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已知点 ${\displaystyle P}$ 是椭圆 ${\displaystyle \frac{x^2}{4}+\frac{y^2}{3}=1}$ 上异于左右顶点的一点, 设 ${\displaystyle \mathrm{\angle }P{F}_1{F}_2=\alpha }$, ${\displaystyle \mathrm{\angle }P{F}_2{F}_1=\beta }$, ${\displaystyle \mathrm{\angle }{F}_{2}P{F}_1=\gamma }$, 求 ${\displaystyle \cos \alpha +\cos \beta +\cos \gamma }$ 的取值范围.

容易得到 ${\displaystyle F_1{F}_2=2}$, ${\displaystyle F_{1}P+F_{2}P=4}$. 不妨设 ${\displaystyle F_{1}P=2+t}$, ${\displaystyle F_{2}P=2-t}$, ${\displaystyle t\in [0,1)}$. 则根据余弦定理$$\begin{array}{rl}& \cos \alpha +\cos \beta +\cos \gamma \\ =& \frac{4+(2+t{)}^2-(2-t{)}^2}{2\cdot 2(2+t)}+\frac{4+(2-t{)}^2-(2+t{)}^2}{2\cdot 2(2-t)}+\frac{(2+t{)}^2+(2-t{)}^2-4}{2(2+t)(2-t)}\\ =& 3-\frac{6}{4-t^2}\in (1,\frac{3}{2}].\end{array}$$

熟知椭圆中的结论 ${\displaystyle e=\frac{\sin \gamma }{\sin \alpha +\sin \beta }=\frac{1}{2}}$. 因此 $$2\sin \gamma =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}.$$ 也即 $$2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha +\beta }{2}=\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2},$$ 从而 $$\cos \frac{\alpha -\beta }{2}=2\cos \frac{\alpha +\beta }{2}.$$
首先确认一下 ${\displaystyle \gamma }$ 的范围, 它在 ${\displaystyle P}$ 在上顶点时达到最大, 为 ${\displaystyle \frac{\pi }{3}}$, 因此 ${\displaystyle 0<\gamma <\frac{\pi }{3}}$. 因此 $$\begin{array}{rl}& \cos \alpha +\cos \beta +\cos \gamma =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}+\cos \gamma \\ =& 4{\cos }^2\frac{\alpha +\beta }{2}+\cos \gamma =4{\sin }^2\frac{\gamma }{2}+\cos \gamma \\ =& 1+2{\sin }^2\frac{\gamma }{2}\in (1,\frac{3}{2}].\end{array}$$