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 t

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

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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: $\sin y = 3\sin t - 4\sin^3 t$. 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 $3(1-4x^2)$.

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

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