Chapter 7.6, Problem 101AYU

To determine

The expression for $\sin \left(2\theta \right)$ as a function of $x$ if $x=2\tan \left(\theta \right)$.

Answer to Problem 101AYU

Solution:

The expression for $\sin \left(2\theta \right)$ as a function of $x$ is $\underset{\_}{\sin \left(2\theta \right)=\frac{4x}{x^2+4}}$.

Explanation of Solution

Given information:

$x=2\tan \left(\theta \right)$

Divide both sides by $2$ in an equation $x=2\tan \left(\theta \right)$.

$\Rightarrow \tan \left(\theta \right)=\frac{x}{2}=\frac{\text{opposite side}}{\text{adjacent side}}$

The rightangled triangle using this information is

Using Pythagoras Theorem, $h^2=x^2+2^2$, where $h$ is the hypotenuse of right triangle.

$\Rightarrow h^2=x^2+4$

Take square root on both sides.

$\Rightarrow \sqrt{h^2}=\sqrt{x^2+4}$

$\Rightarrow h=\sqrt{x^2+4}$

The right triangle is

The double angle identity is $\sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)$.

From above right triangle $\sin \left(\theta \right)=\frac{x}{\sqrt{x^2+4}}$ and $\cos \left(\theta \right)=\frac{2}{\sqrt{x^2+4}}$.

Substitute $\sin \left(\theta \right)=\frac{x}{\sqrt{x^2+4}}$ and $\cos \left(\theta \right)=\frac{2}{\sqrt{x^2+4}}$ in $\sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)$.

$\Rightarrow \sin \left(2\theta \right)=2\frac{x}{\sqrt{x^2+4}}\frac{2}{\sqrt{x^2+4}}$

$\Rightarrow \sin \left(2\theta \right)=\frac{4x}{x^2+4}$

Therefore, the expression for $\sin \left(2\theta \right)$ as a function of $x$ is $\sin \left(2\theta \right)=\frac{4x}{x^2+4}$.