Step 1:
Now $(1+\cos\theta+i\sin\theta)=2\cos^2\large\frac{\theta}{2}$$+2i\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
$\qquad\qquad\;\;\;\;\;\qquad\qquad=2\cos\large\frac{\theta}{2}$$(\cos\large\frac{\theta}{2}$$+i\sin\large\frac{\theta}{2})$
$(1+\cos\theta-i\sin\theta)=(1+\cos\theta+i\sin\theta)=2\cos\large\frac{\theta}{2}$$(\cos\large\frac{\theta}{2}$$-i\sin\large\frac{\theta}{2})$
Step 2:
LHS=$[2\cos\large\frac{\theta}{2}$$(\cos\large\frac{\theta}{2}$$+i\sin\large\frac{\theta}{2}]^n+$$[2\cos\large\frac{\theta}{2}($$\cos\large\frac{\theta}{2}$$-i\sin\large\frac{\theta}{2}]^n$
$\quad=2^n\cos^n\large\frac{\theta}{2}$$[\cos\large\frac{n\theta}{2}+$$i\sin\large\frac{n\theta}{2}+$$\cos\large\frac{n\theta}{2}$$-i\sin\large\frac{n\theta}{2}]$
$\quad=2^n\cos^n\large\frac{\theta}{2}$$.2\cos\large\frac{n\theta}{2}$
$\quad=2^{n+1}\cos^n\large\frac{\theta}{2}$$.\cos\large\frac{n\theta}{2}$
$\quad$=RHS.
Hence proved.