Are these the solutions to sin2x sinx = cosx? pi/2, 3pi/2, 3pi/...

cos2Acos2B + sin²(A-B) - sin²(A+B) = cos(2A+2B)

If sin(θ) = -1/√2, find sin2(θ). (Recall that sin2(θ) means (sin(θ))2.) sin2(θ) =

Find the exactly value of sin2(alpha) if cos(alpha) = 4/5 (alpha in Quadrant I)

ʃ sin² cos² dx

does sin2(x)=sin(2x)?

Did you solve sin2[sin^-1(pi/6)] and sin[2sin^-1(pi/6)] in the same way?

Sin2 theta =-1

Integrate sin² x dx

Sin²(theta)=0.25

Sin²(x)=0.25

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