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

show that cos2thetacos(theta/2)-cos3thetacos((9theta)/2)=sin5thetasin((5theta)/2)

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

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

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

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

Prove that : sin theta/cos(3theta) +sin (3theta)/cos (9theta) + sin(9theta)/cos( 27theta )= 1/2(tan27 theta -tan theta)

2(cos theta+cos2 theta)+(1+2cos theta)sin2 theta=2sin 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=

(sin theta)/(cos(3theta))+(sin(3theta))/(cos(9theta))+(sin(9theta))/(cos(27theta))+(sin(27theta))/(cos(81theta))=