2.3 Inverse trigonometric functions  (Page 5/15)

1. Here, we can directly evaluate the inside of the composition.
   $$\begin{array}{l}\hfill \\ \begin{array}{l}\cos (\frac{13\pi }{6})=\cos (\frac{\pi }{6}+2\pi )\hfill \\ =\cos (\frac{\pi }{6})\hfill \\ =\frac{\sqrt{3}}{2}\hfill \end{array}\hfill \end{array}$$

   Now, we can evaluate the inverse function as we did earlier.

   ${\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{3}$
2. We have $x=\frac{13\pi }{6}\text{,}y=\frac{\pi }{6},$ and
   $$\begin{array}{r}\hfill {\sin }^{-1}\left(\cos \left(\frac{13\pi }{6}\right)\right)=\frac{\pi }{2}-\frac{\pi }{6}\\ \hfill =\frac{\pi }{3}\end{array}$$

Beginning with the inside, we can say there is some angle such that $\theta ={\cos }^{-1}\left(\frac{4}{5}\right),$ which means $\cos \theta =\frac{4}{5},$ and we are looking for $\sin \theta .$ We can use the Pythagorean identity to do this.

Since $\theta ={\cos }^{-1}\left(\frac{4}{5}\right)$ is in quadrant I, $\sin \theta$ must be positive, so the solution is $\frac{3}{5}.$ See [link] .

We know that the inverse cosine always gives an angle on the interval $\left[0,\pi \right],$ so we know that the sine of that angle must be positive; therefore $\sin \left({\cos }^{-1}\left(\frac{4}{5}\right)\right)=\sin \theta =\frac{3}{5}.$

While we could use a similar technique as in [link] , we will demonstrate a different technique here. From the inside, we know there is an angle such that $\tan \theta =\frac{7}{4}.$ We can envision this as the opposite and adjacent sides on a right triangle, as shown in [link] .

Using the Pythagorean Theorem, we can find the hypotenuse of this triangle.

Now, we can evaluate the sine of the angle as the opposite side divided by the hypotenuse.

This gives us our desired composition.

We know there is an angle $\theta$ such that $\sin \theta =\frac{x}{3}.$

Because we know that the inverse sine must give an angle on the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right],$ we can deduce that the cosine of that angle must be positive.