PreCalculus
posted by Peter on .
If θ represents an angle such that sin2θ = tanθ  cos2θ, then sin θ  cosθ =
A. √2
B. 0
C. 1
D. 2√2
What equation can I use to solve this problem?

sin2A + cos2A  sinA/cosA = 0
2sinAcosA + 1  2sin^2 A  sinA/cosA = 0
2sinAcos^2 A + cosA  2sin^2 A cosA  sinA = 0
2sinAcosA(cosA  coA) + (cosA  sinA) =
(cosA  sinA)(2sinAcosA + 1) = 0
cosA  sinA = 0 or ......
got it! That's all they asked for.
cosA  sinA = 0 
Oh...okay. Thanks for the help.