# 2004 AMC 12A Problems/Problem 21

## Problem

If $\sum_{n = 0}^{\infty}{\cos^{2n}}\theta = 5$, what is the value of $\cos{2\theta}$? $\text {(A)} \frac15 \qquad \text {(B)} \frac25 \qquad \text {(C)} \frac {\sqrt5}{5}\qquad \text {(D)} \frac35 \qquad \text {(E)}\frac45$

## Solutions

### Solution 1

This is an infinite 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: $$\cos^{0}\theta=5-5*\cos^{2}\theta$$

Simplifying, we get $cos^{2}\theta=\frac{4}{5}$. Using the formula $\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}$.

## See also

 2004 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

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