Sum ot n terms the series `sin^3theta+sin^3 3theta+sin^3 5 theta+…`

sin theta + sin3 theta + sin5 theta = 0

Prove that sin theta sec3 theta=(1)/(2)[tan3 theta-tan theta] and hence find the sum to n terms of the series sin theta sec3 theta+sin3 theta sec3^(2)theta+sin3^(2)theta sec3^(3)theta+...

Find the sum to n terms of the series sin^3 theta +1/3 sin^3 3theta+1/3^2sin^3 9theta+1/3^3 sin^3 27theta+…

Find the sum to n terms of the series sin^3 theta +1/3 sin^3 3theta+1/3^2sin^3 9theta+1/3^3 sin^3 27theta+…

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

Sum to terms of the series sin theta sin2 theta+sin2 theta sin3 theta+sin3 theta sin4 theta+... is equal to

Sum to n terms the series costheta-sin2theta-cos3theta+sin4theta+cos5theta-……

Sum to n terms the series costheta-sin2theta-cos3theta+sin4theta+cos5theta-……

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