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Math Help - mathematical induction

  1. #1
    Member grgrsanjay's Avatar
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    mathematical induction

    P.T sinx +sin3x+ . . . +sin(2n-1)x = [sin^(2) (nx)] / sinx for any natural number n using mathematical induction
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  2. #2
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    Quote Originally Posted by grgrsanjay View Post
    P.T sinx +sin3x+ . . . +sin(2n-1)x = [sin^(2) (nx)] / sinx for any natural number n using mathematical induction
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    Quote Originally Posted by grgrsanjay View Post
    P.T sinx +sin3x+ . . . +sin(2n-1)x = [sin^(2) (nx)] / sinx for any natural number n using mathematical induction
    So you must prove that \displaystyle{\sum\limits^{n+1}_{k=1}\sin(2k-1)x=\frac{\sin^2(n+1)x}{\sin x}} , as the case n=1 is trivially true and assuming true for n:

    \displaystyle{\sum\limits^{n+1}_{k=1}\sin(2k-1)x=\sum\limits^n_{k=1}\sin(2k-1)x+\sin(2n+1)x=\frac{\sin^2nx}{\sin x}+\sin(2n+1)x}}=

    =\displaystyle{\frac{\sin^2nx+\sin x\sin(2n+1)x}{\sin x}} . Thus, it's enough to show that \sin^2(n+1)x=\sin^2nx+\sin x\sin(2n+1)x.

    For this you'll need identities: in the LHS use the well known \sin(\alpha+\beta)=\sin\alpha\cos\beta+sin\beta\co  s\alpha (watch

    that square!), and on the RHS, second term, use the same identity.

    After the above's done pass all the terms to the LHS, use the trigonometric Pythagoras identity

    and also \cos(2\beta)=1-2\sin^2\beta ...

    Tonio
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    Last edited by tonio; November 23rd 2010 at 09:58 AM. Reason: double post
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