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Math Help - Proof by induction stuck on a sum:help:

  1. #1
    Member ssadi's Avatar
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    Proof by induction stuck on a sum:help:

    Original sum:
    Use the mathematical induction to prove:
    \displaystyle\sum_{r=1}^n sin (2r-1)\theta=\frac{sin^2 n\theta}{sin\theta}

    My problem:
    How do I convert

     \frac{sin^2 k\theta}{sin\theta}+sin (2(k+1)-1)\theta

    =\frac{sin^2 k\theta}{sin\theta}+sin (2k+1)\theta
    to

    \frac{sin^2 (k+1)\theta}{sin\theta} as required by the question. Help me, I am stuck here since yesterday.
    Last edited by ssadi; December 15th 2008 at 04:34 AM.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by ssadi View Post
    My problem:
    How do I convert

     \frac{sin^2 k\theta}{sin\theta}+sin (2(k+1)-1)\theta

    =\frac{sin^2 k\theta}{sin\theta}+sin (2k+1)\theta
    to

    \frac{sin^2 (k+1)\theta}{sin\theta}
    It pays to know your trig formulas! Use these:

    \sin\theta\sin\phi = \tfrac12\bigl(\cos(\theta-\phi) - \cos(\theta+\phi)\bigr),\qquad[1]

    \cos2\theta = 1-2\sin^2\!\theta.\qquad[2]

    Then =\frac{\sin^2 \!k\theta}{\sin\theta}+\sin (2k+1)\theta = \frac{\sin^2 \!k\theta+\sin (2k+1)\theta\sin\theta}{\sin\theta}. The second term in the numerator is

    \tfrac12\bigl(\cos(2k\theta) - \cos(2k+2)\theta\bigr) by [1]

    = \sin^2(k+1)\theta - \sin^2\!k\theta by [2].
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  3. #3
    Member ssadi's Avatar
    Joined
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    Quote Originally Posted by Opalg View Post
    It pays to know your trig formulas! Use these:

    \sin\theta\sin\phi = \tfrac12\bigl(\cos(\theta-\phi) - \cos(\theta+\phi)\bigr),\qquad[1]

    \cos2\theta = 1-2\sin^2\!\theta.\qquad[2]

    Then =\frac{\sin^2 \!k\theta}{\sin\theta}+\sin (2k+1)\theta = \frac{\sin^2 \!k\theta+\sin (2k+1)\theta\sin\theta}{\sin\theta}. The second term in the numerator is

    \tfrac12\bigl(\cos(2k\theta) - \cos(2k+2)\theta\bigr) by [1]

    = \sin^2(k+1)\theta - \sin^2\!k\theta by [2].
    Where are they when I need them?
    Thanks a lot.
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