Let $f:\mathbb{R}\to\mathbb{R}$ be a twice-differentiable function such that $(\sin x\cos y)\bigl[f(2x+2y)-f(2x-2y)\bigr]=(\cos x\sin y)\bigl[f(2x+2y)+f(2x-2y)\bigr]$ for all $x,y\in\mathbb{R}$. If $f'(0)=\tfrac12$, then the value of $24\,f''\bigl(\tfrac{5\pi}{3}\bigr)$ is:

Step 1: Rearrange the given equation

We begin by rearranging the given functional equation to isolate the terms involving $f$. Dividing both sides by $(\sin x\cos y)[f(2x+2y)+f(2x-2y)]$ allows us to express the ratio of the difference and sum of $f$ terms in terms of trigonometric functions.

Step 2: Apply Componendo and Dividendo

Applying the componendo and dividendo rule, which states that if $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$, we can simplify the expression. In our case, we apply it as $\frac{(a+b)+(a-b)}{(a+b)-(a-b)} = \frac{c+d}{c-d}$, which simplifies to $\frac{2a}{2b} = \frac{c+d}{c-d}$.

Step 3: Simplify the expression

We simplify the right-hand side of the equation by expressing $\cot x$ and $\tan y$ in terms of $\sin$ and $\cos$. Then, we use the trigonometric identities for $\sin(A+B)$ and $\sin(A-B)$ to further simplify the expression.

Step 4: Introduce new variables and differentiate

Let $u = 2x+2y$ and $v = 2x-2y$. Then $x = \frac{u+v}{4}$ and $y = \frac{u-v}{4}$. Substituting these into the simplified equation, we get a general form for $f(u)/f(v)$. Differentiating with respect to $u$ while treating $v$ as a constant, we get $f'(u) = C\cos\left(\frac{u}{2}\right)$ for some constant $C$.

Step 5: Find the constant C and the function f(x)

Given $f'(0) = \frac{1}{2}$, we substitute $x=0$ into $f'(x) = C\cos\left(\frac{x}{2}\right)$. This gives $C\cos(0) = C = \frac{1}{2}$. However, upon re-evaluating the differentiation, a simpler approach is to recognize that the form $\frac{f(u)}{f(v)} = \frac{\sin((u+v)/2)}{\sin((u-v)/2)}$ implies $f(x) = k\sin(x/2)$ for some constant $k$. Differentiating this, $f'(x) = k\frac{1}{2}\cos(x/2)$. Using $f'(0) = \frac{1}{2}$, we get $k\frac{1}{2}\cos(0) = \frac{1}{2} \implies k\frac{1}{2} = \frac{1}{2} \implies k=1$. Thus, $f(x) = \sin(x/2)$ and $f'(x) = \frac{1}{2}\cos(x/2)$.

Step 6: Calculate the second derivative

Now we differentiate $f'(x) = \frac{1}{2}\cos\left(\frac{x}{2}\right)$ with respect to $x$ to find the second derivative, $f''(x)$. The derivative of $\cos(ax)$ is $-a\sin(ax)$.

Step 7: Evaluate $24f''\left(\frac{5\pi}{3}\right)$

Finally, we substitute $x = \frac{5\pi}{3}$ into the expression for $f''(x)$ and multiply by 24. We know that $\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

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