If x=sin theta and y=sin(k theta), then (1-x^ | vedclass

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(C) Given $x=\sin 	heta$ and $y=\sin(k 	heta)$.First, find $y_1 = \frac{dy}{dx}$:$\frac{dy}{d	heta} = k \cos(k 	heta)$ and $\frac{dx}{d	heta} = \

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$-k^2$

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(C) Given $x=\sin 	heta$ and $y=\sin(k 	heta)$.First, find $y_1 = \frac{dy}{dx}$:$\frac{dy}{d	heta} = k \cos(k 	heta)$ and $\frac{dx}{d	heta} = \cos 	heta$.$y_1 = \frac{dy/d	heta}{dx/d	heta} = \frac{k \cos(k 	heta)}{\cos 	heta} \implies y_1 \cos 	heta = k \cos(k 	heta)$.Differentiating both sides with respect to $x$:$y_2 \cos 	heta - y_1 \sin 	heta \cdot \frac{d	heta}{dx} = -k^2 \sin(k 	heta) \cdot \frac{d	heta}{dx}$.Since $\frac{d	heta}{dx} = \frac{1}{\cos 	heta}$, we have:$y_2 \cos 	heta - y_1 \sin 	heta \cdot \frac{1}{\cos 	heta} = -k^2 \sin(k 	heta) \cdot \frac{1}{\cos 	heta}$.Multiplying by $\cos 	heta$:$y_2 \cos^2 	heta - y_1 \sin 	heta = -k^2 \sin(k 	heta)$.Since $1-x^2 = 1-\sin^2 	heta = \cos^2 	heta$ and $x = \sin 	heta$:$(1-x^2) y_2 - x y_1 = -k^2 y$.Comparing this with $(1-x^2) y_2 - x y_1 - \alpha y = 0$, we get $\alpha y = -k^2 y$, so $\alpha = -k^2$.

If $x=\sin \theta$ and $y=\sin(k \theta)$, then $(1-x^2) y_2-x y_1-\alpha y=0$, for $\alpha=$

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