prove that
sin (60+x) sin (420-x) = 1+ 2cos2x/2
I am pretty sure your right side is incorrect
let's see ...
LS = (sin60cosx + cos60sinx)(sin420cosx - cos420sinx)
= ( √3/2 cosx + 1/2 sinx)( √3/2 cosx - 1/2 sinx)
= (3/4)cos^2 x - (1/4)sin^2 x
= (1/4)( 3cos^2 x - sin^2 x)
= (1/4)( 3cos^2 x - (1 - cos^2 x) )
= (1/4)(4 cos^2 x - 1)
= cos^2 x - 1/4
testing with some weird angle
let x = 19.78°
LS = sin(79.78) sin(400.22) = appr .63548
My RS = cos^2 19.78 - .25 = appr. .63548
your original RS ≠ LS