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Compute the following sum by using Euler's formula, $ e^{i \theta} = \cos \theta + i \sin \theta$, $$\cos^{2n}\theta-{2n\choose 2}\cos^{2n-2}\theta\sin^2\theta+...+(-1)^{n-1}{2n\choose 2n-2}\cos^2\theta\sin^{2n-2}\theta+(-1)^n\sin^{2n}\theta$$

I have tried to rewrite the expression as:

$$\sum_{k=0}^n{2n\choose 2k}(-1)^k\sin^{2k}\theta\cos^{2n-2k}\theta$$

But I have no certain idea about how to continue. Could you give me some hints? Thanks!

Gibbs
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1 Answers1

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$(-1)^k\sin^{2k}t=(i\sin t)^{2k}$

$$(\cos t+i\sin t)^{2n}+(\cos t-i\sin t)^{2n}=?$$

lab bhattacharjee
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  • Well, that's $(e^{it})^{2n}+(e^{-it})^{2n}$... but I don't see clearly what I am supposed to do. – Gibbs Jan 22 '19 at 18:53
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    @Gibbs, https://math.stackexchange.com/questions/1432568/what-is-the-meaning-of-eulers-identity – lab bhattacharjee Jan 22 '19 at 18:58
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    $$(\cos t + i\sin t)^{2n}=\sum_{k=0}^{2n}{2n\choose k}(\cos t)^{2n-k}i^k(\sin t)^k$$

    $$(\cos t - i\sin t)^{2n}=\sum_{k=0}^{2n}{2n\choose k}(\cos t)^{2n-k}(-i)^k(\sin t)^k$$

    If we sum them both, only the terms with even $k$ survive, so:

    $$(\cos t + i\sin t)^{2n}+(\cos t - i\sin t)^{2n}=\sum_{k=0}^n{2n\choose 2k}(\cos t)^{2n-2k}i^{2k}(\sin t)^{2k}=\sum_{k=0}^n{2n\choose 2k}(\cos t)^{2n-2k}(-1)^k(\sin t)^{2k}$$

     – Gibbs Feb 19 '21 at 17:45