Let $$f : \mathbb{R} \to \mathbb{R}$$ 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 $$x, y \in \mathbb{R}$$. If $$f'(0) = \frac{1}{2}$$, then the value of $$24 f''\left(\frac{5\pi}{3}\right)$$ is:

We are given that $$(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$$ for all $$x, y \in \mathbb{R}$$.

Rearranging: $$\sin x \cos y \cdot f(2x+2y) - \sin x \cos y \cdot f(2x-2y) = \cos x \sin y \cdot f(2x+2y) + \cos x \sin y \cdot f(2x-2y)$$.

Collecting terms with $$f(2x+2y)$$ and $$f(2x-2y)$$:

$$f(2x+2y)(\sin x \cos y - \cos x \sin y) = f(2x-2y)(\sin x \cos y + \cos x \sin y)$$

Using the sine addition and subtraction formulas: $$\sin x \cos y - \cos x \sin y = \sin(x-y)$$ and $$\sin x \cos y + \cos x \sin y = \sin(x+y)$$.

So $$f(2x+2y) \sin(x-y) = f(2x-2y) \sin(x+y)$$.

This gives $$\frac{f(2x+2y)}{\sin(x+y)} = \frac{f(2x-2y)}{\sin(x-y)}$$.

Let $$u = x+y$$ and $$v = x-y$$. Then $$\frac{f(2u)}{\sin u} = \frac{f(2v)}{\sin v}$$ for all valid $$u, v$$. This means $$\frac{f(2u)}{\sin u}$$ is a constant, say $$c$$.

Therefore $$f(t) = c \sin\left(\frac{t}{2}\right)$$ for all $$t$$.

Now $$f'(t) = \frac{c}{2} \cos\left(\frac{t}{2}\right)$$. Given $$f'(0) = \frac{1}{2}$$:

$$\frac{c}{2} \cos(0) = \frac{1}{2}$$, so $$\frac{c}{2} = \frac{1}{2}$$, giving $$c = 1$$.

Therefore $$f(t) = \sin\left(\frac{t}{2}\right)$$, $$f'(t) = \frac{1}{2}\cos\left(\frac{t}{2}\right)$$, and $$f''(t) = -\frac{1}{4}\sin\left(\frac{t}{2}\right)$$.

Now $$f''\left(\frac{5\pi}{3}\right) = -\frac{1}{4}\sin\left(\frac{5\pi}{6}\right) = -\frac{1}{4} \cdot \frac{1}{2} = -\frac{1}{8}$$.

Therefore $$24 f''\left(\frac{5\pi}{3}\right) = 24 \cdot \left(-\frac{1}{8}\right) = -3$$.

Hence, the correct answer is Option B.