Prove that

cos²x + cos² (x+pi/3)+ cos² (x-pi/3) = 3/2

Prove that sin20.sin40.sin60.sin80= 3/16.

1. LHS = Cos²x + Cos²(x+^/3) + Cos²(x-^/3)

= (1+cosx)1/2 + (1+cos(x+^/3))1/2 + (1+cos(x-^/3)) 1/2

= 1/2 ( 1+cosx + 1 +Cos(x+^/3) + 1 + Cos(x-^/3) )

= 1/2 ( 3 + cosx + Cos( x + 60 ) + Cos( x - 60)

. = 1/2 ( 3 + Cosx + 2 cosx cos120)

= 1/2 ( 3 + cosx + 2 cosx . -1/2) (since cos 120 = -1/2)

= 1/2 ( 3 + cosx - cosx)

= 3/2 = RHS

here , ^ means "pi"

2. L.H.S

= sin20.sin40.√3/2.sin80

=√3/4.sin20(2sin40.sin80)

=√3/4.sin20{cos40-cos120}

=√3/4.sin20{cos40+1/2}

=√3/8.sin20{2cos40+1}

=√3/8.{2cos40sin20+sin20}

=√3/8.{sin60-sin20+sin20}

=√3/8.sin60=√3/8.√3/2

=3/16=R.H.S