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 $\sin(x)=-\sin(-x)$

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.