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

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===Solution 1===
 
===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 1a====
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We can more directly solve this with superficially less work. Again, applying the formula for an infinite geometric series,
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<cmath>\sum_{i=0}^{\infty}\cos^{2i}\theta=\dfrac1{1-\cos^2\theta}=\dfrac1{\sin^2\theta}=5.</cmath>
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Thus, <math>\sin^2\theta=\dfrac15</math>, so <math>\cos(2\theta)=1-2\sin^2\theta=1-\dfrac25=\dfrac35.</math> QED.
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~Technodoggo
  
 
===Solution 2===
 
===Solution 2===
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<cmath>\cos^{0}\theta=5-5*\cos^{2}\theta</cmath>
 
<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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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>.
  
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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}}

Latest revision as of 00:18, 23 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 $\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}$.

See also

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

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