2004 AMC 12A Problems/Problem 21: Difference between revisions
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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 == | == Solutions == | ||
===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>. | ||
===Solution 2=== | |||
<cmath>\sum_{n = 0}^{\infty}{\cos^{2n}}\theta = \cos^{0}\theta + \cos^{2}\theta + \cos^{4}\theta + ... = 5</cmath> | |||
Multiply both sides by <math>\cos^{2}\theta</math> to get: | |||
<cmath>\cos^{2}\theta + \cos^{4}\theta + \cos^{6}\theta + ... = 5*\cos^{2}\theta</cmath> | |||
Subtracting the two equations, we get: | |||
<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>. | |||
== 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 
, what is the value of 
?
Solutions
Solution 1
This is an infinite geometric series, which sums to 
. Using the formula 
.
Solution 2
![\[\sum_{n = 0}^{\infty}{\cos^{2n}}\theta = \cos^{0}\theta + \cos^{2}\theta + \cos^{4}\theta + ... = 5\]](http://latex.artofproblemsolving.com/1/2/e/12ec554fccc37c1deb6cf7079a0c91039098e697.png)
Multiply both sides by 
to get:
![\[\cos^{2}\theta + \cos^{4}\theta + \cos^{6}\theta + ... = 5*\cos^{2}\theta\]](http://latex.artofproblemsolving.com/9/4/9/9490d6446fce03624c048ad7195270634def7715.png)
Subtracting the two equations, we get:
![\[\cos^{0}\theta=5-5*\cos^{2}\theta\]](http://latex.artofproblemsolving.com/5/e/b/5eb6bf302469c6990405ea0ca6c34934a34185d9.png)
Simplifying, we get 
. Using the formula .
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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