Question

Prove that ${(1+\cos \theta +i\sin \theta )}^n+{(1+\cos \theta -i\sin \theta )}^n=2^{n+1}{\cos }^n\frac{\theta }{2}.\cos \frac{n\theta }{2}$, here $n\in N$

Answer 1

${(1+\cos \theta +i\sin \theta )}^n={[2{\cos }^2\frac{\theta }{2}+i2\sin \frac{\theta }{2}\cos \frac{\theta }{2}]}^n={\left[2\cos \frac{\theta }{2}(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})\right]}^n$
$=2^n{\cos }^n\frac{\theta }{2}(\cos \frac{n\theta }{2}+i\sin \frac{n\theta }{2})$ ....(1)
Replacing $i$ by $-i$ we get,
${(1+\cos \theta -i\sin \theta )}^n=2^n{\cos }^n\frac{\theta }{2}(\cos \frac{n\theta }{2}-i\sin \frac{n\theta }{2})$ ....(2)
Adding equations (1) and (2), we get

${(1+\cos \theta +i\sin \theta )}^n+{(1+\cos \theta -i\sin \theta )}^n=2^n{\cos }^n\frac{\theta }{2}\left(2\cos \frac{n\theta }{2}\right)$
$=2^{n+1}{\cos }^n\frac{\theta }{2}.\cos \frac{n\theta }{2}$
Hence, proved.