show that sin3theta-cos2theta =(1-sin theta)(4sin squared theta + 2sin theta -1)
I will use x instead of Ø
sin 3x - cos 2x = (1-sinx)(4sin^2 x + 2sinx - 1)
RS = 4sin^2 x + 2sinx - 1 - 4sin^3 x + 2sin^2 x + sinx
= -4sin^3 x + 2sin^2 x + 3sinx - 1
LS = sin(2x+x) - cos 2x
= (sin 2x)(cosx) + (cos 2x)(sinx) - (1 - 2sin^2 x)
= 2sinxcosxcosx + sinx(1-2sin^2 x) - 1 + 2sin^2 x
= 2sinxcos^2 x + sinx - 2sin^3 x - 1 + 2sin^2 x
= 2sinx(1 - sin^2 x) + sinx - 2sin^3 x - 1 + 2sin^2 x
= 2sinx - 2sin^3 x + sinx - 2sin^3 x - 1 + 2sin^2 x
= -4sin^3 x + 2sin^2 x + 3sinx - 1
= RS
YEAHHH