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Math Help - Conversion (Trigonometry Variable To Standard Variable)

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    Conversion (Trigonometry Variable To Standard Variable)

    I came across an example when understanding Fourier Series. There was one part where they converted a trigonometry function to standard variable. They did this:

    \frac{2}{n^2\pi^2}(1-\cos n \pi) = \frac{2}{n^2\pi^2}(1-(-1)^n).

    What have they done here? Can this be done to sine too? Thanks in advance for the help.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Air View Post
    I came across an example when understanding Fourier Series. There was one part where they converted a trigonometry function to standard variable. They did this:

    \frac{2}{n^2\pi^2}(1-\cos n \pi) = \frac{2}{n^2\pi^2}(1-(-1)^n).

    What have they done here? Can this be done to sine too? Thanks in advance for the help.
    If n\in\mathbb{Z}, we can generate a pattern for values of \cos\left(n\pi\right)...

    ...
    \cos\left(-3\pi\right)=\cos\left(3\pi\right)=-1
    \cos\left(-2\pi\right)=\cos\left(2\pi\right)=1
    \cos\left(-\pi\right)=\cos\left(\pi\right)=-1
    \cos\left(0\right)=1

    So, when n is even (or zero), we see that \cos\left(n\pi\right)=1. When n is odd, we see that \cos\left(n\pi\right)=-1

    So, we can rewrite \cos\left(n\pi\right) as \left(-1\right)^n;~n\in\mathbb{Z}.


    A similar thing can be done with the sine function:

    I'll let you verify that \forall n\in\mathbb{Z},~\sin\left(\frac{\left(2n-1\right)\pi}{2}\right)=\left(-1\right)^{n+1}

    Does this make sense?
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    Yes, that makes sense. Thanks.

    But, does that mean that this genralisation only works with \pi because it's a periodic function?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Air View Post
    But, does that mean that this genralisation only works with \pi because it's a periodic function?
    From what I understand, Yes.
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    Quote Originally Posted by Air View Post
    Yes, that makes sense. Thanks.

    But, does that mean that this genralisation only works with \pi because it's a periodic function?
    Quote Originally Posted by Chris L T521 View Post
    From what I understand, Yes.
    Kind of. It depends how you interpret that question. If you'll notice \cos(3\pi n) gives a similar set of values and in fact \cos((2z+1)\pi n)~~z\in\mathbb{Z} gives the same set of values.
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    Hello,

    Here is a better way than induction :

    \cos(n\pi)=\frac{e^{in\pi}+e^{-in\pi}}{2}=\frac{\left(e^{i\pi}\right)^n+\left(e^{-i\pi}\right)^n}{2}

    But e^{i \pi}=-1=e^{-i\pi}

    So we have \cos(n\pi)=\frac 12 \cdot \left((-1)^n+(-1)^n\right)
    If n is even, then it's equal to 1. If n is odd, then it's equal to -1 (just take n=2k or n=2k+1)


    As for the sine thing Chris gave, you can use this identity :
    \cos(x)=\sin \left(\frac \pi 2-x\right)=-\sin \left(x-\frac \pi 2\right)
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