Chapter 7.2, Problem 31AYU

To determine

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

Answer to Problem 31AYU

The exact value is $-\frac{3\sqrt{10}}{10}$ .

Explanation of Solution

Given information:

  $\sin \left[{\tan }^{-1}\left(-3\right)\right]$ .

Formula used:

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

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 ,

  $\sin \left[{\tan }^{-1}\left(-3\right)\right]$

Let

  $\begin{array}{l}\theta ={\tan }^{-1}\left(-3\right)\\ \tan \theta =\left(-3\right)\end{array}$

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

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

This implies $\theta$ lies in quadrant IV.

Now,

  $\tan \theta =\frac{y}{x}=\frac{-3}{1}$

  $=y=-3;x=1$

Using equation of circle we have,

  $\begin{array}{l}=x^2+y^2=r^2\\ ={(}^1+{\left(-3\right)}^2=r^2\\ =1+9=r^2\\ =10=r^2\\ =r=\pm \sqrt{10}\end{array}$

Thus we have,

  $=x=1;y=-3;r=\sqrt{10}$ .

Now,

  $\begin{array}{l}\sin \left[{\tan }^{-1}\left(-3\right)\right]=\sin \theta =\frac{-3}{\sqrt{10}}\\ \sin \theta =-\frac{3}{\sqrt{2}}\times \frac{\sqrt{10}}{\sqrt{10}}=-\frac{3\sqrt{10}}{10}\end{array}$ ( where $\sin \frac{y}{r}$ )

Hence , the exact value of $\sin \left[{\tan }^{-1}\left(-3\right)\right]$ is $-\frac{3\sqrt{10}}{10}$ .