Difference between revisions of "2006 Cyprus Seniors Provincial/2nd grade/Problem 3"
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So <math>\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{sin^2\theta}{2^2} + \frac{cos^2\theta}{2^2}= \frac{1}{4}</math> | So <math>\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{sin^2\theta}{2^2} + \frac{cos^2\theta}{2^2}= \frac{1}{4}</math> | ||
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Revision as of 04:48, 12 November 2006
Problem
If $\Alpha=\frac{1-cos\theta}{sin\theta}$ (Error compiling LaTeX. Unknown error_msg) and $\Beta=\frac{1-sin\theta}{cos\theta}$ (Error compiling LaTeX. Unknown error_msg), prove that $\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{1}{4}$ (Error compiling LaTeX. Unknown error_msg).
Solution
$\frac{\Alpha}{1+\Alpha^2} = \frac{\frac{1-cos\theta}{sin\theta}}{1+(\frac{1- cos\theta}{sin\theta})^2} = \frac{\frac{1-cos\theta}{sin\theta}}{\frac{sin^2\theta+ cos^2\theta-2cos\theta+1}{sin^2\theta}} = \frac{\frac{1-cos\theta}{sin\theta}}{\frac{2(1-cos\theta)}{sin^2\theta}} = \frac{sin\theta}{2}$ (Error compiling LaTeX. Unknown error_msg)
Similarly $\frac{\Beta}{1+\Beta^2} = \frac{cos\theta}{2}$ (Error compiling LaTeX. Unknown error_msg)
So $\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{sin^2\theta}{2^2} + \frac{cos^2\theta}{2^2}= \frac{1}{4}$ (Error compiling LaTeX. Unknown error_msg)
