1603

在 ${\displaystyle \mathrm{\Delta }ABC}$ 中, 角 ${\displaystyle A,B,C}$ 所对的边分别为 ${\displaystyle a,b,c}$, 若 ${\displaystyle B=\frac{2\pi }{3}}$, ${\displaystyle b^2=6ac}$, 则 ${\displaystyle \cos A+\cos C=}$ _____.

因为 ${\displaystyle B=\frac{2\pi }{3}}$, 所以根据余弦定理 $$-\frac{1}{2}=\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+c^2-6ac}{2ac},$$ 整理得 ${\displaystyle a^2+c^2=5ac}$. 从而 ${\displaystyle (a+c{)}^2=7ac}$, ${\displaystyle (a-c{)}^2=3ac}$. 这样$$\begin{array}{rl}& \cos A+\cos C=\frac{c^2+b^2-a^2}{2bc}+\frac{a^2+b^2-c^2}{2ab}\\ =& \frac{(a+c)(b^2-(a-c{)}^2)}{2abc}=\frac{\sqrt{7ac}(6ac-3ac)}{2ac\sqrt{6ac}}=\frac{\sqrt{42}}{4}.\end{array}$$

根据正弦定理, $$\sin A\sin C=\frac{1}{6}{\sin }^{2}B=\frac{1}{8}.$$ 而 $$\frac{1}{2}=\cos (A+C)=\cos A\cos C-\sin A\sin C,$$ 所以 $$\cos (A-C)=\cos A\cos C+\sin A\sin C=\frac{3}{4}.$$ 这样 $$\begin{array}{rl}\cos A+\cos C& =2\cos \frac{A+C}{2}\cos \frac{A-C}{2}\\ & =\sqrt{3}\cos \frac{A-C}{2}=\sqrt{3}\cdot \sqrt{\frac{1+\cos (A-C)}{2}}=\frac{\sqrt{42}}{4}.\end{array}$$