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De Moivre's formula gives the following identity $$(1)\,\,\,\,\cos(n\,\theta)+i\sin(n\,\theta)=\sum_\limits{k=0}^{n}{n \choose k}\cos^{n-k}(\theta)\left(i\sin(\theta)\right)^k\qquad \text{for } n\in\mathbb{N}.$$ And I am asked to prove the following equation $$(2)\,\,\,\,\,\cos(n\,\theta)=\sum_\limits{k=0}^m{n\choose2k}(-1)^k\cos^{n-2k}(\theta)\sin^{2k}(\theta)\qquad \text{for } n\in\mathbb{N}.$$ with $$m=\begin{cases} n/2, & \text{if $n$ is even,} \\ (n-1)/2, & \text{if $n$ is odd}. \end{cases}$$

I am well aware, that one can prove this by induction, however, I was wondering if there is an algebraic proof to this.

I can clearly see that $(2)$ follows from equating the real parts of $(1)$, which is done by seeing when $i$ is a real number (so every even $k$ and will have a value of $(-1)^k$). But I can't go from here, I mean I cannot formalize it.

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    $\begingroup$ That's basically it. Equating real and imaginaries and you will see that all odd $k$ terms are imaginary. For the even ones, we will get $1,i^2,i^4,i^6,...$ = $1,-1,1,-1$ which is where the $(-1)^k$ comes from. To formalise it, I would split the original summation into odd and even $k$ and then you can simplify the $i^{2k}$ term using the above ideas. $\endgroup$ – videlity Feb 8 '18 at 5:55
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\begin{align} \cos(n\,\theta)+i\sin(n\,\theta) &=\sum_\limits{k=0}^{n}{n \choose k}\cos^{n-k}(\theta)\left(i\sin(\theta)\right)^k \qquad \text{for } n\in\mathbb{N} \tag1 \\ &= \underbrace{\sum_\limits{k = 0 \\ k \text{ even}}^{n}{n \choose k}\cos^{n-k}(\theta)\left(i\sin(\theta)\right)^k}_{\text{real part}} + \underbrace{\sum_\limits{k = 1 \\ k \text{ odd}}^{n}{n \choose k}\cos^{n-k}(\theta)\left(i\sin(\theta)\right)^k}_{\text{imaginary part}} \\ \cos(n\,\theta) &= \sum_\limits{k = 0}^{\large\left\lfloor n/2 \right\rfloor}{n \choose 2k}\cos^{n-2k}(\theta)\left(i\sin(\theta)\right)^{2k} \qquad \text{for } n\in\mathbb{N} \tag2 \\ &= \sum_\limits{k = 0}^{\large\left\lfloor n/2 \right\rfloor}{n \choose 2k}(-1)^k\cos^{n-2k}(\theta)\sin^{2k}(\theta) \\ \text{with } \left\lfloor n/2 \right\rfloor &= \begin{cases} n/2, & \text{if $n$ is even,} \\ (n-1)/2, & \text{if $n$ is odd}. \end{cases} \end{align}

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