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Math Help - Not sure about this equation

  1. #1
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    Not sure about this equation

    The 'a' is supposed to be theta. I'm not sure why it's comming out as 'a'.

    \frac{2sin^2(\frac{1}{2}\theta)}{\theta} = (\frac{2sin^2(\frac{1}{2}\theta)}{\frac{1}{2}\thet  a})^2 * (\frac{1}{2}\theta)

    I can't see how these equate.
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  2. #2
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    Re: Not sure about this equation

    Quote Originally Posted by alyosha2 View Post
    The 'a' is supposed to be theta. I'm not sure why it's comming out as 'a'.

    \frac{2sin^2(\frac{1}{2}\theta)}{\theta} = (\frac{2sin^2(\frac{1}{2}\theta)}{\frac{1}{2}\thet  a})^2 * (\frac{1}{2}\theta)

    I can't see how these equate.
    it's not an identity, but does have solutions ...

    let u = \frac{\theta}{2}

    \frac{\sin^2{u}}{u} = \frac{4\sin^4{u}}{u}

    ... note that u \ne 0 \implies \theta \ne 0

    4\sin^4{u} - \sin^2{u} = 0

    \sin^2{u}(4\sin^2{u} - 1) = 0

    \sin^2{u} = 0 ... u = k\pi ; k \ne 0 \implies \theta = 2k \pi ; k \ne 0

    \sin{u} = \frac{1}{2} ... u = \frac{\pi}{6} \pm 2k\pi ; \frac{5\pi}{6} \pm 2k\pi  \implies \theta = \frac{\pi}{3} \pm 2k\pi ; \frac{5\pi}{3} \pm 2k\pi
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