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+…`

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+...

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

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

Find the sum oif the series sintheta- (sin2theta)/(2!)+(sin3theta)/(3!)-… to oo

Prove that sin theta cos^(3)theta - cos theta sin^(3) theta = (1)/(4) sin 4 theta .

(sin3 theta-cos3 theta) / (sin theta + cos theta) + 1 =

sin theta cos ^ (3) theta-cos theta sin ^ (3) theta = (1) / (4) sin4 theta

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

(sin theta+cos theta)(1-sin theta cos theta)=sin^3 theta+cos^3 theta