`(1+"cos"theta + "i sin"theta)^(n) + (1+"cos"theta - "i sin"theta)^(n)` का मान है

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

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) .

(1+cos theta+i sin theta)^(10)+(1+cos theta-i sin theta)^(10)=

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

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

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)

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

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

" (i) "(cos theta+sin theta)/(cos theta-sin theta)