Chapter 7.3, Problem 43E

To determine

To write: the given expression as an algebraic expression in $x$ .

Answer to Problem 43E

  $\sin \left(2{\tan }^{-1}x\right)=\frac{2x}{1+x^2}$

Explanation of Solution

Given information:

Given expression

  $\sin \left(2{\tan }^{-1}x\right)$

Calculation:

We need to express the given expression $\sin \left(2{\tan }^{-1}x\right)$ as an algebraic expression in $x$ .

Let $\theta ={\tan }^{-1}x$ , then , $\tan \theta =x$

Now, we have the double angle formula for sine, as

  $\sin 2\theta =2\sin \theta \cos \theta$

  

From the triangle, we get

  $\sin \theta =\frac{x}{\sqrt{1+x^2}}$

And

  $\cos \theta =\frac{1}{\sqrt{1+x^2}}$

Substituting the value of $\sin \theta$ and $\cos \theta$ in $\sin 2\theta$ , to get

  $\begin{array}{c}\sin 2\theta =2\times \frac{x}{\sqrt{1+x^2}}\times \frac{1}{\sqrt{1+x^2}}\\ =\frac{2x}{1+x^2}\end{array}$

Substituting $\theta ={\tan }^{-1}x$ , we get

  $\sin \left(2{\tan }^{-1}x\right)=\frac{2x}{1+x^2}$