Difference between revisions of "1972 AHSME Problems/Problem 20"
(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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Revision as of 15:53, 29 June 2017
We start by letting 
so that sour equation is now: ![\[\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}\]](http://latex.artofproblemsolving.com/2/3/0/23001ccb364ad21ed172d8aeeb4146f44045ef3c.png)
Multiplying through and rearranging gives us the equation: ![\[\cos x = \frac{a^2-b^2}{2ab} * \sin x\]](http://latex.artofproblemsolving.com/6/6/6/666fac0a1773dc1fd8b7a898144b989982226acd.png)
We now apply the Pythagorean identity 
, using our substitution: ![\[\left(\frac{a^2-b^2}{2ab} * \sin x \right)^2 + \sin ^2 x =1\]](http://latex.artofproblemsolving.com/3/c/c/3cc02365b94a102b58a672c173598e89603f2d31.png)
We can isolate 
without worrying about division by 
since 
our final answer is 
