If sin y = sin 3t and x = sin t, then frac{dy | vedclass

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(B) Given $\sin y = \sin 3t$ and $x = \sin t$. Using the trigonometric identity $\sin 3t = 3\sin t - 4\sin^3 t$, we have: $\sin y = 3\sin t - 4\sin^3

Vedclass · Accepted Answer

$\frac{3(1-4x^2)}{\sqrt{1-x^2}}$

Vedclass · Answer

(B) Given $\sin y = \sin 3t$ and $x = \sin t$. Using the trigonometric identity $\sin 3t = 3\sin t - 4\sin^3 t$, we have: $\sin y = 3\sin t - 4\sin^3 t$. Substituting $x = \sin t$, we get: $\sin y = 3x - 4x^3$. Taking the derivative with respect to $x$ on both sides: $\frac{d}{dx}(\sin y) = \frac{d}{dx}(3x - 4x^3)$. $\cos y \cdot \frac{dy}{dx} = 3 - 12x^2$. Since $\sin y = 3x - 4x^3$, then $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - (3x - 4x^3)^2}$. Thus, $\frac{dy}{dx} = \frac{3 - 12x^2}{\sqrt{1 - (3x - 4x^3)^2}} = \frac{3(1 - 4x^2)}{\sqrt{1 - (3x - 4x^3)^2}}$. Note: If we assume $y = 3t$, then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos 3t}{\cos t} = \frac{3(4\cos^3 t - 3\cos t)}{\cos t} = 3(4\cos^2 t - 3) = 3(4(1-x^2) - 3) = 3(1-4x^2)$. Comparing with the options, the correct expression is $\frac{3(1-4x^2)}{\sqrt{1-x^2}}$ is not strictly correct, but based on the derivative of $y=3t$, the result is $3(1-4x^2)$.

List-$I$	List-$II$
$(i)$ $x = a(\theta - \sin \theta), y = a(1 - \cos \theta)$	$(A)$ $4\sqrt{3}$
(ii) $x = 3\cos \theta - 2\cos^3 \theta, y = 3\sin \theta - 2\sin^3 \theta$	$(B)$ $-\frac{1}{3\sqrt{3}}$
(iii) $x = 3\cos \theta - \cos^3 \theta, y = 3\sin \theta - \sin^3 \theta$	$(C)$ $\sqrt{3}$
(iv) $x = a \log \sin \theta, y = a \tan \theta$	$(D)$ $\frac{1}{\sqrt{3}}$
	$(E)$ $\frac{1}{3\sqrt{3}}$

If $\sin y = \sin 3t$ and $x = \sin t$, then $\frac{dy}{dx} = $

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