The angle made by the tangent at theta = pi / | vedclass

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(A) Given the parametric equations of the curve: $x = a(	heta + \sin 	heta)$ and $y = a(1 - \cos 	heta)$. First,we find the derivatives with respec

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$\pi / 6$

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(A) Given the parametric equations of the curve: $x = a(	heta + \sin 	heta)$ and $y = a(1 - \cos 	heta)$. First,we find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = a(1 + \cos 	heta)$ $\frac{dy}{d	heta} = a \sin 	heta$ Now,the slope of the tangent is given by: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \sin 	heta}{a(1 + \cos 	heta)} = \frac{\sin 	heta}{1 + \cos 	heta}$ Using the trigonometric identity $\sin 	heta = 2 \sin(	heta/2) \cos(	heta/2)$ and $1 + \cos 	heta = 2 \cos^2(	heta/2)$: $\frac{dy}{dx} = \frac{2 \sin(	heta/2) \cos(	heta/2)}{2 \cos^2(	heta/2)} = 	an(	heta/2)$ At $	heta = \pi / 3$: $\frac{dy}{dx} = 	an(\frac{\pi/3}{2}) = 	an(\pi / 6)$ Since the slope $m = 	an \psi$,where $\psi$ is the angle with the $X$-axis: $	an \psi = 	an(\pi / 6) \Rightarrow \psi = \pi / 6$.

The angle made by the tangent at $\theta = \pi / 3$ on the curve $x = a(\theta + \sin \theta)$,$y = a(1 - \cos \theta)$ with the $X$-axis is

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