Difference between revisions of "2006 Cyprus Seniors Provincial/2nd grade/Problem 3"
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== Problem == | == Problem == | ||
| − | If <math>\Alpha=\frac{1-cos\theta}{sin\theta}</math> and <math>\Beta=\frac{1-sin\theta}{cos\theta}</math>, prove that | + | If <math>\Alpha=\frac{1-\cos\theta}{\sin\theta}</math> and <math>\Beta=\frac{1-\sin\theta}{\cos\theta}</math>, prove that |
<math>\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{1}{4}</math>. | <math>\frac{\Alpha^2}{(1+\Alpha^2)^2} + \frac{\Beta^2}{(1+\Beta^2)^2} = \frac{1}{4}</math>. | ||
== Solution == | == Solution == | ||
| − | <math>\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- | + | <math>\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-2\cos\theta+1}{\sin^2\theta}} = \frac{\frac{1-\cos\theta}{\sin\theta}}{\frac{2(1-\cos\theta)}{\sin^2\theta}} = \frac{\sin\theta}{2}</math> |
| − | Similarly <math>\frac{\Beta}{1+\Beta^2} = \frac{cos\theta}{2}</math> | + | Similarly <math>\frac{\Beta}{1+\Beta^2} = \frac{\cos\theta}{2}</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> | + | 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> |
Revision as of 16:36, 7 March 2007
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
$\frac{\Alpha}{1+\Alpha^2} = \frac{\frac{1-\cos\theta}{\sin\theta}}{1+(\frac{1- \cos\theta}{\sin\theta})^2} = \frac{\frac{1-\cos\theta}{\sin\theta}}{\frac{\sin^2\theta+ \cos^2\theta-2\cos\theta+1}{\sin^2\theta}} = \frac{\frac{1-\cos\theta}{\sin\theta}}{\frac{2(1-\cos\theta)}{\sin^2\theta}} = \frac{\sin\theta}{2}$ (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)
