De Moivre's formula proof step ($\cos \phi + i\sin \phi)^n = \cos (n \phi) + i\sin (n \phi)$
Saw this in the De Moivre's formula proof and some other calculations involving complex numbers, but I do not understand why the equation is true.
 A: This is DeMoivre's formula:
$$(\cos\phi+i\sin\phi)^n=\cos n\phi+i\sin n\phi$$
which may be proven by induction:
First of all, this is true for $n=1$.  Now, we prove that if it is true for $k$, then it is true for $k+1$:
$$\begin{align}(\cos\phi+i\sin\phi)^{k+1}&=(\cos\phi+i\sin\phi)(\cos\phi+i\sin\phi)^k\\&=(\cos\phi+i\sin\phi)(\cos k\phi+i\sin k\phi)\\&=\cos\phi\cos k\phi-\sin\phi\sin k\phi+i(\sin\phi\cos k\phi+\cos\phi\sin k\phi)\\&\stackrel{(*)}=\cos(k+1)\phi+i\sin(k+1)\phi\end{align}$$
That is, if it holds for $k=1$, it holds true for $k+1=2$, and if it holds for $k=2$, it holds true for $k+1=3$, etc.
$(*)$ we used the sum of angles formula.
Likewise, this extends to negative $n$:
$$\begin{align}(\cos\phi+i\sin\phi)^{-n}&=((\cos\phi+i\sin\phi)^n)^{-1}\\&=(\cos n\phi+i\sin n\phi)^{-1}\\&=\frac1{\cos n\phi+i\sin n\phi}\\&=\frac1{\cos n\phi+i\sin n\phi}\frac{\cos n\phi-i\sin n\phi}{\cos n\phi-i\sin n\phi}\\&=\frac{\cos n\phi-i\sin n\phi}{\cos^2n\phi+\sin^2n\phi}\\&\stackrel{(**)}=\cos n\phi-i\sin n\phi\\&\stackrel{(***)}=\cos(-n\phi)+i\sin(-n\phi)\end{align}$$
$(**)$ we used the pythagorean identity $\sin^2+\cos^2=1$
$(***)$ we used symmetry formulas $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$
A: DeMoivre's formula.
$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
De Moivre's formula can easily be derived from Euler's formula.
$ e^{i \theta} =(\cos \theta + i \sin \theta)$
$ e^{i n \theta} =(\cos n \theta + i \sin n \theta)$
And you can use induction to prove it. Here's link with full details.
A: The "identity is wrong.  Consider $n=2$.  Then the left hand side is 
$$
\cos^2 \theta + \sin^2 \theta + 2\cos \theta \sin\theta = 1 + 2\cos \theta \sin\theta = 1 + \sin(2\theta)
$$ 
And the right hand side is 
$$
\cos (2\theta) + \sin(2\theta) 
$$
So unless $\cos (2\theta) = 1$ the equality fails.
