China Mathematical Olympiad

Proof Clearly, ① and ② are equivalent to $$\sin {\theta }_1\sin {\theta }_2-\cos {\theta }_1\cos {\theta }_2\le x\le \sin {\theta }_1\sin {\theta }_2+\cos {\theta }_1\cos {\theta }_2,④$$ $$\sin {\theta }_3\sin {\theta }_4-\cos {\theta }_3\cos {\theta }_4\le x\le \sin {\theta }_3\sin {\theta }_4+\cos {\theta }_3\cos {\theta }_4,⑤$$ respectively. It is easy to show that there exists $x\in \mathbb{R}$ such that ④ and ⑤ hold simultaneously if and only if $$\sin {\theta }_1\sin {\theta }_2+\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_3\sin {\theta }_4+\cos {\theta }_3\cos {\theta }_4\ge 0,⑥$$ $$\sin {\theta }_3\sin {\theta }_4+\cos {\theta }_3\cos {\theta }_4-\sin {\theta }_1\sin {\theta }_2+\cos {\theta }_1\cos {\theta }_2\ge 0.⑦$$

$$\begin{gathered} {\cos }^2{\theta }_1{\cos }^2{\theta }_2+2\cos {\theta }_1\cos {\theta }_2\cos {\theta }_3\cos {\theta }_4+{\cos }^2{\theta }_3{\cos }^2{\theta }_4 \\ -{\sin }^2{\theta }_1{\sin }^2{\theta }_2+2\sin {\theta }_1\sin {\theta }_2\sin {\theta }_3\sin {\theta }_4-{\sin }^2{\theta }_3{\sin }^2{\theta }_4\ge 0, \\ \text{or} \end{gathered}$$ $$(\cos {\theta }_1\cos {\theta }_2+\cos {\theta }_3\cos {\theta }_4{)}^2-(\sin {\theta }_1\sin {\theta }_2-\sin {\theta }_3\sin {\theta }_4{)}^2\ge 0.$$ That is, $$(\sin {\theta }_1\sin {\theta }_2+\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_3\sin {\theta }_4+\cos {\theta }_3\cos {\theta }_4)\cdot$$ $$(\sin {\theta }_3\sin {\theta }_4+\cos {\theta }_3\cos {\theta }_4-\sin {\theta }_1\sin {\theta }_2+\cos {\theta }_1\cos {\theta }_2)\ge 0.⑧$$ If there exists $x\in \mathbb{R}$ such that ④ and ⑤ hold simultaneously, then from ⑥ and ⑦ we can get ⑧ immediately. It follows that ③ holds.

Conversely, if ③ holds, or equivalently ⑧ holds, but ⑥ and ⑦ do not hold, then we have $$\sin {\theta }_1\sin {\theta }_2+\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_3\sin {\theta }_4+\cos {\theta }_3\cos {\theta }_4<0,\text{ and}$$ $$\sin {\theta }_3\sin {\theta }_4+\cos {\theta }_3\cos {\theta }_4-\sin {\theta }_1\sin {\theta }_2+\cos {\theta }_1\cos {\theta }_2<0.$$ Adding the last two equations, we obtain $$2(\cos {\theta }_1\cos {\theta }_2+\cos {\theta }_3\cos {\theta }_4)<0,$$ which contradicts to the fact ${\theta }_i\in (-\frac{\pi }{2},\frac{\pi }{2})$, $i=1,2,3,4$. So ⑥ and ⑦ hold simultaneously. Hence there exists $x\in \mathbb{R}$ such that ④ and ⑤ hold simultaneously.