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By definition we have $$\cos(x):= \frac{1}{2}\left(e^{ix}+e^{-ix} \right) $$

Using this definition and applying the binomial expansion to the right hand side I would get $$\cos^n(x)= \left(\frac{1}{2}\right)^n\sum_{k=0}^n\left(\begin{matrix} n \\ k\end{matrix}\right) e^{ixk} e^{(-ix)(n-k)} = \left(\frac{1}{2} \right)^n \sum_{k=0}^n\left(\begin{matrix} n \\ k\end{matrix}\right) e^{(2k-n)ix}$$ which looks promising, but I can't continue from there, apparently to this Wikipedia entry here there is a much more elegant form. I do of course with the above already obtain the $\cos(2k-n)$ part but there is still the $i\sin(2k-n)$ part present. Namely: $$\cos^n(x) = \frac{1}{2^n}\sum_{k=0}^n \left(\begin{matrix} n \\ k\end{matrix}\right)\left(\cos((2k-n)x)+i \sin((2k-n)x)\right) $$

It feels like an invalid argument to say that $\cos^n(x)$ is real and therefore the $i \sin(2k-n)$ part will vanish.

Spaced
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    Do you know DeMoivre Formula? $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$. – DiegoMath Nov 15 '14 at 17:31
  • absolutely, and I have written it down too on my calculations, but I didn't see an appropriate substitution. – Spaced Nov 15 '14 at 17:32
  • You seem to have forgotten the binomial coefficients. – Lucian Nov 15 '14 at 17:34
  • True @Lucian, I will fix them in, I did forget the LaTeX on the way. – Spaced Nov 15 '14 at 17:35
  • @DiegoMath, so you would use DeMoivre as a substitution in the last step of my calculations? Or compute $(\cos x + i \sin x)^n$ also through the binomial expansion? – Spaced Nov 15 '14 at 17:45
  • @Spaced I think you can argue that, since the imaginary part of the LHS is zero, the one on the RHS must also be zero. It's like having an equality $1=a+bi$ which certainly implies that $a=1$ and $b=0$. – Nocturne Nov 15 '14 at 17:48
  • Thanks @Nocturne, I considered that already as I've written in my post above, it just feels rather vague to me, but that is of course due to the fact that I don't have much experience with complex numbers. Written the right way it does sound reasonable though. I could split the last sum up into two sums and say that the 2nd one is purely imaginary and therefore doesn't contribute to the solution over the reals. – Spaced Nov 15 '14 at 17:52
  • Here you can find an answer to your question http://math.stackexchange.com/questions/125774/how-to-expand-cos-nx-with-cos-x – mfl Nov 15 '14 at 18:12
  • @Spaced The argument that you just had in the last sentence of your comment is a perfectly valid and rigorous way to prove the desired identity. – Spenser Nov 15 '14 at 19:01

1 Answers1

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For the power of trigonometric function we have the following identities: \begin{eqnarray} \cos^{2n}x &=& \frac{1}{2^{2n}}\binom{2n}{n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\cos[2(n-k)x] \tag 1\\ \cos^{2n+1}x &=& \frac{1}{4^n}\sum_{k=0}^{n}\binom{2n+1}{k}\cos[(2n+1-2k)x]\tag 2\\ \sin^{2n}x &=& \frac{1}{2^{2n}}\binom{2n}{n}+\frac{(-1)^n}{2^{2n-1}}\sum_{k=0}^{n}(-1)^k\binom{2n}{k}\cos[2(n-k)x] \tag 3 \\ \sin^{2n+1}x &=& \frac{(-1)^n}{4^n}\sum_{k=0}^{n-1}\binom{2n+1}{k}\sin[(2n+1-2k)x] \tag 4\\ \end{eqnarray} We prove the identity (1). We have $$ \cos x=\frac{(\cos x+i\sin x)+(\cos x-i\sin x)}{2}. $$ Put $\cos x+i\sin x=z$ so that $\cos x-i\sin x=z^{-1}$. Then we have $$ \cos^{2n}x=\left (\frac{z+z^{-1}}{2}\right)^{2n}=\frac{1}{2^{2n}}\sum_{k=0}^{2n}\binom{2n}{k}z^{-k}z^{2n-k}. $$ Further $$ 2^{2n}\cos^{2n}x=\sum_{k=0}^{n-1}\binom{2n}{k}z^{2(n-k)}+\binom{2n}{n}+\sum_{k=n+1}^{2n}\binom{2n}{k}z^{2(n-k)}. $$ In the second sum put $n-k=-(n-k')$. Then this sum is rewritten as $$ \sum_{k'=n-1}^{0}\binom{2n}{2n-k'}z^{-2(n-k')}=\sum_{k=0}^{n-1}\binom{2n}{k}z^{-2(n-k)} $$ and so $$ 2^{2n}\cos^{2n}x=\sum_{k=0}^{n-1}\binom{2n}{k}\left[z^{2(n-k)}+z^{-2(n-k)}\right]+\binom{2n}{n}. $$ However, $z^{2(n-k)}+z^{-2(n-k)}=2\cos[2(n-k)x]$, and therefore $$ \cos^{2n}x =\frac{1}{2^{2n}}\binom{2n}{n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\cos[2(n-k)x]. $$

Replacing in (1) $x$ by $\frac{\pi}{2}-x$, we get formula (2). Formulas (3) and (4) are deduced as (1) and (2).

alexjo
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