If $\sum_{n = 0}^{\infty}{\cos^{2n}}\theta = 5$, what is the value of $\cos{2\theta}$?
$\text {(A)} \frac15 \qquad \text {(B)} \frac25 \qquad \text {(C)} \frac {\sqrt5}{5}\qquad \text {(D)} \frac35 \qquad \text {(E)}\frac45$
Solution 1
This is an infinite [geometric series](https://artofproblemsolving.com/wiki/index.php/Geometric_series), which sums to $\frac{\cos^0 \theta}{1 - \cos^2 \theta} = 5 \Longrightarrow 1 = 5 - 5\cos^2 \theta \Longrightarrow \cos^2 \theta = \frac{4}{5}$. Using the formula $\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}$.

Solution 2
\[\sum_{n = 0}^{\infty}{\cos^{2n}}\theta = \cos^{0}\theta + \cos^{2}\theta + \cos^{4}\theta + ... = 5\]
Multiply both sides by $\cos^{2}\theta$ to get:
\[\cos^{2}\theta + \cos^{4}\theta + \cos^{6}\theta + ... = 5*\cos^{2}\theta\]
Subtracting the two equations, we get:
\[\cos^{0}\theta=5-5*\cos^{2}\theta\]
After simplification, we get $cos^{2}\theta=\frac{4}{5}$. Using the formula $\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}$.