Question

sin3θ=cos^2θsinθ-sin^3θ verify the identity

Answer 1

This identity as written is false.

The right hand side, RHS =
cos^2θ sinθ-sin^3θ = sinθ ( cos^2 θ - sin^2 θ)
= sinθ . cos 2θ

which doesn't equal the LHS = sin 3θ.

For example, set θ = 60 degrees (or pi/3)

then sin (3θ) = sin (180) = 0

but sinθ . cos 2θ is not = 0.

Answer 2

ahh - thats why i could'nt figure it out!

Answer 3

Same :(

Answer 4

Because I'm lazy, I'm doing this proof with S for sintheta and C for cos,
S3=S(2+1)=S*C2+S2*C
C2=C^2-S^2 and S2=2S*C
Subbing in gives S3=S*(C^2-S^2)+2S*C^2
sin3θ=3cos^2θsinθ-sin^3θ