题目内容
1.已知sin(30°-α)=$\frac{1}{3}$,求$\frac{1}{tan(30°-α)}$+$\frac{cos(60°+α)}{1+sin(60°+α)}$的值.分析 由题意和同角三角函数基本关系可得cos(30°-α)和$\frac{1}{tan(30°-α)}$的值,再由诱导公式可得原式=$\frac{1}{tan(30°-α)}$+$\frac{sin(30°-α)}{1+cos(30°-α)}$,代值计算可得.
解答 解:∵sin(30°-α)=$\frac{1}{3}$,∴cos(30°-α)=±$\sqrt{1-si{n}^{2}(30°-α)}$=±$\frac{2\sqrt{2}}{3}$,
当cos(30°-α)=$\frac{2\sqrt{2}}{3}$,时,$\frac{1}{tan(30°-α)}$=$\frac{cos(30°-α)}{sin(30°-α)}$=2$\sqrt{2}$,
∴$\frac{1}{tan(30°-α)}$+$\frac{cos(60°+α)}{1+sin(60°+α)}$=2$\sqrt{2}+$$\frac{cos[90°-(30°-α)]}{1+sin[90°-(30°-α)]}$
=2$\sqrt{2}$+$\frac{sin(30°-α)}{1+cos(30°-α)}$=2$\sqrt{2}$+3-2$\sqrt{2}$=3;
当cos(30°-α)=-$\frac{2\sqrt{2}}{3}$,时,$\frac{1}{tan(30°-α)}$=$\frac{cos(30°-α)}{sin(30°-α)}$=-2$\sqrt{2}$,
∴$\frac{1}{tan(30°-α)}$+$\frac{cos(60°+α)}{1+sin(60°+α)}$=2$\sqrt{2}+$$\frac{cos[90°-(30°-α)]}{1+sin[90°-(30°-α)]}$
=-2$\sqrt{2}$+$\frac{sin(30°-α)}{1+cos(30°-α)}$=-2$\sqrt{2}$+3+2$\sqrt{2}$=3;
综上可得$\frac{1}{tan(30°-α)}$+$\frac{cos(60°+α)}{1+sin(60°+α)}$=3
点评 本题考查三角函数化简求值,涉及同角三角函数基本关系和诱导公式,涉及分类讨论的思想,属中档题.
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