MathematicsHard63×since 2002Q3695If x=2sinθ−sin2θx = 2\sin \theta - \sin 2\thetax=2sinθ−sin2θ and y=2cosθ−cos2θy = 2\cos \theta - \cos 2\thetay=2cosθ−cos2θ, θ∈[0,2π]\theta \in \left[ {0,2\pi } \right]θ∈[0,2π], then d2ydx2{{{d^2}y} \over {d{x^2}}}dx2d2y at θ\thetaθ = π\piπ is :A38{3 \over 8}83B32{3 \over 2}23C34{3 \over 4}43D-34{3 \over 4}43Check answerSkip