" 4."sin^(2)n theta-sin^(2)(n-1)theta=sin^(2)theta

sin^(2)n theta-sin^(2)(n-1)theta=sin^(2)theta where n is constant and n!=0,1

Solve the equation sin^(2)ntheta - sin^(2)(n-1)theta = sin^(2)theta

Solve the equation sin^(2)ntheta - sin^(2)(n-1)theta = sin^(2)theta

sin^2 n theta- sin^2 (n-1)theta= sin^2 theta where n is constant and n != 0,1

sin^2 n theta- sin^2 (n-1)theta= sin^2 theta where n is constant and n != 0,1

Prove that sin^(2)theta+sin^(2)2 theta+sin^(2)3 theta+....+sin^(2)n theta=(n)/(2)-(sin n theta cos(n+1)theta)/(2sin theta)

If 4n theta=pi, then sin^(2)theta+sin^(2)3 theta+sin^(2)5 theta+....2n terms is equal to: (1)n,(2)2n*(3)3n(4)n^(2)

Prove that sin theta+sin3 theta+sin5 theta+....+sin(2n-1)theta=(sin^(2)n theta)/(sin theta)

1+sin^(2)theta,sin^(2)theta,sin^(2)thetacos^(2)theta,1+cos^(2)theta,cos^(2)theta4sin4 theta,4sin4 theta,1+4sin4 theta]|=0