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 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=== | ||
| + | <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> | ||
| + | |||
| + | 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>. | ||
== See also == | == See also == | ||
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,
![\[\sum_{i=0}^{\infty}\cos^{2i}\theta=\dfrac1{1-\cos^2\theta}=\dfrac1{\sin^2\theta}=5.\]](http://latex.artofproblemsolving.com/4/8/b/48b18f1eea93b46e4e819b30c7515c7981b41431.png)
Thus, 
, so 
QED.
~Technodoggo
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)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
