1972 AHSME Problems/Problem 20: Difference between revisions
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== Problem 20 == | |||
If <math>\tan x=\dfrac{2ab}{a^2-b^2}</math> where <math>a>b>0</math> and <math>0^\circ <x<90^\circ</math>, then <math>\sin x</math> is equal to | |||
<math>\textbf{(A) }\frac{a}{b}\qquad | |||
\textbf{(B) }\frac{b}{a}\qquad | |||
\textbf{(C) }\frac{\sqrt{a^2-b^2}}{2a}\qquad | |||
\textbf{(D) }\frac{\sqrt{a^2-b^2}}{2ab}\qquad | |||
\textbf{(E) }\frac{2ab}{a^2+b^2} </math> | |||
[[1972 AHSME Problems/Problem 20|Solution]] | |||
==Solution== | |||
We start by letting <math>\tan x = \frac{\sin x}{\cos x}</math> so that our equation is now: <cmath>\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}</cmath> Multiplying through and rearranging gives us the equation: <cmath>\cos x = \frac{a^2-b^2}{2ab} * \sin x</cmath> We now apply the Pythagorean identity <math>\sin ^2 x + \cos ^2 x =1</math>, using our substitution: <cmath>\left(\frac{a^2-b^2}{2ab} * \sin x \right)^2 + \sin ^2 x =1</cmath> We can isolate <math>\sin x</math> without worrying about division by <math>0</math> since <math>a \neq b \neq 0</math> our final answer is <math>(E) \frac{2ab}{a^2+b^2}</math> | We start by letting <math>\tan x = \frac{\sin x}{\cos x}</math> so that our equation is now: <cmath>\frac{\sin x}{\cos x} = \frac{2ab}{a^2-b^2}</cmath> Multiplying through and rearranging gives us the equation: <cmath>\cos x = \frac{a^2-b^2}{2ab} * \sin x</cmath> We now apply the Pythagorean identity <math>\sin ^2 x + \cos ^2 x =1</math>, using our substitution: <cmath>\left(\frac{a^2-b^2}{2ab} * \sin x \right)^2 + \sin ^2 x =1</cmath> We can isolate <math>\sin x</math> without worrying about division by <math>0</math> since <math>a \neq b \neq 0</math> our final answer is <math>(E) \frac{2ab}{a^2+b^2}</math> | ||
Latest revision as of 18:14, 2 January 2024
Problem 20
If 
where 
and 
, then 
is equal to
Solution
We start by letting 
so that our 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 
