Chapter 12.9, Problem 23PPS

To determine

To find the value of the given expression round to the nearest hundredth if necessary.

Answer to Problem 23PPS

  $\frac{1}{\sqrt{2}}$

Explanation of Solution

Given: $\sin \left[{\cos }^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right]$

Calculation: Let

  $\begin{array}{l}{\cos }^{-1}\left(-\frac{\sqrt{2}}{2}\right)=x\\ \Rightarrow -\frac{\sqrt{2}}{2}=\cos x\\ \Rightarrow \cos x=-\frac{\sqrt{2}}{2}\\ \Rightarrow \cos x=-\frac{1}{\sqrt{2}}\\ \Rightarrow x=\frac{3\pi }{4}\\ \sin \left[{\cos }^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right]\\ =\sin x\\ =\sin \frac{3\pi }{4}\\ =\frac{1}{\sqrt{2}}\end{array}$

Thus, obtained value is equal to the given expression.