Chapter 7.2, Problem 9AYU

To determine

To calculate: The exact value of expression $\cos \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)$ .

Answer to Problem 9AYU

The exact value is $\frac{\sqrt{2}}{2}$ .

Explanation of Solution

Given information:

  $\cos \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)$ .

Formula used:

Equation $y={\sin }^{-1}x$ means $\sin y=x$ .

Theorem :- For any angle $\theta$ , in standard position, let $P=(x,y)$ be the point on the terminal side of $\theta$ , that is on the circle $x^2+y^2=r^2$ . Then

  $\begin{array}{l}\sin \theta =\frac{y}{r}\\ \cos \theta =\frac{x}{r}\\ \tan \theta =\frac{y}{x},x\ne 0\end{array}$  $\begin{array}{l}\cos ec\theta =\frac{r}{y},y\ne 0\\ \sec \theta =\frac{r}{x},x\ne 0\\ \cot \theta =\frac{x}{y},y\ne 0\end{array}$ .

Calculation:

Consider,

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

Let

  $\begin{array}{l}\theta ={\sin }^{-1}\frac{\sqrt{2}}{2}\\ \sin \theta =\frac{\sqrt{2}}{2}\end{array}$

where $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$ .

Since $\sin \theta >0$ , it follows $0<\theta <\frac{\pi }{2}$ .

This implies $\theta$ lies in quadrant I.

Now,

  $\sin \theta =\frac{y}{r}=\frac{\sqrt{2}}{2}$

  $=y=\sqrt{2};r=2$ .

Using equation of circle we have,

  $\begin{array}{l}=x^2+y^2=r^2\\ =x^2+{\left(\sqrt{2}\right)}^2={\left(2\right)}^2\\ =x^2+2=4\\ =x^2=4-2\\ =x^2=2\\ =x=\pm \sqrt{2}\end{array}$

Thus we have,

  $=x=\sqrt{2};y=\sqrt{2};r=2$ .

Now,

  $\cos \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)=\cos \theta =\frac{\sqrt{2}}{2}$ ( where $\cos \frac{x}{r}$ )

Hence , the exact value of $\cos \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)$ is $\frac{\sqrt{2}}{2}$ .