सिद्ध करे की
`cos (npi + theta)=(-1)^(n) cos theta` जहाँ n कोई पूर्णांक 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) .

Prove that sin (n + 1) theta sin (n - 1) theta + cos (n + 1) theta cos (n - 1) theta = cos 2 theta , n in ZZ .

Statement-1: sin (pi+theta)=-sin theta Statement-2: sin(npi+theta)=(-1)^(n) sin theta , n in N .

Statement-1: sin (pi+theta)=-sin theta Statement-2: sin(npi+theta)=(-1)^(n) sin theta , n in N .

Prove that : (2 cos 2^n theta + 1)/(2 cos theta +1) = (2 cos theta -1) (2 cos 2theta -1) (2 cos 2^2 theta -1) … (2 cos 2^(n-1) theta -1)

Prove that : (2 cos 2^n theta + 1)/(2 cos theta +1) = (2 cos theta -1) (2 cos 2theta -1) (2 cos 2^2 theta -1) … (2 cos 2^(n-1) theta -1)

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

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