每日一题_190930

在\(\triangle ABC\)中,若\(\dfrac{b}{a}+\dfrac{a}{b}=4\cos C\),\(\cos (A-B)=\dfrac{1}{6}\),则\(\cos C=\underline{\qquad\qquad}\).
解析:
法一 由题有$$ \cos C=\dfrac{a^2+b2}{4ab}=\dfrac{a^2+b2-c^2}{2ab}.$$因此\(a^2+b^2=2c^2\).于是在$\triangle ABC $中应用正弦定理可得

\[\dfrac{c^2}{\sin^2C}=\dfrac{a^2}{\sin^2A}=\dfrac{b^2}{\sin^2B}=\dfrac{a^2+b^2}{\sin^2A+\sin^2B}.$$即有 \]

\begin{split}
\dfrac{1}{\sin^2C}& =\dfrac{4}{1-\cos 2A+1-\cos 2B}\
&=\dfrac{2}{1-\cos (A+B)\cos(A-B)}\
&=\dfrac{2}{1+\dfrac{1}{6}\cos C}.
\end{split}$$
解得\(\cos C=\dfrac{2}{3}\)或\(-\dfrac{3}{4}\),显然\(\cos C>0\),因此\(\cos C=\dfrac{2}{3}\).

法二 由题有$$ \cos C=\dfrac{a^2+b2}{4ab}=\dfrac{a^2+b2-c^{2}{2ab}=\dfrac{c}2}{2ab}=\dfrac{\sin^2C}{2\sin A \sin B}.$$
又因为$$ \dfrac{1}{6}+\cos C=\cos (A-B)-\cos\left(A+B\right)=2\sin A\sin B.$$
两式联立消去\(\sin A\sin B\)可得关于\(\cos C\)的方程,解得\(\cos C=\dfrac{2}{3}\)或\(-\dfrac34\),显然\(\cos C>0\),因此\(\cos C=\dfrac{2}{3}\).