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OpenStudy (anonymous):

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

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OpenStudy (jamesj):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Same :(

OpenStudy (anonymous):

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

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