Let f: R rightarrow R be a twice differentiab | vedclass

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(B) Given the equation: $(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$.Rearranging the terms,we get: $f(2x+2y)(\sin x \

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$-3$

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(B) Given the equation: $(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$.Rearranging the terms,we get: $f(2x+2y)(\sin x \cos y - \cos x \sin y) = f(2x-2y)(\sin x \cos y + \cos x \sin y)$.Using the trigonometric identities $\sin(A-B) = \sin A \cos B - \cos A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we have: $f(2x+2y)\sin(x-y) = f(2x-2y)\sin(x+y)$.This can be written as: $\frac{f(2x+2y)}{\sin(x+y)} = \frac{f(2x-2y)}{\sin(x-y)}$.Let $2x+2y = m$ and $2x-2y = n$. Then $x+y = m/2$ and $x-y = n/2$.The equation becomes: $\frac{f(m)}{\sin(m/2)} = \frac{f(n)}{\sin(n/2)} = K$ (a constant).Thus,$f(x) = K \sin(x/2)$.Differentiating with respect to $x$,we get $f'(x) = \frac{K}{2} \cos(x/2)$.Given $f'(0) = 1/2$,we have $\frac{K}{2} \cos(0) = 1/2$,which implies $K = 1$.So,$f(x) = \sin(x/2)$.Then $f'(x) = \frac{1}{2} \cos(x/2)$ and $f''(x) = -\frac{1}{4} \sin(x/2)$.We need to find $24f''\left(\frac{5\pi}{3}ight) = 24 \left(-\frac{1}{4} \sin\left(\frac{5\pi}{6}ight)ight)$.Since $\sin(5\pi/6) = 1/2$,the value is $24 \left(-\frac{1}{4} 	imes \frac{1}{2}ight) = 24 \left(-\frac{1}{8}ight) = -3$.

Let $f: R \rightarrow R$ be a twice differentiable function such that $(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$ for all $x, y \in R$. If $f'(0) = \frac{1}{2}$,then the value of $24f''\left(\frac{5\pi}{3}\right)$ is:

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