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 == | + | == 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>. | ||
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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 , what is the value of ?
Solutions
Solution 1
This is an infinite geometric series, which sums to . Using the formula .
Solution 2
Multiply both sides by to get:
Subtracting the two equations, we get:
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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