Question

If $x=2\tan \theta$, express $\sin \left(2\theta \right)$as a function ofx.

Short answer

On expressing $\sin \left(2\theta \right)$ as a function of xis $\sin \left(2\theta \right)=\frac{4x}{4+x^2}$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$x=2\tan \theta$

From the double-angle formula,

$\sin \left(2\theta \right)=2\sin \theta \cos \theta$

From inverse trigonometry,

$$\begin{gathered} \theta ={\tan }^{-1}\frac{x}{2} \\ ={\sin }^{-1}\left(\frac{x}{\sqrt{4+x^2}}\right) \\ ={\cos }^{-1}\left(\frac{2}{\sqrt{4+x^2}}\right) \end{gathered}$$

Step 2. Use inverse trigonometry and solve the equation.

Consider the given question,

$x=2\tan \theta$

From inverse trigonometry,

$\theta ={\tan }^{-1}\frac{x}{2}$

Then,

$$\begin{gathered} \theta ={\tan }^{-1}\frac{x}{2} \\ \sin \left(2\theta \right)=\sin \left(2{\tan }^{-1}\frac{x}{2}\right) \\ \sin \left(2\theta \right)=2\sin \left({\tan }^{-1}\frac{x}{2}\right)\cos \left({\tan }^{-1}\frac{x}{2}\right) \\ \sin \left(2\theta \right)=2\sin \left({\sin }^{-1}\left(\frac{x}{\sqrt{4+x^2}}\right)\right)\cos \left(\left({\cos }^{-1}\left(\frac{2}{\sqrt{4+x^2}}\right)\right)\right) \\ \sin \left(2\theta \right)=\frac{4x}{4+x^2} \end{gathered}$$