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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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    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.2011-12-19
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    Otherwise, induction on $n$ and addition formulas.2011-12-19
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    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?2011-12-19
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    Let me suggest you try @André's hint.2011-12-19
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    But how? Then you have to write $cos(nx)$ in terms of $cos(\alpha)$ and $i sin(\alpha)$ in order to apply the Binomial Theorem?2011-12-19
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    Where did your $\alpha$ come from? In any event, $\cos\,nx=\Re\exp(inx)$; you can apply that after expanding out $(\cos\,x+i\sin\,x)^n$.2011-12-19
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    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.2011-12-19
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    Ah :), thanks! That will do the job I guess ;)!2011-12-19
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    I gave that suggestion in case you did not know about the Euler formula that André suggested.2011-12-19
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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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    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 $[\frac{n+1}{2}]$?2011-12-19
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    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).2011-12-19
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    is $[\cdot]$ the nearest integer function or the ceiling function?2018-12-10
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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