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Math Help - Challenge: The nastiest trig. identity that you have ever met!

  1. #1
    Junior Member
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    Challenge: The nastiest trig. identity that you have ever met!

    Prove the identities

    <br />
\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}<br />

    and

    <br />
\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)}<br />

    Good luck!
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  2. #2
    MHF Contributor red_dog's Avatar
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    1.
    \displaystyle\frac{\cos(n+1)\theta\cos\theta-\cos\theta-\cos(n+1)\theta+1+\sin\theta\sin(n+1)\theta}{2-2\cos\theta}=
    \displaystyle=\frac{\cos n\theta-\cos(n+1)\theta+1-\cos\theta}{2(1-\cos\theta)}=\frac{2\sin\frac{\theta}{2}\sin\frac{  (2n+1)\theta}{2}+2\sin^2\frac{\theta}{2}}{4\sin^2\  frac{\theta}{2}}=
    \displaystyle=\frac{\sin\frac{(2n+1)\theta}{2}+\si  n\frac{\theta}{2}}{2\sin\frac{\theta}{2}}=\frac{\s  in\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}}{\sin  \frac{\theta}{2}}

    2.
    \displaystyle\frac{\sin n\theta-\sin(n+1)\theta+\sin\theta}{2-2\cos\theta}=\frac{-2\sin\frac{\theta}{2}\cos\frac{(2n+1)\theta}{2}+2\  sin\frac{\theta}{2}\cos\frac{\theta}{2}}{4\sin^2\f  rac{\theta}{2}}=
    \displaystyle=\frac{\cos\frac{\theta}{2}-\cos\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}}  =\frac{\sin\frac{n\theta}{2}\sin\frac{(n+1)\theta}  {2}}{\sin\frac{\theta}{2}}
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