Step 1:
Let $(1+i)=r(\cos\theta+i\sin\theta)$
Equating real and imaginary parts separately
$r\cos \theta=1$
$r\sin \theta=1$
Squaring and adding,
$r^2=2$
$r=\sqrt 2$
Step 2:
Here $\alpha=\tan^{-1}1=\large\frac{\pi}{4}$
Since $1+i$ is represented by a point in the quadrant 1,$\theta=\alpha=\large\frac{\pi}{4}$
$1+i=\cos\large\frac{\pi}{4}$$+i\sin\large\frac{\pi}{4}$
$1-i=1+i=\cos\large\frac{\pi}{4}$$-i\sin\large\frac{\pi}{4}$
Step 3:
LHS=$[\sqrt 2(\cos\large\frac{\pi}{4}$$+i\sin\large\frac{\pi}{4})^n]$$+[\sqrt 2(\cos\large\frac{\pi}{4}$$-i\sin\large\frac{\pi}{4})^n],$$n\in N$
$\quad\;\;=2^{\large\frac{n}{2}}[\cos\large\frac{n\pi}{4}$$+i\sin\frac{n\pi}{4}+\cos\large\frac{n\pi}{4}$$-i\sin\frac{n\pi}{4}]$
$\quad\;\;=2^{\large\frac{n}{2}}2\cos\large\frac{n\pi}{4}$
$\quad\;\;=2^{\large\frac{n+2}{2}}\cos\large\frac{n\pi}{4}$=RHS.
Hence proved.