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

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<math>\text {(A)} \frac15 \qquad \text {(B)} \frac25 \qquad \text {(C)} \frac {\sqrt5}{5}\qquad \text {(D)} \frac35 \qquad \text {(E)}\frac45</math>
 
<math>\text {(A)} \frac15 \qquad \text {(B)} \frac25 \qquad \text {(C)} \frac {\sqrt5}{5}\qquad \text {(D)} \frac35 \qquad \text {(E)}\frac45</math>
  
== Solution ==
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== Solutions ==
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===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>.
 
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>.
  
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===Solution 2===
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<cmath>\sum_{n = 0}^{\infty}{\cos^{2n}}\theta = \cos^{0}\theta + \cos^{2}\theta + \cos^{4}\theta + ... = 5</cmath>
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Multiply both sides by <math>\cos^{2}\theta</math> to get:
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<cmath>\cos^{2}\theta + \cos^{4}\theta + \cos^{6}\theta + ... = 5*\cos^{2}\theta</cmath>
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Subtracting the two equations, we get:
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<cmath>\cos^{0}\theta=5-5*\cos^{2}\theta</cmath>
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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>.
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2004|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2004|ab=A|num-b=20|num-a=22}}

Revision as of 16:55, 12 July 2017

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 (ProblemsAnswer KeyResources)
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

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