# 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] ).

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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