China Mathematical Competition

Let $g(x)=\frac{f(x)+f(-x)}{2}$, and $h(x)=\frac{f(x)-f(-x)}{2}$. Then $f(x)=g(x)+h(x)$, $g(x)$ is an even function, $h(x)$ is an odd function, and $g(x+2\pi )=g(x)$, $h(x+2\pi )=h(x)$ for any $x\in \mathbb{R}$.

Define $$f_1(x)=\frac{g(x)+g(x+\pi )}{2},$$ $$f_2(x)=\{\begin{array}{ll}\frac{g(x)-g(x+\pi )}{2\cos x},& x\ne k\pi +\frac{\pi }{2},\\ 0,& x=k\pi +\frac{\pi }{2},\end{array}$$ $$f_3(x)=\{\begin{array}{ll}\frac{h(x)-h(x+\pi )}{2\sin x},& x\ne k\pi ,\\ 0,& x=k\pi ,\end{array}$$ $$f_4(x)=\{\begin{array}{ll}\frac{h(x)+h(x+\pi )}{2\sin 2x},& x\ne \frac{k\pi }{2},\\ 0,& x=\frac{k\pi }{2},\end{array}$$ where $k$ is an arbitrary integer. It is easy to check that $f_i(x)$ ($i=1,2,3,4$) satisfy (1).

Next we prove that $f_1(x)+f_2(x)\cos x=g(x)$ for any $x\in \mathbb{R}$. When $x\ne k\pi +\frac{\pi }{2}$, it is obviously true. When $x=k\pi +\frac{\pi }{2}$, we have $$f_1(x)+f_2(x)\cos x=f_1(x)=\frac{g(x)+g(x+\pi )}{2},$$ and $$\begin{array}{rl}g(x+\pi )& =g\left(k\pi +\frac{3\pi }{2}\right)=g\left(k\pi +\frac{3\pi }{2}-2(k+1)\pi \right)\\ & =g\left(-k\pi -\frac{\pi }{2}\right)=g\left(k\pi +\frac{\pi }{2}\right)=g(x).\end{array}$$ The proof is complete.

Further we prove that $f_3(x)\sin x+f_4(x)\sin 2x=h(x)$ for any $x\in \mathbb{R}$. When $x\ne \frac{k\pi }{2}$, it is obviously true. When $x=k\pi$, we have $$h(x)=h(k\pi )=h(k\pi -2k\pi )=h(-k\pi )=-h(k\pi ).$$ That means $h(x)=h(k\pi )=0$. In this case, $$f_3(x)\sin x+f_4(x)\sin 2x=0.$$ $$\text{Therefore }h(x)=f_3(x)\sin x+f_4(x)\sin 2x.$$ When $x=k\pi +\frac{\pi }{2}$, we have $$\begin{array}{rl}h(x+\pi )& =h\left(k\pi +\frac{3\pi }{2}\right)=h\left(k\pi +\frac{3\pi }{2}-2(k+1)\pi \right)\\ & =h\left(-k\pi -\frac{\pi }{2}\right)=-h\left(k\pi +\frac{\pi }{2}\right)=-h(x).\end{array}$$ So $$f_3(x)\sin x=\frac{h(x)-h(x+\pi )}{2}=h(x).$$ Furthermore, $f_4(x)\sin 2x=0$. Therefore $$h(x)=f_3(x)\sin x+f_4(x)\sin 2x.$$ This completes the proof.

In conclusion, $f_i(x)$ ($i=1,2,3,4$) satisfy (2).