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1972 AHSME Problems/Problem 20

Revision as of 16:53, 29 June 2017 by Zhenqi (talk | contribs) (Created page with "We start by letting <math>\tan x = \frac{\sin x}{\cos x}</math> so that sour equation is now: <cmath>\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}</cmath> Multiplying through an...")
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We start by letting $\tan x = \frac{\sin x}{\cos x}$ so that sour equation is now: \[\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}\] Multiplying through and rearranging gives us the equation: \[\cos x = \frac{a^2-b^2}{2ab} * \sin x\] We now apply the Pythagorean identity $\sin ^2 x + \cos ^2 x =1$, using our substitution: \[\left(\frac{a^2-b^2}{2ab} * \sin x \right)^2 + \sin ^2 x =1\] We can isolate $\sin x$ without worrying about division by $0$ since $a \neq b \neq 0$ our final answer is $(E) \frac{2ab}{a^2+b^2}$

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