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

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

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

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

cos alpha cos 2 alpha cos 4 alpha cos 8 alpha = (1)/(16), if alpha = 24 ^(@)

1/(cos alpha + cos 3 alpha)+ 1/(cos alpha + cos 5 alpha)+ 1/(cos alpha + cos 7 alpha) +....+ 1/(cos alpha +cos(2n+1) alpha) = 1/2 "cosec" alpha[tan (n+1) alpha - tan alpha ]