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Math Help - help need on trigonomtery function

  1. #1
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    help need on trigonomtery function

    Solve : sin^2nƟ - sin^2(n-1) Ɵ = sin^2 Ɵ
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    Re: help need on trigonomtery function

    Quote Originally Posted by rlaydab View Post
    Solve : sin^2nƟ - sin^2(n-1) Ɵ = sin^2 Ɵ
    I can't tell whether your equation is \displaystyle \begin{align*} \sin^{2n}{\theta} - \sin^{2(n-1)}{\theta} = \sin^2{\theta} \end{align*} or \displaystyle \begin{align*} \sin^2{(n\theta)} - \sin^2{[(n-1)\theta]} = \sin^2{\theta} \end{align*}...
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    Re: help need on trigonomtery function

    The second bottom one is correct. N or n-1 or theta is not in the exponent/power. Squared power is for sine only.
    Last edited by rlaydab; July 14th 2012 at 08:47 AM.
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    Re: help need on trigonomtery function

    Quote Originally Posted by rlaydab View Post
    The second bottom one is correct. N or n-1 or theta is not in the exponent/power. Squared power is for sine only.
    \sin^2(n\theta) - \sin^2[(n-1)\theta] = \sin^2{\theta}

    it's true for n = 1 ... are you to prove that this is an identity? looks like a candidate for induction.
    Last edited by skeeter; July 14th 2012 at 08:50 AM.
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    Re: help need on trigonomtery function

    Can you help to solve it. Can LHS be proven to be = RHS
    Last edited by rlaydab; July 14th 2012 at 09:10 AM.
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    Re: help need on trigonomtery function

    the equation is not an identity, however (after playing with the calculator) ...

    for n = 2

    \sin^2(2\theta) - \sin^2{\theta} = \sin^2{\theta} for \theta equal to integral multiples of \pi and odd multiples of \frac{\pi}{4}

    for n = 3

    \sin^2(3\theta) - \sin^2{2\theta} = \sin^2{\theta} for \theta equal to integral multiples of \pi and odd multiples of \frac{\pi}{6}

    for n = 4

    \sin^2(4\theta) - \sin^2{3\theta} = \sin^2{\theta} for \theta equal to integral multiples of \pi and odd multiples of \frac{\pi}{8}

    hmmm ...

    conjecture that \sin^2(n\theta) - \sin^2[(n-1)\theta] = \sin^2{\theta} for \theta equal to integral multiples of \pi and odd multiples of \frac{\pi}{2n}

    ... ???
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    Re: help need on trigonomtery function

    Thanks but I thought we have to prove the LHS = RHS?
    May be I am wrong.
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  8. #8
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    Re: help need on trigonomtery function

    It needs the use of three trig identities.

     \sin A + \sin B=2\sin \left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)..................(1)

     \sin A - \sin B=2\cos \left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)..................(2)

    \sin 2A = 2\sin A \cos A.................................................  .......(3)

    Briefly, take a difference of two squares on the LHS, simplify the two brackets using (1) and (2) and then use (3) (twice) to simplify the resulting expression. Move the resulting expression to the RHS, remove \sin \theta as a common factor and then use (2) again on the expression within the brackets.
    That leads to

    2\sin \theta \cos n\theta \sin(n-1)\theta =0 from which \sin\theta=0, or \cos n\theta=0 or \sin (n-1)\theta = 0.
    Last edited by BobP; July 14th 2012 at 03:30 PM. Reason: sorry, lost the last line
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