根式方程：结构联想巧用三角代换

解方程组：

{x(1−y)+y(1−x)=12,xy+(1−x)(1−y)=32. \left\{\begin{array}{l} \sqrt{x(1-y)}+\sqrt{y(1-x)}=\dfrac{1}{2},\\ \sqrt{x y}+\sqrt{(1-x)(1-y)}=\dfrac{\sqrt{3}}{2}. \end{array}\right. ⎩ ⎨ ⎧x(1−y) +y(1−x) =21,xy +(1−x)(1−y) =23 .

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分析

由题中结构和 0⩽x⩽1,0⩽y⩽1，0\leqslant x\leqslant 1, 0\leqslant y\leqslant 1，0⩽x⩽1,0⩽y⩽1，

联想到用 x=sin⁡2α,y=sin⁡2βx=\sin^{2}\alpha, y=\sin^{2}\betax=sin2α,y=sin2β 代换解决．

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解

依题意有 0⩽x⩽1,0⩽y⩽10\leqslant x\leqslant 1, 0\leqslant y\leqslant 10⩽x⩽1,0⩽y⩽1，

可设 x=sin⁡2α,y=sin⁡2β，x=\sin^{2}\alpha, y=\sin^{2}\beta，x=sin2α,y=sin2β，

且 0⩽α,β⩽π2，0\leqslant \alpha,\beta\leqslant\frac{\pi}{2}，0⩽α,β⩽2π，

则原方程组可化为
{sin⁡αcos⁡β+sin⁡βcos⁡α=12,sin⁡αsin⁡β+cos⁡αcos⁡β=32, \left\{\begin{array}{l} \sin\alpha\cos\beta+\sin\beta\cos\alpha=\dfrac{1}{2},\\ \sin\alpha\sin\beta+\cos\alpha\cos\beta=\dfrac{\sqrt{3}}{2}, \end{array}\right. ⎩ ⎨ ⎧sinαcosβ+sinβcosα=21,sinαsinβ+cosαcosβ=23 ,

即
{sin⁡(α+β)=12,cos⁡(α−β)=32. \left\{\begin{array}{l}\sin(\alpha+\beta)=\dfrac{1}{2},\\\cos(\alpha-\beta)=\dfrac{\sqrt{3}}{2}.\end{array}\right. ⎩ ⎨ ⎧sin(α+β)=21,cos(α−β)=23 .

因为 0⩽α+β⩽π,−π2⩽α−β⩽π2，0\leqslant\alpha+\beta\leqslant\pi, -\frac{\pi}{2}\leqslant\alpha-\beta\leqslant\frac{\pi}{2}，0⩽α+β⩽π,−2π⩽α−β⩽2π，

所以
{α+β=π6 或 5π6,α−β=π6 或 −π6, \left\{\begin{array}{l}\alpha+\beta=\dfrac{\pi}{6}\text{ 或 }\dfrac{5\pi}{6},\\\alpha-\beta=\dfrac{\pi}{6}\text{ 或 }-\dfrac{\pi}{6},\end{array}\right. ⎩ ⎨ ⎧α+β=6π 或 65π,α−β=6π 或 −6π,

故
(α,β)=(π6,0),(π2,π3),(0,π6),(π3,π2)，(\alpha,\beta) = (\dfrac{\pi}{6},0),(\dfrac{\pi}{2},\dfrac{\pi}{3}),(0,\dfrac{\pi}{6}),(\dfrac{\pi}{3},\dfrac{\pi}{2})，(α,β)=(6π,0),(2π,3π),(0,6π),(3π,2π)，

所以原方程组的解：
(x,y)=(14,0),(1,34),(0,14),(34,1)．(x,y) = (\dfrac{1}{4},0),(1,\dfrac{3}{4}),(0,\dfrac{1}{4}),(\frac{3}{4},1)．(x,y)=(41,0),(1,43),(0,41),(43,1)．