frac{d^2}{dx^2}(2cos x cos 3x) = | vedclass

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(D) Given $y = 2\cos x \cos 3x$. Using the trigonometric identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$,we have: $y = \cos(x+3x) + \cos(x-3x) = \co

Vedclass · Accepted Answer

$-2^2(\cos 2x + 2^2\cos 4x)$

Vedclass · Answer

(D) Given $y = 2\cos x \cos 3x$. Using the trigonometric identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$,we have: $y = \cos(x+3x) + \cos(x-3x) = \cos 4x + \cos 2x$. Now,differentiate with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\cos 4x + \cos 2x) = -4\sin 4x - 2\sin 2x$. Differentiate again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(-4\sin 4x - 2\sin 2x) = -16\cos 4x - 4\cos 2x$. Factoring out $-4$: $\frac{d^2y}{dx^2} = -4(\cos 2x + 4\cos 4x) = -2^2(\cos 2x + 2^2\cos 4x)$.

$\frac{d^2}{dx^2}(2\cos x \cos 3x) = $

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