If x = 2sinθ - 2θ and y = 2cosθ - 2θ, θ ∈ [0, 2π], then d ^ 2 y d x...

$\dfrac{dx}{d\theta} = 2cos\theta - 2\cos 2\theta$

$\dfrac{dy}{d\theta} = - 2sin\theta + 2\sin 2\theta$

$∴ \dfrac{dy}{dx} = \dfrac{\sin 2\theta - sin\theta}{cos\theta - \cos 2\theta}$

$= \dfrac{2sin\dfrac{\theta}{2} . cos\dfrac{3\theta}{2}}{2sin\dfrac{\theta}{2} . sin\dfrac{3\theta}{2}} = cot\dfrac{3\theta}{2}$

$\dfrac{{d}^{2}y}{d{x}^{2}} = \dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right)\dfrac{d\theta}{dx} = - \dfrac{3}{2}{cosec}^{2}\dfrac{3\theta}{2} . \dfrac{d\theta}{dx}$

$⇒ \dfrac{{d}^{2}y}{d{x}^{2}} = \dfrac{- \dfrac{3}{2}{cosec}^{2}\dfrac{3\theta}{2}}{2\left(cos\theta - \cos 2\theta\right)}$

⇒ {\left\dfrac{{d}^{2}y}{d{x}^{2}}\right|}_{\theta = \pi} = \dfrac{3}{4\left(- 1 - 1\right)} = \dfrac{3}{8}