" 6."cos alpha cos(60^(@)-alpha)cos(60^(@)+alpha)=(1)/(4)cos3 alpha" ."

4sin(420^(@)-alpha)cos(60^(@)+alpha)=

cos alpha cos(60^(@) - alpha ) cos (60^(@) + alpha ) = 1/4 cos 3 alpha.

Prove that cos alpha cos (120^(@) + alpha) cos (240^(@) + alpha) = 1/4 cos 3 alpha

cos alpha*cos60-alpha*cos60+alpha-(1)/(4)cos3 alpha

cos4 alpha=1-8cos^(2)alpha+8cos^(4)alpha

4 sin (420^(@) - alpha ) cos (60^@ + alpha ) =

Prove that, cos^(3) alpha + cos^(3) (120^(@) + alpha) + cos^(3) (240^(@) + alpha) = 3/4 cos 3 alpha

( cos 2 alpha)/( cos^(4) alpha - sin^(4) alpha ) -(cos^(4) alpha + sin^(4) alpha )/(2-sin^(2) 2alpha ) =

(1)/(cos alpha+cos3 alpha)+(1)/(cos alpha+cos5 alpha)...(.1)/(cos alpha+cos(2n+1)alpha)

Assertion (A):cos^(2)theta+cos^(2)(60^(@)-theta)+cos^(2)(60^(@)+theta)=(3)/(2)sin alpha-sin(120^(@)-alpha)+sin(120^(@)+alpha)=0