Write $\sin[2\tan^{-1} (x)]$ as an algebraic expression.

Question

I need to write $\sin[2\tan^{-1} (x)]$ as an algebraic expression. Unfortunately, I have no idea what my textbook even really means by "algebraic expression".

The stated answer is $\frac{2x}{1 + x^2}$

What are the steps to solve this sort of problem?

Answer 1

Let $\arctan x = \theta$. Then $\theta$ is the unique angle in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ such that $\tan\theta = x$. Use the identity $$\sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta}$$ to obtain the result.

Answer 2

I think it always helps to draw a picture. Let $\theta = \arctan x$.

Then $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$.

Note that the sign of $x$ agrees with the sign of $\theta$, which agrees with the sign of $\sin \theta$.

Note that $\cos \theta > 0$ for all $x$.

So

\begin{align} \sin(2 \arctan x) &= \sin(2 \theta) \\ &= 2 \sin(\theta) \cos(\theta) \\ &= \dfrac{2x}{1+x^2} \end{align}