Chapter 7.2, Problem 18AYU

To determine

To calculate: The exact value of expression $\cos \left[{\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$ .

Answer to Problem 18AYU

The exact value is $\frac{1}{2}$ .

Explanation of Solution

Given information:

  $\cos \left[{\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$ .

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,

  $\cos \left[{\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$

Let

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

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

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

This implies $\theta$ lies in quadrant IV.

Now,

  $\sin \theta =\frac{y}{r}=\frac{-\sqrt{3}}{2}$

  $=y=-\sqrt{3};r=2$

Using equation of circle we have,

  $\begin{array}{l}=x^2+y^2=r^2\\ $ =x^2+{\left(-\sqrt{3}\right)}^2={(2)}^2$=x^2+3=4\\ =x^2=4-3\\ =x=\pm 1\end{array}$

Thus we have,

  $=x=1;y=-\sqrt{3};r=2$ .

Now,

  $\cos \left[{\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=\cos \theta =\frac{1}{2}$ ( where $\cos \frac{x}{r}$ )

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