If n is a positive integer, prove the following `(1+cos theta+i sin theta)+(1+cos theta-i sin theta)^n=2^(n+1) cos^n (theta/2) cos ((n theta)/2)`

The value of (1+cos theta+i sin theta)^(n)+(1+cos theta-i sin theta)^n=

If n is an integer then show that (1 + cos theta + i sin theta)^(n) + (1 + cos theta - i sin theta )^(n) = 2^(n+1) cos ^(n) ( theta//2) cos ((n theta)/2) .

Prove that ((1+cos theta+i sin theta)/(1+cos theta-i sin theta))^(n)=cos n theta+i sin n theta

The value of (1+cos theta+i sin theta)^(n)+(1+cos theta-i sin theta)^(n)=

(cos theta + i sin theta) ^ (n) = cos n theta + i sin n theta

If ((1+cos theta+i sin theta)/(sin theta+i+i cos theta))^(4)=cos n theta+i sin n theta , then n=

[(1 + cos theta + i sin theta) / (sin theta + i (1 + cos theta))] ^ (4) = cos n theta + i sin n theta

Using De Moivre 's theorem prove that : ((1+cos theta +i sin theta)/(1+cos theta - i sin theta))^n= cos n theta + i sin n theta , where i = sqrt(-1)

cos theta + cos 2 theta + cos 3 theta + …. + cos { (n-1) theta}+ cos n theta=

If n is an integer, prove that, (i) cos(n pi + theta) = (-1)^(n) cos theta