Question

sin2xcos2x=1/2sin4x"
"cos^2x/2-sin^2x/2=cosx

Answer 1

Sin2α = 2sinα • cosα
sin2x • cos2x = (2sin2x • cos2x)/2 = (sin(2 • 2x))/2 = (1/2) • sin4x
(1/4) • sin4x = (1/4) • sin4x

cos2α = (cosα)²– (sinα)²
(cos(x/2))²– (sin (x/2))² = cos(2 • (x/2)) = cosx
cosx = cosx