# Prove by Induction

For $n\in \mathbb{N}$ and $z\in \mathbb{C}$:

1. $\sin{(nz)}=\sum _{ k=0 }^{ n }{ \binom{n}{k} }\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n-k}(\sin{z})^k$
2. $\cos{(nz)}=\sum _{ k=0 }^{ n }{ \binom{n}{k} }\frac{1}{2}(i^k+(-i)^k)(\cos{z})^{n-k}(\sin{z})^k$

Formula: $\forall_{z,w\in \mathbb{C}} \ \sin{(z+w)}=\sin{z}\cos{w}+\cos{z}\sin{w}$

Hear is the solution:

Basis for $n=0$

• $\sin(0)=0=\frac{1}{2i}(i^0+(-i)^0)(\cos{z})^0(\sin{z})^0$
• $\cos(0)=1)\frac{1}{2}(i^0+(-i)^0)(\cos{z})^0(\sin{z})^0$

Induction hypothesis:

Inductive step: $n \rightarrow n+1$

$\sin{((n+1)z)}=\sin{(nz+z)}=\sin{(nz)}\cos{(z)}+\cos{(nz)}\sin{(z)}= \\ =(\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{(z)})^{n-k}(\sin{(z)})^k)(\cos{(z)}) \\ +(\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2i}(i^k+(-i)^k)(\cos{z})^{n-k}(\sin{z})^k)(\sin{z})= \\ =\sum_{k=0}^{n}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k + \\ +\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{1}{2}(i^{k-1}+(-i)^{k-1})(\cos{z})^{n-(k-1)}(\sin{z})^k= \\ =\binom{1}{0}\frac{1}{2i}(i^0-(-i)^0)(\cos{z})^{n+1}(\sin{z})^0 + \\ + \sum_{k=1}^{n}[\binom{n}{k}+\binom{n}{k-1}]\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k + \\ + \binom{n}{n}\frac{1}{2i}(i^{n+1}-(-i)^{n+1})(\cos{z})^0(\sin{z})^{n+1}= \\ =\sum_{k=0}^{n+1}\binom{n+1}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k$

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Hear is the solution:

Basis for $n=0$

• $\sin(0)=0=\frac{1}{2i}(i^0+(-i)^0)(\cos{z})^0(\sin{z})^0$
• $\cos(0)=1)\frac{1}{2}(i^0+(-i)^0)(\cos{z})^0(\sin{z})^0$

Induction hypothesis:

Inductive step: $n \rightarrow n+1$

$\sin{((n+1)z)}=\sin{(nz+z)}=\sin{(nz)}\cos{(z)}+\cos{(nz)}\sin{(z)}= \\ =(\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{(z)})^{n-k}(\sin{(z)})^k)(\cos{(z)}) \\ +(\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2i}(i^k+(-i)^k)(\cos{z})^{n-k}(\sin{z})^k)(\sin{z})= \\ =\sum_{k=0}^{n}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k + \\ +\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{1}{2}(i^{k-1}+(-i)^{k-1})(\cos{z})^{n-(k-1)}(\sin{z})^k= \\ =\binom{1}{0}\frac{1}{2i}(i^0-(-i)^0)(\cos{z})^{n+1}(\sin{z})^0 + \\ + \sum_{k=1}^{n}[\binom{n}{k}+\binom{n}{k-1}]\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k + \\ + \binom{n}{n}\frac{1}{2i}(i^{n+1}-(-i)^{n+1})(\cos{z})^0(\sin{z})^{n+1}= \\ =\sum_{k=0}^{n+1}\binom{n+1}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k$

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This is just my attempt on this problem...

Can we try to split this sum:

$\sin{(nz+z)}=\sum _{ k=0 }^{ n+1 }{ \binom{n+1}{k} }\frac{1}{2i}(i^k+(-i)^k)(\cos{z})^{(n+1)-k}(\sin{z})^k$

into a sum going from 0 to 1, and another sum going from 0 to n ? Then we can use the inductive hypothesis and the base case, to say that the sum, up to n+1 steps, holds.

After some attempts, everything works until I'm getting stuck at trying to split the combination term. I forgot on how to do it >_<

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You mean to split it in $0\rightarrow n$ and + $n+1$ ,right? –  Devid Feb 3 '13 at 19:29
Yes, a sum up to n terms, and isolate the n+1 term out. –  Cecile Feb 3 '13 at 20:36
Ok so you mean this: $$\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n-k}(\sin{z})^k+ \binom{n+1}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k$$ and now ? –  Devid Feb 3 '13 at 20:44