# 5.6 Inverse trigonometric functions  (Page 6/15)

 Page 6 / 15

Find a simplified expression for $\text{\hspace{0.17em}}\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(4x\right)\right)\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}-\frac{1}{4}\le x\le \frac{1}{4}.$

$\frac{4x}{\sqrt{16{x}^{2}+1}}$

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## Key concepts

• An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.
• Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.
• For any trigonometric function $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x={f}^{-1}\left(y\right),\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}f\left(x\right)=y.\text{\hspace{0.17em}}$ However, $\text{\hspace{0.17em}}f\left(x\right)=y\text{\hspace{0.17em}}$ only implies $\text{\hspace{0.17em}}x={f}^{-1}\left(y\right)\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in the restricted domain of $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ See [link] .
• Special angles are the outputs of inverse trigonometric functions for special input values; for example, $\text{\hspace{0.17em}}\frac{\pi }{4}={\mathrm{tan}}^{-1}\left(1\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{\pi }{6}={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right).$ See [link] .
• A calculator will return an angle within the restricted domain of the original trigonometric function. See [link] .
• Inverse functions allow us to find an angle when given two sides of a right triangle. See [link] .
• In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, $\text{\hspace{0.17em}}\mathrm{sin}\left({\mathrm{cos}}^{-1}\left(x\right)\right)=\sqrt{1-{x}^{2}}.\text{\hspace{0.17em}}$ See [link] .
• If the inside function is a trigonometric function, then the only possible combinations are $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-x\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}0\le x\le \pi \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-x\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}-\frac{\pi }{2}\le x\le \frac{\pi }{2}.$ See [link] and [link] .
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function. See [link] .
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides. See [link] .

## Verbal

Why do the functions $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{sin}}^{-1}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{cos}}^{-1}x\text{\hspace{0.17em}}$ have different ranges?

The function $\text{\hspace{0.17em}}y=\mathrm{sin}x\text{\hspace{0.17em}}$ is one-to-one on $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right];\text{\hspace{0.17em}}$ thus, this interval is the range of the inverse function of $\text{\hspace{0.17em}}y=\mathrm{sin}x,$ $f\left(x\right)={\mathrm{sin}}^{-1}x.\text{\hspace{0.17em}}$ The function $\text{\hspace{0.17em}}y=\mathrm{cos}x\text{\hspace{0.17em}}$ is one-to-one on $\text{\hspace{0.17em}}\left[0,\pi \right];\text{\hspace{0.17em}}$ thus, this interval is the range of the inverse function of $\text{\hspace{0.17em}}y=\mathrm{cos}x,f\left(x\right)={\mathrm{cos}}^{-1}x.\text{\hspace{0.17em}}$

Since the functions $\text{\hspace{0.17em}}y=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y={\mathrm{cos}}^{-1}x\text{\hspace{0.17em}}$ are inverse functions, why is $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(-\frac{\pi }{6}\right)\right)\text{\hspace{0.17em}}$ not equal to $\text{\hspace{0.17em}}-\frac{\pi }{6}?$

Explain the meaning of $\text{\hspace{0.17em}}\frac{\pi }{6}=\mathrm{arcsin}\left(0.5\right).$

$\frac{\pi }{6}\text{\hspace{0.17em}}$ is the radian measure of an angle between $\text{\hspace{0.17em}}-\frac{\pi }{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\frac{\pi }{2}$ whose sine is 0.5.

Most calculators do not have a key to evaluate $\text{\hspace{0.17em}}{\mathrm{sec}}^{-1}\left(2\right).\text{\hspace{0.17em}}$ Explain how this can be done using the cosine function or the inverse cosine function.

Why must the domain of the sine function, $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ be restricted to $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\text{\hspace{0.17em}}$ for the inverse sine function to exist?

In order for any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\text{\hspace{0.17em}}$ so that it is one-to-one and possesses an inverse.

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