$$\begin{gathered} \text{ A function }f:R\to R\text{ satisfies }\sin x\cos y\left[f\right(2x+2y)-f(2x-2y\left)\right]=\cos x\sin y\left[f\right(2x+2y)+f(2x-2y\left)\right]\text{.} \\ \text{If }{f}^{\mathrm{\prime }}\left(0\right)=\frac{1}{2}\text{, then f(x) =} \end{gathered}$$

 $$\begin{gathered} \begin{array}{l}\frac{\sin x\cos y}{\cos x\sin y}=\frac{f(2x+2y)+f(2x-2y)}{f(2x+2y)-f(2x-2y)} \\ Apply C \& D\\ \frac{\sin (x+y)}{\sin (x-y)}=\frac{f(2x+2y)}{f(2x-2y)}=k,\text{ say}\end{array} \end{gathered}$$
$$\begin{gathered} \Rightarrow f\left(x\right)=k\sin \frac{x}{2}{f}^{\mathrm{\prime }}\left(x\right)=\frac{k}{2}\cos \frac{x}{2}\text{ and }{f}^{\prime \prime }\left(x\right)=-\frac{k}{4}\sin \frac{x}{2} \\ {f}^{\prime \prime }\left(x\right)=-\frac{1}{4}f\left(x\right) \end{gathered}$$