1746

已知 ${\displaystyle x\in (0,1)}$, ${\displaystyle y\in (0,+\mathrm{\infty })}$, 满足 $$y^2+\frac{4}{y^2}+2y\cos (\pi x)-\frac{4\sin (\pi x)}{y}=0,$$ 求 ${\displaystyle xy}$ 的值.

由辅助角公式 $$\begin{array}{}\text{(1)}& y^2+\frac{4}{y^2}=\frac{4\sin (\pi x)}{y}-2y\cos (\pi x)\le \sqrt{\frac{16}{y^2}+4y^2}.\end{array}$$ 而 $$\begin{array}{rl}\text{(2)}& & (y^2-2{)}^2+{(\frac{4}{y^2}-2)}^2\ge 0\Rightarrow & y^4+\frac{16}{y^4}+8\ge 4y^2+\frac{16}{y^2}\\ \Rightarrow & y^2+\frac{4}{y^2}\ge \sqrt{4y^2+\frac{16}{y^2}},\end{array}$$ 所以只有 ${\displaystyle y^2+\frac{4}{y^2}=\sqrt{4y^2+\frac{16}{y^2}}}$, 且由 (2) 看出取等条件为 ${\displaystyle y=\sqrt{2}}$. 再由 (1) $$\sin (\pi x)-\cos (\pi x)=\sqrt{2}\Rightarrow \sin (\pi x-\frac{\pi }{4})=1\Rightarrow x=\frac{3}{4}.$$
综上, ${\displaystyle xy=\frac{3\sqrt{2}}{4}}$.