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Math Help - Trig Power Series Question

  1. #1
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    Trig Power Series Question

    Any help on this would be greatly appreciated! Thank You!


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  2. #2
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    Let r = 2\theta, then we have:

    \displaystyle\cos^2{\theta} = \frac{1}{2}\left(1+\cos{r}\right) = \frac{1}{2}\left(1+\sum_{n=0}^{\infty}\frac{(-1)^n(r)^{2n}}{(2n)!}\right)[/Math] <br /> <br />
 [Math]\displaystyle = \frac{1}{2}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n(r)^{2n}}{(2n)!} = \frac{1}{2}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n(2\theta)^{2n}}{(2n)!} .
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  3. #3
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    Thank You very much! I didn't realize it was that simple!
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  4. #4
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    Quote Originally Posted by ihatemath View Post
    Thank You very much! I didn't realize it was that simple!
    It wasn't! The machine made it simple. You're welcome!
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