Chapter 7.6, Problem 102AYU

To determine

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

Answer to Problem 102AYU

Solution:

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

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 right angle 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 $\cos \left(2\theta \right)=2{\cos }^2\left(\theta \right)-1$.

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

Substitute $\cos \left(\theta \right)=\frac{2}{\sqrt{x^2+4}}$ in $\cos \left(2\theta \right)=2{\cos }^2\left(\theta \right)-1$.

$\Rightarrow \cos \left(2\theta \right)=2{\left(\frac{2}{\sqrt{x^2+4}}\right)}^2-1$

$\Rightarrow \cos \left(2\theta \right)=2\cdot \frac{2^2}{{\left(\sqrt{x^2+4}\right)}^2}-1$

$\Rightarrow \cos \left(2\theta \right)=2\cdot \frac{4}{x^2+4}-1$.

$\Rightarrow \cos \left(2\theta \right)=\frac{8}{x^2+4}-1$

$\Rightarrow \cos \left(2\theta \right)=\frac{8}{x^2+4}-\frac{x^2+4}{x^2+4}$

$\Rightarrow \cos \left(2\theta \right)=\frac{8-\left(x^2+4\right)}{x^2+4}$

$\Rightarrow \cos \left(2\theta \right)=\frac{8-x^2-4}{x^2+4}$

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

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