題目 4. When x=2sinθ−sin2θx=2\sin\theta-\sin2\thetax=2sinθ−sin2θ and y=2cosθ−cos2θy=2\cos\theta-\cos2\thetay=2cosθ−cos2θ, determine d2ydx2\dfrac{d^2y}{dx^2}dx2d2y at θ=π/3\theta=\pi/3θ=π/3. 解答
