Chapter 7.2, Problem 74AYU

To determine

To calculate: The exact value of expression $f\left(h^{-1}\left(\frac{5}{12}\right)\right)$ .

Answer to Problem 74AYU

The exact value is $\frac{5}{13}$ .

Explanation of Solution

Given information:

  $f\left(h^{-1}\left(\frac{5}{12}\right)\right)$ where $h(x)=\tan x,-\frac{\pi }{2}<x<\frac{\pi }{2}$ ; $f(x)=\sin x,-\frac{\pi }{2}\le x\le \frac{\pi }{2}$

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 ,

  $f\left(h^{-1}\left(\frac{5}{12}\right)\right)=\sin \left({\tan }^{-1}\left(\frac{5}{12}\right)\right)$

Let

  $\begin{array}{l}\theta ={\tan }^{-1}\frac{5}{12}\\ \tan \theta =\frac{5}{12}\end{array}$

where $-\pi <\theta <\pi$ .

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

This implies $\theta$ lies in quadrant I.

Now,

  $\tan \theta =\frac{y}{x}=\frac{5}{12}$

  $=x=12;y=5$

Using equation of circle we have,

  $\begin{array}{l}=x^2+y^2=r^2\\ ={(}^{12}+{(}^5=r^2\\ =144+25=r^2\\ =r^2=169\\ =r=\pm 13\end{array}$

Thus we have,

  $=x=12;y=5;r=13$ .

Now,

  $\sin \left({\tan }^{-1}\left(\frac{5}{12}\right)\right)=\sin \theta =\frac{5}{13}$ ( where $\sin \frac{y}{r}$ )

Hence , the exact value of $\sin \left({\tan }^{-1}\left(\frac{5}{12}\right)\right)$ is $\frac{5}{13}$ .