高中数学必修 第一册（648题）

---

已知\(f(x)=2\sin {\left ( {x+\frac {π} {6}} \right )}\cos {x}-\frac {1} {2}\)，且\(f\left ( {α} \right )=\frac {1} {3}\)，\(α\in \left ( {\frac {π} {6}，\frac {5π} {12}} \right )\)，求cos2α的值.

知识点：第五章 三角函数

参考答案：\(\because f(x)=2\sin {\left ( {x+\frac {π} {6}} \right )}\cos {x}-\frac {1} {2}=\sqrt {3}\sin {x\cos {x}}+\cos^{2} {x}-\frac {1} {2}=\frac {\sqrt {3}} {2}\sin {2x}+\frac {1} {2}\cos {2x}=\sin {\left ( {2x+\frac {π} {6}} \right )}\)，\(f\left ( {α} \right )=\frac {1} {3}\)，可得\(\sin {\left ( {2α+\frac {π} {6}} \right )}=\frac {1} {3}\)，\(\because α\in \left ( {\frac {π} {6}，\frac {5π} {12}} \right )\)，可得\(2α+\frac {π} {6}\in \left ( {\frac {π} {2}，π} \right )\)，

\(\therefore \cos {\left ( {2α+\frac {π} {6}} \right )}=-\sqrt {1-\sin^{2} {\left ( {2α+\frac {π} {6}} \right )}}=-\frac {2\sqrt {2}} {3}\)，\(\therefore \cos {2α}=\cos {\left ( {2α+\frac {π} {6}-\frac {π} {6}} \right )}=\cos {\left ( {2α+\frac {π} {6}} \right )}\cos {\frac {π} {6}}+\sin {\left ( {2α+\frac {π} {6}} \right )}\sin {\frac {π} {6}}=-\frac {2\sqrt {2}} {3}\times \frac {\sqrt {3}} {2}+\frac {1} {3}\times \frac {1} {2}=\frac {1-2\sqrt {6}} {6}\)．