $\theta$ such that $\text{cos}\left(\theta \right)=\text{−}\sqrt{2}\text{/}2$ and $0\le \theta \le \pi .$ Since $3\pi \text{/}4$ satisfies both these conditions, we have $\text{cos}\left({\text{cos}}^{-1}\left(5\pi \text{/}4\right)\right)=\text{cos}\left({\text{cos}}^{-1}\left(\text{−}\sqrt{2}\text{/}2\right)\right)=3\pi \text{/}4.$
4. Since $\text{cos}\left(2\pi \text{/}3\right)=-1\text{/}2,$ we need to evaluate ${\text{sin}}^{-1}\left(-1\text{/}2\right).$ That is, we need to find the angle $\theta$ such that $\text{sin}\left(\theta \right)=-1\text{/}2$ and $\text{−}\pi \text{/}2\le \theta \le \pi \text{/}2.$ Since $\text{−}\pi \text{/}6$ satisfies both these conditions, we can conclude that ${\text{sin}}^{-1}\left(\text{cos}\left(2\pi \text{/}3\right)\right)={\text{sin}}^{-1}\left(-1\text{/}2\right)=\text{−}\pi \text{/}6.$

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