Use induction to prove trignometric identity with imaginary number Prove by induction that if $i^2 = -1 $, then for every integer $n >= 1$, $[\cos(x) + i\sin(x)]^n = \cos(nx) + i\sin(nx)$.
My solution so far:
1. It can be easily shown that it is true for n = 1.
2. Then it has to be proven that if  $$[\cos(x) + i\sin(x)]^n = \cos(nx) + i\sin(nx)$$ it follows that  $$[\cos(x) + i\sin(x)]^{n+1} = \cos[(n+1)x] + i\sin[(n+1)x]$$ 
$$[\cos(x) + i\sin(x)]^n [\cos(x) + i\sin(x)]  = \cos[(n+1)x] + i\sin[(n+1)x]$$
Using the induction hypothesis to substitute $[\cos(x) + i\sin(x)]^n$ we get 
$$[\cos(nx) + i\sin(nx)][\cos(x) + i\sin(x)] = \cos[(n+1)x] + i\sin[(n+1)x]$$
This can be rewritten as
$$\cos(x)\cos(nx)+i\sin(x)\cos(nx) +i\sin(nx)\cos(x)-\sin(x)\sin(nx) = \cos[(n+1)x] + i\sin[(n+1)x]$$
I am not sure how to proceed at this point, any help is appreciated.
 A: Use the fact that $$\cos(x+y) = \cos x \cos y - \sin x \sin y$$For example, $$\cos\left((n+1)x\right)=\cos(nx + x) = \cos nx \cos x - \sin nx \sin x$$

A stylistic point:
When trying to prove that $2$ expressions are equal, say $f(x)$ and $g(x)$, it doesn't make mathematical sense to write from the beginning $f(x)=g(x)$ and then manipulate them. At the beginning you don't know they are equal, since that's what you're  trying to prove!!
Rather, you should write something along the lines of $$\begin{align}f(x) &=\ldots\\&=\ldots\\&=g(x) \end{align}$$
For example, in your case, it makes no sense to write $$[\cos(nx)+i\sin(nx)][\cos(x)+i\sin(x)]=\cos[(n+1)x]+i\sin[(n+1)x]$$ at the beginning since you haven't proved it yet! Rather you should write
$$\begin{align} (\cos(x)+i\sin(x))^{n+1} &=(\cos(x)+i\sin(x))^n(\cos(x)+i\sin(x))\\&=(\cos(nx)+i\sin(nx))(\cos(x) + i\sin(x))\qquad\text{(by induction)}\\&=(\cos(nx)\cos (x)-\sin(nx)\sin (x))+i(\sin(nx)\cos (x)+\cos(nx)\sin (x))\\&=\ldots\end{align}$$where at each step you get closer to what you are trying to get to - but this way every step is mathematically rigourous.
A: The way you write is not good.
For the $n+1$ step, 
$$\begin{align}(\cos x+i\sin x)^{n+1}&=\color{red}{(\cos x+i\sin x)^n}(\cos x+i\sin x)\\&=\color{red}{\{\cos(nx)+i\sin(nx)\}}(\cos x+i\sin x)\\&=(\cos(nx)\cos x-\sin(nx)\sin x)+i(\sin(nx)\cos x+\cos(nx)\sin x)\\&=\cos(nx+x)+i\sin(nx+x)\\&=\cos\{(n+1)x\}+i\sin\{(n+1)x\}.\end{align}$$
