Chapter 7, Problem 2T

To determine

To simplify: The expression $\frac{x}{\sqrt{4-x^2}}$ for $x=2\sin \theta$ and $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$.

Explanation of Solution

Formula used:

The trigonometric identity ${\sin }^{2}x+{\cos }^{2}x=1$.

Calculation:

Substitute $x=2\sin \theta$ in $\frac{x}{\sqrt{4-x^2}}$ and proceed as follows.

$\begin{array}{c}\frac{x}{\sqrt{4-x^2}}=\frac{{\left(2\sin \theta \right)}^2}{\sqrt{4-{\left(2\sin \theta \right)}^2}}\\ =\frac{4\sin \theta }{\sqrt{4\left(1-{\sin }^2\theta \right)}}\\ =2\frac{\sin \theta }{\cos \theta }\left[\because 1-{\sin }^2\theta ={\cos }^2\theta \right]\\ =2\tan \theta \end{array}$

Therefore, $\frac{x}{\sqrt{4-x^2}}=2\tan \theta$ for $x=2\sin \theta$ and $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$.