Difference between revisions of "2011 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
We can rewrite the given expression as | We can rewrite the given expression as | ||
| − | < | + | <cmath>\sqrt{24^3\sin^3 x}=24\cos x</cmath> |
Square both sides and divide by <math>24^2</math> to get | Square both sides and divide by <math>24^2</math> to get | ||
| − | < | + | <cmath>24\sin ^3 x=\cos ^2 x</cmath> |
| − | Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math> | + | Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math> |
| − | < | + | <cmath>24\sin ^3 x=1-\sin ^2 x</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>\Arcsin \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. Thus < | + | Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root. <math>\Arcsin \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. Thus |
| + | <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>. | ||
Revision as of 11:53, 19 March 2011
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. $\Arcsin \frac{1}{3}$ (Error compiling LaTeX. Unknown error_msg) does fall in the first quadrant so it satisfies the interval. Thus
![\[\sin ^2 x=\frac{1}{9}\]](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. 
.
