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How do I prove the following formulas?

Let $n \in \mathbb{N}, x \in \mathbb{R}$. Prove that:

$\cos(nx)=\sum_{j=0}^{[n/2]} (-1)^j {n \choose 2j} (\cos x)^{n-2j} (\sin x)^{2j}$

$\sin(nx)=\sum_{j=0}^{[(n-1)/2]} (-1)^j {n \choose 2j+1} (\cos x)^{n-2j-1} (\sin x)^{2j+1}$

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    @Kevin: If you want to give an inductive proof (i.e. the hard way), you need two items. The first is the Addition Laws for sine and cosine. The other is (essentially) the Binomial Theorem. The key observation there is that $\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}$.2011-12-19

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Since $n\in\mathbb{N}$ and $x\in\mathbb{R}$, the we can start a derivation for both at once using DeMoivre's formula and Binomial Theorem as @André Nicolas says: $ \cos(nx)+i\sin(nx) =e^{inx} =(\cos x+i\sin x)^n =\sum_{k=0}^{n}i^k\binom{n}{k}\cos^{n-k}x\sin^{k}x $ Next we must observe that even and odd $k\in\{0,1,\dots,n\}$ give rise to the real and imaginary part, respectively, and can be parametrized by $k=2j$ for $0 \leq j \leq [\frac{n}{2}]$ and by $k=2j+1$ for $0 \leq j \leq [\frac{n-1}{2}]$.

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    clathratus: the square bracket $\left[\cdot\right]$ notation refers to the "floor" or "greatest integer" function (less than or equal to its argument), which truncates any fractional part and moves non-integers to the left on the number line (and is thus asymmetric, not an odd function).2018-12-19