Difference between revisions of "2004 AMC 12A Problems/Problem 21"

Latest revision as of 23:18, 22 October 2024

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 1a

We can more directly solve this with superficially less work. Again, applying the formula for an infinite geometric series,

$\sum_{i=0}^{\infty}\cos^{2i}\theta=\dfrac1{1-\cos^2\theta}=\dfrac1{\sin^2\theta}=5.$

Thus, $\sin^2\theta=\dfrac15$ , so $\cos(2\theta)=1-2\sin^2\theta=1-\dfrac25=\dfrac35.$ QED.

~Technodoggo

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 $\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}$ .

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 7: / Line 7: @@
 ===Solution 1===
 This is an infinite [[geometric series]], which sums to <math>\frac{\cos^0 \theta}{1 - \cos^2 \theta} = 5 \Longrightarrow 1 = 5 - 5\cos^2 \theta \Longrightarrow \cos^2 \theta = \frac{4}{5}</math>. Using the formula <math>\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}</math>.
+====Solution 1a====
+We can more directly solve this with superficially less work. Again, applying the formula for an infinite geometric series,
+<cmath>\sum_{i=0}^{\infty}\cos^{2i}\theta=\dfrac1{1-\cos^2\theta}=\dfrac1{\sin^2\theta}=5.</cmath>
+Thus, <math>\sin^2\theta=\dfrac15</math>, so <math>\cos(2\theta)=1-2\sin^2\theta=1-\dfrac25=\dfrac35.</math> QED.
+~Technodoggo
 ===Solution 2===
@@ Line 19: / Line 29: @@
 <cmath>\cos^{0}\theta=5-5*\cos^{2}\theta</cmath>
-Simplifying, we get <math>cos^{2}\theta=\frac{4}{5}</math>. Using the formula <math>\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}</math>.
+After simplification, we get <math>cos^{2}\theta=\frac{4}{5}</math>. Using the formula <math>\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}</math>.
 == See also ==
 {{AMC12 box|year=2004|ab=A|num-b=20|num-a=22}}

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