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 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))

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

Solve (1)/(cos theta-i sin theta)

The argument of 1-cos theta-i sin theta is

1 + cos theta + i sin theta | =?

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

If theta=(pi)/(6) then the 10^(th) term of the series : 1+(cos theta+i sin theta)^(1)+(cos theta+i sin theta)^(2)+

If theta in R - {(4n + 1) pi/4 , n in I} , then range of (cos 3 theta + sin 3 theta)/(cos theta - sin theta) is :