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

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

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

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

(cos alpha + (cos ((2 pi) / (3) + alpha)) + cos ((4 pi) / (3) + 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 ]

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

If alpha = (pi)/(13), Prove that cos alpha cos 2alpha cos 3alpha cos 4alpha cos 5alpha cos 6alpha = (1)/(64) .