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`cos 2 theta cos .(theta)/(2)- cos 3theta .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))

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)

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

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

(cos theta + sin theta )^(2) + (cos theta - sin theta )^(2) =

(sin theta-cos theta)^(2)*(sin theta+cos theta)^(2)=

(cos theta + sin theta) ^ (2) + (cos theta-sin theta) ^ (2) =

2 cos theta - cos 3 theta - cos 5 theta - 16 cos^(3) theta sin^(2) theta =