Chapter 7.2, Problem 12AYU

To determine

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

Answer to Problem 12AYU

The exact value is $-\frac{1}{\sqrt{3}}$ .

Explanation of Solution

Given information:

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

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 ,

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

Let

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

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,

  $\sin \theta =\frac{y}{r}=-\frac{1}{2}$

  $=y=-1;r=2$

Using equation of circle we have,

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

Thus we have,

  $=x=\sqrt{3};y=-1;r=2$ .

Now,

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

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