The sum of the series `sintheta+sin((n-4)/(n-2))theta+sin((n-6)/(n-2))theta+........n` terms is equal to

sin((n+1)/(2))theta-sin((n-1)/(2))theta=sin theta

(sintheta+sin3theta+.......n terms)/(costheta+cos3theta+.......n terms) is equal to

Sum to terms of the series sintheta sin2theta + sin2thetasin3theta + sin3theta sin4theta +..... is equal to

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

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)

If n is even then sin theta + sin ( pi + theta ) + sin (2 pi + theta) + ...........n terms =

The sum S= sin theta + sin 2 theta + …+ sin n theta, equals

The sum S= sin theta + sin 2 theta + …+ sin n theta, equals

The value of sintheta + sin3theta +sin5theta + "….."+ sin(2n-1)theta is

The sum of the series, sin theta sec(3theta)+sin 3theta sec(3^(2)theta)+sin(3^(2)theta)sec(3^(3)theta)+……. upto n terms, is