5. `sin m theta=sin n theta`

Solve that equation : sin m theta+sin n theta=0

cos m theta-sin n theta=0

cos m theta=sin n theta then theta=

(sin4 theta+sin5 theta)/(cos3 theta+1)=sin(m theta)+sin(n theta) where m,n in N, then (m+n) equals.

If m and n are positive integers satisfying 1+cos2 theta+cos4 theta+cos6 theta+cos8 theta+cos10 theta=(cos m theta*sin n theta)/(sin theta) then (m+n) is equal to

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

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

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

Solution of the equation sin (sqrt(1+sin 2 theta))= sin theta + cos theta is (n in Z)