If y = cos^2left(frac{5x}{2}right) - sin^2lef | vedclass

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(A) We know the trigonometric identity $\cos^2 	heta - \sin^2 	heta = \cos(2	heta)$.Given $y = \cos^2\left(\frac{5x}{2}ight) - \sin^2\left(\frac{

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$-25y$

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(A) We know the trigonometric identity $\cos^2 	heta - \sin^2 	heta = \cos(2	heta)$.Given $y = \cos^2\left(\frac{5x}{2}ight) - \sin^2\left(\frac{5x}{2}ight)$,we can substitute $	heta = \frac{5x}{2}$.Thus,$y = \cos\left(2 	imes \frac{5x}{2}ight) = \cos(5x)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = -\sin(5x) 	imes \frac{d}{dx}(5x) = -5\sin(5x)$.Differentiate again with respect to $x$ to find the second derivative:$\frac{d^2y}{dx^2} = -5 	imes \cos(5x) 	imes \frac{d}{dx}(5x) = -5 	imes 5 \cos(5x) = -25\cos(5x)$.Since $y = \cos(5x)$,we have $\frac{d^2y}{dx^2} = -25y$.

If $y = \cos^2\left(\frac{5x}{2}\right) - \sin^2\left(\frac{5x}{2}\right)$,then $\frac{d^2y}{dx^2} =$

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