高中数学必修 第一册（648题）

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已知\(\tan {\left ( {α-β} \right )}=\frac {1} {2}\)，\(\tan {β=-\frac {1} {7}}\)，且\(α，β\in \left ( {0，π} \right )\)，求\(2α-β\)的值．

知识点：第五章 三角函数

参考答案：\(\because 2α-β=2\left ( {α-β} \right )+β\)，又\(\tan {\left ( {α-β} \right )}=\frac {1} {2}\)，\(\therefore \tan {2\left ( {α-β} \right )}=\frac {2\tan {\left ( {α-β} \right )}} {1-\tan^{2} {\left ( {α-β} \right )}}=\frac {4} {3}\)，

故\(\tan {\left ( {2α-β} \right )}=\tan {\left [ {2\left ( {α-β} \right )+β} \right ]}=\frac {\tan {2\left ( {α-β} \right )+\tan {β}}} {1-\tan {2\left ( {α-β} \right )\tan {β}}}=\frac {\frac {4} {3}-\frac {1} {7}} {1+\frac {4} {3}\times \frac {1} {7}}=1\)，又\(\because \tan {α=\tan {\left [ {\left ( {α-β} \right )+β} \right ]}}=\frac {\tan {\left ( {α-β} \right )+\tan {β}}} {1-\tan {\left ( {α-β} \right )+\tan {β}}}=\frac {1} {3}<1\)，

且\(0<α<\pi \)，\(\therefore 0<α<\frac {\pi } {4}\)，\(\therefore 0<2α<\frac {\pi } {2}\)，

又\(\tan {β=-\frac {1} {7}}\)，且\(β\in \left ( {0，\pi } \right )⇒β\in \left ( {\frac {π} {2}，π} \right )⇒-β\in \left ( {-π，-\frac {π} {2}} \right )\)．

\(\therefore 2α-β\in \left ( {-π，0} \right )\)．又\(\tan {\left ( {2α-β} \right )}=1\)，\(\therefore 2α-β=-\frac {3π} {4}\)．