Prove by Induction 
For $n\in \mathbb{N}$ and $z\in \mathbb{C}$:
  
  
*
  
*$\sin{(nz)}=\sum _{ k=0 }^{ n }{ \binom{n}{k} }\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n-k}(\sin{z})^k $
  
*$\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: $n \rightarrow n+1$
$\sin{((n+1)z)}=\sin{(nz+z)}=\sin{(nz)}\cos{(z)}+\cos{(nz)}\sin{(z)}= \\ =(\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{(z)})^{n-k}(\sin{(z)})^k)(\cos{(z)}) \\ +(\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2i}(i^k+(-i)^k)(\cos{z})^{n-k}(\sin{z})^k)(\sin{z})= \\ =\sum_{k=0}^{n}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k + \\ +\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{1}{2}(i^{k-1}+(-i)^{k-1})(\cos{z})^{n-(k-1)}(\sin{z})^k= \\ =\binom{1}{0}\frac{1}{2i}(i^0-(-i)^0)(\cos{z})^{n+1}(\sin{z})^0 + \\ + \sum_{k=1}^{n}[\binom{n}{k}+\binom{n}{k-1}]\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k + \\ + \binom{n}{n}\frac{1}{2i}(i^{n+1}-(-i)^{n+1})(\cos{z})^0(\sin{z})^{n+1}= \\ =\sum_{k=0}^{n+1}\binom{n+1}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k$
 A: 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: $n \rightarrow n+1$
$\sin{((n+1)z)}=\sin{(nz+z)}=\sin{(nz)}\cos{(z)}+\cos{(nz)}\sin{(z)}= \\ =(\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{(z)})^{n-k}(\sin{(z)})^k)(\cos{(z)}) \\ +(\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2i}(i^k+(-i)^k)(\cos{z})^{n-k}(\sin{z})^k)(\sin{z})= \\ =\sum_{k=0}^{n}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k + \\ +\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{1}{2}(i^{k-1}+(-i)^{k-1})(\cos{z})^{n-(k-1)}(\sin{z})^k= \\ =\binom{1}{0}\frac{1}{2i}(i^0-(-i)^0)(\cos{z})^{n+1}(\sin{z})^0 + \\ + \sum_{k=1}^{n}[\binom{n}{k}+\binom{n}{k-1}]\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k + \\ + \binom{n}{n}\frac{1}{2i}(i^{n+1}-(-i)^{n+1})(\cos{z})^0(\sin{z})^{n+1}= \\ =\sum_{k=0}^{n+1}\binom{n+1}{k}\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n+1-k}(\sin{z})^k$
A: 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 >_<
