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Math Help - soluiton required up to n terms

  1. #1
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    soluiton required up to n terms

    Find Sina - Sin2a + Sin3a - Sin4a . n terms.
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  2. #2
    MHF Contributor red_dog's Avatar
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    First of all we have two cases: when n is even or n is odd.
    I'll do the case when n is even. I think you can do the other case.
    S=\sin\alpha-\sin 2\alpha+\sin 3\alpha-\sin 4\alpha +\ldots+\sin(2n-1)\alpha-\sin 2n\alpha
    Split the sum in two other sums:
    \displaystyle S_1=\sin\alpha+\sin 3\alpha+\ldots+\sin(2n-1)\alpha=\frac{\sin^2n\alpha}{\sin\alpha}
    \displaystyle S_2=\sin 2\alpha+\sin 4\alpha+\ldots+\sin 2n\alpha=\frac{\sin n\alpha\sin(n+1)\alpha}{\sin\alpha}
    To calculate each sum, multiply both sides by \sin\alpha and transform the products of sines in differences of cosines.
    Now \displaystyle S=S_1-S_2=\frac{\sin n\alpha(\sin n\alpha-\sin(n+1)\alpha)}{\sin\alpha}\displaystyle=-\frac{\sin n\alpha\cos\frac{(2n+1)\alpha}{2}}{\cos\frac{\alph  a}{2}}
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  3. #3
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    Use complex numbers. And the fact that i\sin x = \sinh (ix)  = (e^{ix}-e^{-ix})/2. Now use geometric series.
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