If $x = 2\sin \theta - \sin 2\theta$ and $y = 2\cos \theta - \cos 2\theta$,
$\theta \in \left[ {0,2\pi } \right]$, then ${{{d^2}y} \over {d{x^2}}}$ at $\theta$ = $\pi$ is :

$x = 2\sin \theta - \sin 2\theta$ <br><br>$\Rightarrow$ ${{dx} \over {d\theta }}$ = $2\cos \theta - 2\cos 2\theta$ <br><br>$y = 2\cos \theta - \cos 2\theta$ <br><br>$\Rightarrow$ ${{dy} \over {d\theta }}$ = –2sin$\theta$ + 2sin2$\theta$ <br><br>${{dy} \over {dx}} = {{{{dy} \over {d\theta }}} \over {{{dx} \over {d\theta }}}}$ = ${{\sin 2\theta - \sin \theta } \over {\cos \theta - \cos 2\theta }}$ <br/><br/>This expression is simplified by recognizing that $\sin 2 \theta = 2\sin \theta \cos \theta$ and $\cos 2 \theta = 2\cos^2 \theta -1$, leading to the expression: <br/><br/>$$ \frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}$$ <br/><br/>This is the rate of change of y with respect to x, as a function of θ. <br/><br/>Next, we differentiate this function with respect to θ to find $\frac{d^2 y}{dx^2}$, yielding : <br/><br/>$$ \frac{d^2 y}{dx^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x}=-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}$$ <br/><br/>Here, $\frac{d \theta}{d x}$ is the reciprocal of $\frac{dx}{d\theta}$, so the equation becomes : <br/><br/>$$ \frac{d^2 y}{dx^2}=\frac{-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2}}{2\left(\cos \theta-\cos2 \theta\right)}$$ <br/><br/>Finally, we evaluate this expression at $\theta = \pi$, yielding : <br/><br/>$\frac{d^2 y}{dx^2}(\pi)=\frac{-3}{4(-1-1)}=\frac{3}{8}$ <br/><br/>Therefore, the correct answer is (A) $\frac{3}{8}$. <br/><br/><b>Alternate Method :</b> <br/><br/>First, let's find the derivatives of x and y with respect to θ : <br/><br/>1) $\frac{dx}{d\theta} = 2\cos \theta - 2\cos 2\theta$ <br/><br/>2) $\frac{dy}{d\theta} = -2\sin \theta + 2\sin 2\theta$ <br/><br/>We know that $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$ <br/><br/>So, we substitute 1) and 2) into this equation : <br/><br/>We have <br/><br/>$$\frac{dy}{dx} = \frac{-2\sin \theta + 2\sin 2\theta}{2\cos \theta - 2\cos 2\theta} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$$ <br/><br/>For simplification, let's denote the numerator as $N = \sin 2\theta - \sin \theta$ and the denominator as $D = \cos \theta - \cos 2\theta$. <br/><br/>We have to compute $\frac{d}{d\theta}(\frac{dy}{dx})$ which is $\frac{d}{d\theta}(\frac{N}{D})$. <br/><br/>We can use the quotient rule for differentiation, which states that if we have a function of the form $\frac{u}{v}$, then its derivative is given by $\frac{vu' - uv'}{v^2}$. <br/><br/>So here, $N' = 2\cos 2\theta - \cos \theta$ and $D' = -\sin \theta + 2\sin 2\theta$. <br/><br/>Applying the quotient rule : <br/><br/>$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{D N' - N D'}{D^2}$ <br/><br/>Substituting the expressions for $N'$, $D'$, $N$, and $D$ we get : <br/><br/>$$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$ <br/><br/>Now $$ \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x} $$ <br/><br/>= $$\frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$$\times \frac{1}{(2 \cos \theta-2 \cos 2 \theta)}$ <br/><br/>$$ \begin{aligned} \left.\therefore \frac{d^2 y}{d x^2}\right|_{\theta=\pi} & =\frac{(-1-1)(2+1)-(0-0)(-0+0)}{2(-1-1)^3} \\\\ & =\frac{-2 \times 3}{-2 \times 8}=\frac{3}{8} \end{aligned} $$