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Difference between revisions of "2011 AIME I Problems/Problem 9"

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Square both sides and divide by <math>24^2</math> to get
 
Square both sides and divide by <math>24^2</math> to get
 
<math>24\sin ^3 x=\cos ^2 x</math>
 
<math>24\sin ^3 x=\cos ^2 x</math>
 +
Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math>
 +
<math>24\sin ^3 x=1-\sin ^2 x</math>
 +
<math>24\sin ^3 x+\sin ^2 x - 1=0</math>
 +
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 <math>\sin ^2 x=\frac{1}{9}</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 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}$.

Revision as of 12:50, 19 March 2011

Problem

Suppose $x$ is in the interval $[0, \pi/2]$ and $\log_(24\sin x) (24\cos x)=\frac{3}{2}$. Find $24\cot^2 x$.

Solution

We can rewrite the given expression as $\sqrt{24^3\sin^3 x}=24\cos x$. Square both sides and divide by $24^2$ to get $24\sin ^3 x=\cos ^2 x$ Rewrite $\cos ^2 x$ as $1-\sin ^2 x$ $24\sin ^3 x=1-\sin ^2 x$ $24\sin ^3 x+\sin ^2 x - 1=0$ Testing values using the rational root theorem gives $\sin x=\frac{1}{3}$ 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 $\sin ^2 x=\frac{1}{9}$. Using the Pythagorean Identity gives us $\cos ^2 x=\frac{8}{9}$. Then we use the definition of $\cot ^2 x to compute our final answer.$24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}$.

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