Difference between revisions of "2011 AIME I Problems/Problem 9"
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<cmath>24\sin ^3 x=1-\sin ^2 x</cmath> | <cmath>24\sin ^3 x=1-\sin ^2 x</cmath> | ||
<cmath>24\sin ^3 x+\sin ^2 x - 1=0</cmath> | <cmath>24\sin ^3 x+\sin ^2 x - 1=0</cmath> | ||
| − | Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root, <math>\sin^{-1} \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. | + | Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root, <math>\sin^{-1} \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. |
| + | There are now two ways to finish this problem. | ||
| + | |||
| + | '''First way:''' Since <math>\sin x=\frac{1}{3}</math>, we have | ||
<cmath>\sin ^2 x=\frac{1}{9}</cmath> | <cmath>\sin ^2 x=\frac{1}{9}</cmath> | ||
Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}</math>. | Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}</math>. | ||
| + | |||
| + | '''Second way:''' Multiplying our old equation <math>24\sin ^3 x=\cos ^2 x</math> by <math>\dfrac{24}{\sin^2x}</math> gives | ||
| + | <cmath>576\sin x = 24\cot^2x</cmath> | ||
| + | So, <math>24\cot^2x=576\sin x=576\cdot\frac{1}{3}=\boxed{192}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=8|num-a=10}} | {{AIME box|year=2011|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:00, 28 February 2014
Problem
Suppose 
is in the interval ![$[0, \pi/2]$](http://latex.artofproblemsolving.com/3/7/2/37278732e6da75cdf01fab8f86657b7b3feb2006.png)
and 
. Find 
.
Solution
We can rewrite the given expression as
![\[\sqrt{24^3\sin^3 x}=24\cos x\]](http://latex.artofproblemsolving.com/2/9/a/29a6f2faf8ae9cf593ff0be066b49f980b0cee19.png)
Square both sides and divide by 
to get
![\[24\sin ^3 x=\cos ^2 x\]](http://latex.artofproblemsolving.com/e/7/5/e753337da12027e288d4c5085b196513d6ecfeed.png)
Rewrite 
as 
![\[24\sin ^3 x=1-\sin ^2 x\]](http://latex.artofproblemsolving.com/c/c/7/cc70fbcab3789aece0c0b5b5aed945d818c976e0.png)
![\[24\sin ^3 x+\sin ^2 x - 1=0\]](http://latex.artofproblemsolving.com/7/b/3/7b34ecf4cf49979719e6b2ef5c87fcd996777a30.png)
Testing values using the rational root theorem gives 
as a root, 
does fall in the first quadrant so it satisfies the interval.
There are now two ways to finish this problem.
First way: Since , we have
![\[\sin ^2 x=\frac{1}{9}\]](http://latex.artofproblemsolving.com/c/d/9/cd944850744c2ee107a9c334cf91d381b89855a2.png)
Using the Pythagorean Identity gives us 
. Then we use the definition of 
to compute our final answer. 
.
Second way: Multiplying our old equation 
by 
gives
![\[576\sin x = 24\cot^2x\]](http://latex.artofproblemsolving.com/6/6/0/6604608065c09acf27208364c7e325f80a320e20.png)
So, 
.
See also
| 2011 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
