2004 AMC 12A Problems/Problem 21: Difference between revisions
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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>. | ||
====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=== | ===Solution 2=== | ||
Latest revision as of 23:18, 22 October 2024
Problem
If
, what is the value of
?
Solutions
Solution 1
This is an infinite geometric series, which sums to
. Using the formula
.
Solution 1a
We can more directly solve this with superficially less work. Again, applying the formula for an infinite geometric series,
Thus,
, so
QED.
~Technodoggo
Solution 2
Multiply both sides by
to get:
Subtracting the two equations, we get:
After simplification, 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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