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I have problem with the sum:

$$ \sum_{k=0}^n \dbinom{n}{k}(\cos \alpha)^k(i\sin \alpha)^{n-k}\,\, $$ Apparantly, I have an imaginary unit therefore I need to distinguish even and odd powers of $i$ to do so I need to introduce $2k$ as in: $$ \sum_{k=0}^n f(k) = \sum_{k=0}^{n/2} g(2k) $$ and eventually find $g$ starting from $f$

The goal of the exercise is to separate the real part and an imaginary part of this sum to find real expressions of $\sin (n\alpha)$ and $\cos (n\alpha)$

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    $\begingroup$ Yes, you'll need to separate terms into those where $k = 2m$ and those where $k = 2m + 1$. Note that we are actually concerned with the parity of $n - k$, not $k$, so that which is real/imaginary depends on the parity of $n$. You can avoid this by performing the binomial expansion with $i \sin{\alpha}$ as the first term in the product, rather than the second. $\endgroup$ Nov 21, 2014 at 0:09

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Follow Christopher's suggestion. Suppose that $n=2m+1$ then $$A_{2m+1}=\sum_{k=0}^m \dbinom{2m+1}{2k}(\cos \alpha)^{2m+1-2k}(i\sin \alpha)^{2k}$$ $$+\sum_{k=0}^m \dbinom{2m+1}{2k+1}(\cos \alpha)^{2m+1-(2k+1)}(i\sin \alpha)^{2k+1}$$

$$Re(A_{2m+1})=\sum_{k=0}^m \dbinom{2m+1}{2k}(\cos \alpha)^{2m+1-2k}(-1)^k(\sin \alpha)^{2k}$$ $$Im(A_{2m+1})=\sum_{k=0}^m \dbinom{2m+1}{2k+1}(\cos \alpha)^{2m+1-(2k+1)}(-1)^k(\sin \alpha)^{2k+1}$$

Similarly let $n=2m$ then

$$A_{2m}=\sum_{k=0}^m \dbinom{2m}{2k}(\cos \alpha)^{2m-2k}(i\sin \alpha)^{2k}$$ $$+\sum_{k=0}^{m-1} \dbinom{2m}{2k+1}(\cos \alpha)^{2m+1-(2k+1)}(i\sin \alpha)^{2k+1}$$

$$Re(A_{2m})=\sum_{k=0}^m \dbinom{2m}{2k}(\cos \alpha)^{2m-2k}(-1)^k(\sin \alpha)^{2k}$$ $$Im(A_{2m})=\sum_{k=0}^{m-1} \dbinom{2m}{2k+1}(\cos \alpha)^{2m-(2k+1)}(-1)^k(\sin \alpha)^{2k+1}$$

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Hint: Using a slightly extended version of the binomial coefficient (see e.g. Wilf p.15) with

$$\binom{n}{k}=0\qquad k>n$$

the calculation can be written more compactly:

\begin{align*} (\cos \alpha + i \sin \alpha)^n&= \sum_{k=0}^n\binom{n}{k}(i\sin\alpha)^k(\cos\alpha)^{n-k}\\ &=\sum_{k=0}^{n}\binom{n}{2k}(-1)^k(\sin\alpha)^{2k} (\cos\alpha)^{n-2k}\tag{1}\\ &\qquad + i\sum_{k=0}^{n}\binom{n}{2k+1}(-1)^k(\sin\alpha)^{2k+1} (\cos\alpha)^{n-(2k+1)}\tag{2} \end{align*}

Observe that $\binom{n}{2k}=0$ if $2k > n$ in (1) and a similar argument applies to (2).

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