高中数学必修 第一册（648题）

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已知α，β∈（0，π），\(\cos {α=\frac {4} {5}}\)，\(\sin {\left ( {α-β} \right )}=\frac {5} {13}\)，求sin（α+β）的值．

知识点：第五章 三角函数

参考答案：\(\because \cos {α=\frac {4} {5}}\)，
\(\therefore \cos {2α=2\cos^{2} {α-1}}=\frac {7} {25}\)．
∵α，β∈（0，π），
\(\cos {α}=\frac {4} {5}>0\)，
\(\therefore 0<α<\frac {π} {2}\)，
\(-π<α-β<\frac {π} {2}\)．
又\(\because \sin {\left ( {α-β} \right )}=\frac {5} {13}\)，
\(\therefore α-β\in \left ( {0，\frac {π} {2}} \right )\)，
\(\therefore \cos {\left ( {α-β} \right )}=\frac {12} {13}\)．
\(\therefore \sin {2α}=2\sin {α}\cos {α}=2\times \frac {3} {5}\times \frac {4} {5}=\frac {24} {25}\)．
\(\therefore \sin {\left ( {α+β} \right )}=\sin {\left [ {2α-\left ( {α-β} \right )} \right ]}=\sin {2α}\cos {\left ( {α-β} \right )}-\cos {2α}\sin {\left ( {α-β} \right )}=\frac {24} {25}\times \frac {12} {13}-\frac {7} {25}\times \frac {5} {13}=\frac {253} {325}\)．