1.4 Inverse functions  (Page 4/11)

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] ).

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

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

For the second one, we have

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

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

and

Similarly, for the cosine function,

and

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