" 5."cos2 theta*cos(theta)/(2)-cos3 theta*cos(9 theta)/(2)=sin5 theta*sin(5 theta)/(2)

cos2 theta*cos(theta/(2))-cos3 theta*cos((9theta)/2)=sin5 theta*sin((5theta)/(2))

show that cos2 theta cos((theta)/(2))-cos3 theta cos(9(theta)/(2))=sin5 theta sin(5(theta)/(2))

Prove that: cos 2theta"cos"(theta)/(2)-cos 3theta cos((9theta)/2)="sin" 5theta "sin"((5theta)/2) .

Prove that: cos2theta.costheta/2-cos3theta.cos(9theta)/(2)=sin5theta.sin(5theta)/(2)

Prove that : cos2 theta cos frac (theta)(2)- cos3 theta cos frac (9 theta)(2)= sin 5 theta sin frac (5theta)(2).

cos (2 theta) cos ((theta) / (2)) - cos (3 theta) cos ((9 theta) / (2)) = sin (5 theta) sin ((5 theta) / (2))

cos theta-cos3 theta+cos5 theta-cos7 theta=4sin theta sin4 theta cos2 theta

(cos3 theta+2cos5 theta+cos7 theta)/(cos theta+2cos3 theta+cos5 theta)=cos2 theta-sin2 theta tan3 theta

2cos theta-cos3 theta-cos5 theta-16cos^(3)theta sin^(2)theta=

2cos theta-cos3 theta-cos5 theta-16cos^(3)theta sin^(2)theta=