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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    $\begingroup$ Can you use $\cos\theta +i\sin\theta=e^{i\theta}$? If so, expanding $(\cos x+i\sin x)^n$ using the Binomial Theorem will do the job. $\endgroup$ – André Nicolas Dec 19 '11 at 8:01
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    $\begingroup$ Otherwise, induction on $n$ and addition formulas. $\endgroup$ – AD. Dec 19 '11 at 8:05
  • $\begingroup$ Thanks, I'm trying induction. But now I come to the point $\cos(kx)\cos(x)-\sin(kx)\sin(x)$ and I need to prove that that formula equals $\sum_{j=0}^{[(k+1)/2]} (-1)^j {k+1 \choose 2j} cos(x)^{k+1-2j} sin(x)^2j$, but how? That $\sin(kx)$ won't fit in here, will it? $\endgroup$ – www.data-blogger.com Dec 19 '11 at 8:56
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    $\begingroup$ Explicit hint: $$(\cos n x+ \sin n x) = e^{n i x} = \left( e^{i x} \right)^n = (\cos x+ i \sin x)^n .$$ Now apply the binomial theorem, and then separate out the real and imaginary parts. $\endgroup$ – Srivatsan Dec 19 '11 at 9:20
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    $\begingroup$ @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 $\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}$. $\endgroup$ – André Nicolas Dec 19 '11 at 16:06

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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  • $\begingroup$ Thanks, but the only thing I don't get is $j \le [\frac{n-1}{2}]$. $j$ must loop through all odd parts, so I thought it must loop untill $[\frac{n+1}{2}]$? $\endgroup$ – www.data-blogger.com Dec 19 '11 at 19:38
  • $\begingroup$ For odd $k=2j+1\leq n$ (with $j\in\mathbb{Z}$), the inequality implies (requires) that $j\leq\frac{n-1}{2}$. However the highest such $k$ occurs for $j=[\frac{n-1}{2}]$ (rounding the fraction down when $n$ is even). $\endgroup$ – bgins Dec 19 '11 at 23:35
  • $\begingroup$ is $[\cdot]$ the nearest integer function or the ceiling function? $\endgroup$ – clathratus Dec 10 '18 at 21:47
  • $\begingroup$ 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). $\endgroup$ – bgins Dec 19 '18 at 5:41

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