If x = 2 sin theta - sin 2 theta and y = 2 co | vedclass

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(D) Given $x = 2 \sin 	heta - \sin 2 	heta$ and $y = 2 \cos 	heta - \cos 2 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}

Vedclass · Accepted Answer

$\frac{3}{8}$

Vedclass · Answer

(D) Given $x = 2 \sin 	heta - \sin 2 	heta$ and $y = 2 \cos 	heta - \cos 2 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = 2 \cos 	heta - 2 \cos 2 	heta = 2(\cos 	heta - (2 \cos^2 	heta - 1)) = 2(1 + \cos 	heta - 2 \cos^2 	heta) = 2(1 - \cos 	heta)(1 + 2 \cos 	heta)$. Alternatively,using sum-to-product formulas: $\frac{dx}{d	heta} = 2(\cos 	heta - \cos 2 	heta) = 4 \sin \frac{3	heta}{2} \sin \frac{	heta}{2}$. $\frac{dy}{d	heta} = -2 \sin 	heta + 2 \sin 2 	heta = 2(\sin 2 	heta - \sin 	heta) = 2(2 \sin 	heta \cos 	heta - \sin 	heta) = 2 \sin 	heta (2 \cos 	heta - 1)$. Using sum-to-product: $\frac{dy}{d	heta} = 2(\sin 2 	heta - \sin 	heta) = 4 \cos \frac{3	heta}{2} \sin \frac{	heta}{2}$. Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{4 \cos(3	heta/2) \sin(	heta/2)}{4 \sin(3	heta/2) \sin(	heta/2)} = \cot \frac{3	heta}{2}$. Now,differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{d	heta} \left( \cot \frac{3	heta}{2} ight) \cdot \frac{d	heta}{dx} = -\frac{3}{2} \csc^2 \frac{3	heta}{2} \cdot \frac{1}{dx/d	heta}$. At $	heta = \pi$,$\frac{dx}{d	heta} = 2 \cos \pi - 2 \cos 2 \pi = 2(-1) - 2(1) = -4$. $\frac{d^2y}{dx^2} = -\frac{3}{2} \csc^2 \frac{3\pi}{2} \cdot \frac{1}{-4} = -\frac{3}{2} (-1)^2 \cdot \left( -\frac{1}{4} ight) = \frac{3}{8}$.

If $x = 2 \sin \theta - \sin 2 \theta$ and $y = 2 \cos \theta - \cos 2 \theta$ where $\theta \in [0, 2 \pi]$,then find the value of $\frac{d^{2} y}{dx^{2}}$ at $\theta = \pi$.

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