Chapter 13.2, Problem 58HP

To determine

To find : $f(x)$ in terms of a single trigonometric function of $\theta$ .

Answer to Problem 58HP

  $\pm \frac{1}{2}\sin \theta$

Explanation of Solution

Given information :

We know that, $\text{Let }x=\frac{1}{2}\tan \theta ,-\frac{\pi }{2}<\theta <\frac{\pi }{2}andf(x)=\frac{x}{\sqrt{1+4x^2}}$

Calculation :

  $\begin{array}{l}f(x)=\frac{x}{\sqrt{1+4x^2}}\\ f\left(\frac{1}{2}\tan \theta \right)=\frac{\frac{1}{2}\tan \theta }{\sqrt{1+4{\left(\frac{1}{2}\tan \theta \right)}^2}}=\frac{\frac{1}{2}\tan \theta }{\sqrt{1+4\cdot \frac{1}{4}{\tan }^2\theta }}\\ f\left(\frac{1}{2}\tan \theta \right)=\frac{\frac{1}{2}\tan \theta }{\sqrt{1+{\tan }^2\theta }}=\frac{1}{2}\frac{\tan \theta }{\sqrt{{\sec }^2\theta }}=\frac{1}{2}\frac{\tan \theta }{\sec \theta }\\ f\left(\frac{1}{2}\tan \theta \right)=\frac{1}{2}\frac{\sin \theta }{\cos \theta }\cdot \cos \theta =\frac{1}{2}\sin \theta \\ f\left(\frac{1}{2}\tan \theta \right)=\frac{1}{2}\sin \theta \\ \because -\frac{\pi }{2}<\theta <\frac{\pi }{2}\\ I.When\theta \in \left(-\frac{\pi }{2},0\right)\\ f\left(\frac{1}{2}\tan \theta \right)=-\frac{1}{2}\sin \theta [\because \text{ in IV quadrant sin }\theta \text{ is negative}]\\ II.When\theta \in \left(0,\frac{\pi }{2}\right)\\ f\left(\frac{1}{2}\tan \theta \right)=\frac{1}{2}\sin \theta \text{[}\because \text{In}\text{I quadrant sin }\theta \text{ is positive]}\end{array}$