If y = sin x sin 3x,then find the n^{th} deri | vedclass

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(B) Given $y = \sin x \sin 3x$. Using the trigonometric identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$,we have: $y = \frac{1}{2} [\cos(3x - x)

Vedclass · Accepted Answer

$\frac{1}{2} \left[ 2^n \cos \left( 2x + \frac{n\pi}{2} \right) - 4^n \cos \left( 4x + \frac{n\pi}{2} \right) \right]$

Vedclass · Answer

(B) Given $y = \sin x \sin 3x$. Using the trigonometric identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$,we have: $y = \frac{1}{2} [\cos(3x - x) - \cos(3x + x)] = \frac{1}{2} [\cos 2x - \cos 4x]$. We know that the $n^{th}$ derivative of $\cos(ax + b)$ is $a^n \cos(ax + b + \frac{n\pi}{2})$. Applying this to each term: $y_n = \frac{d^n}{dx^n} \left( \frac{1}{2} \cos 2x - \frac{1}{2} \cos 4x \right) = \frac{1}{2} [2^n \cos(2x + \frac{n\pi}{2}) - 4^n \cos(4x + \frac{n\pi}{2})]$. Thus,the correct option is $B$.

If $y = \sin x \sin 3x$,then find the $n^{th}$ derivative $y_n$.

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