Chapter 7.2, Problem 70AYU

To determine

To calculate: The exact value of expression $f^{-1}\left[g\left(\frac{5\pi }{6}\right)\right]$ .

Answer to Problem 70AYU

The exact value is $\frac{2\pi }{3}$ .

Explanation of Solution

Given information:

  $f^{-1}\left[g\left(\frac{5\pi }{6}\right)\right]$ where $g(x)=\cos x,0\le x\le \pi$ ; $f(x)=\sin x,-\frac{\pi }{2}\le x\le \frac{\pi }{2}$

Formula used:

Trigonometric Ratios :

  $\cos \left(\pi +\theta \right)=-\cos \theta$ .

Calculation:

Consider ,

  $f^{-1}\left(g\left(\frac{5\pi }{6}\right)\right)={\sin }^{-1}\left[\cos \left(\frac{5\pi }{6}\right)\right]$

Let

  $\theta =\cos \left(\frac{5\pi }{6}\right)$

where $-\pi <\theta <\pi$ .

It follows $\frac{\pi }{2}<\theta \le \pi$ .

This implies $\theta$ lies in quadrant II.

Now,

  $\begin{array}{l}\cos \left(\frac{5\pi }{6}\right)=\cos \left(\pi -\frac{\pi }{6}\right)=-\cos \frac{\pi }{6}\\ =-\cos \frac{\pi }{6}=-\left(\frac{\sqrt{3}}{2}\right)\end{array}$

Where ${\sin }^{-1}\theta$ lies between $\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

And the negative value of ${\sin }^{-1}\theta$ is between zero and $\frac{\pi }{2}$ .

  $\begin{array}{l}\sin \theta =-\frac{\sqrt{3}}{2}\\ \sin \theta =-\frac{\pi }{3}\end{array}$

  $\begin{array}{l}={\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\theta \\ =\sin \theta =-\frac{\sqrt{3}}{2}\end{array}$

  $=\left(\pi -\frac{\pi }{3}\right)=\frac{2\pi }{3}$

Hence , the exact value of ${\sin }^{-1}\left[\cos \left(\frac{5\pi }{6}\right)\right]$ is $\frac{2\pi }{3}$ .