高中数学必修 第一册（648题）

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知识点：第五章 三角函数

参考答案：若\(f(\frac{\pi }{3} - \alpha ) = \frac{1}{3}\)，则\(\cos (\frac{\pi }{3} - \alpha ) = \frac{1}{3}\)，

\(\therefore \)\({\cos^{2}}(\frac {\pi } {6}+\alpha )={\cos^{2}}\left [ {\frac {\pi } {2}-(\frac {\pi } {3}-\alpha )} \right ]={\sin^{2}}(\frac {\pi } {3}-\alpha )=1-{\cos^{2}}(\frac {\pi } {3}-\alpha )=1-\frac {1} {9}=\frac {8} {9}\)，

\(\cos {(}\frac {2\pi } {3}+\alpha )=\cos {\left [ {\pi -(\frac {\pi } {3}-\alpha )} \right ]}=-\cos {(}\frac {\pi } {3}-\alpha )=-\frac {1} {3}\)，

则 \({\cos ^2}(\frac{\pi }{6} + \alpha ) + \cos (\frac{{2\pi }}{3} + \alpha ) = \frac{8}{9} - \frac{1}{3} = \frac{5}{9}\)．

解析：