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Challenge: The nastiest trig. identity that you have ever met!
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June 14th, 2008, 09:33 PM
mathwizard
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Challenge: The nastiest trig. identity that you have ever met!
Prove the identities
and
Good luck!
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Challenge: The nastiest trig. identity that you have ever met! ![\frac{cos(n\theta) sin[\frac{(n+1)\theta}{2}]}{sin(\theta/2)}](http://www.mathhelpforum.com/math-help/latex2/img/89e896cf42f0a41e3efa927b2ddae71b-1.gif "\frac{cos(n\theta) sin[\frac{(n+1)\theta}{2}]}{sin(\theta/2)}")
![= \frac{cos[(n+1)\theta] cos\theta - cos\theta - cos[(n+1)\theta] + 1 + sin\theta sin[(n+1)\theta]}{2-2cos\theta}](http://www.mathhelpforum.com/math-help/latex2/img/a1402874a758ed4d99b231b1696e8024-1.gif "= \frac{cos[(n+1)\theta] cos\theta - cos\theta - cos[(n+1)\theta] + 1 + sin\theta sin[(n+1)\theta]}{2-2cos\theta}")
![\frac{sin(n\theta) - sin[(n+1)\theta] + sin\theta}{2-2cos\theta} = \frac{sin(n\theta/2)sin[(n+1)\theta/2]}{sin(\theta/2)}](http://www.mathhelpforum.com/math-help/latex2/img/adb53573cf65002372f7a798db08c771-1.gif "\frac{sin(n\theta) - sin[(n+1)\theta] + sin\theta}{2-2cos\theta} = \frac{sin(n\theta/2)sin[(n+1)\theta/2]}{sin(\theta/2)}")