高中数学必修 第一册（648题）

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已知\(\cos {\left ( {\frac {π} {4}-α} \right )}=\frac {3} {5}\)，\(\sin {\left ( {\frac {3π} {4}+β} \right )}=-\frac {12} {13}\)，\(α\in \left ( {\frac {π} {4}，\frac {3π} {4}} \right )\)，\(β\in \left ( {\frac {π} {2}，\frac {3π} {4}} \right )\)．求\(\sin {\left ( {α+β} \right )}\)的值．

知识点：第五章 三角函数

参考答案：\(\because \frac {3} {4}π+β-\left ( {\frac {π} {4}-α} \right )=\frac {π} {2}+α+β\)，\(\therefore \sin {\left ( {α+β} \right )}=\sin {\left [ {\left ( {\frac {3} {4}π+β} \right )-\left ( {\frac {π} {4}-α} \right )-\frac {π} {2}} \right ]}=\cos {\left [ {\left ( {\frac {3} {4}π+β} \right )-\left ( {\frac {\pi } {4}-α} \right )} \right ]}=-\cos {\left ( {\frac {3} {4}\pi +β} \right )\cos {\left ( {\frac {π} {4}-α} \right )}}-\sin {\left ( {\frac {3} {4}\pi +β} \right )}\sin {\left ( {\frac {π} {4}-α} \right )}\)，

\(\because \frac {π} {4}<α<\frac {3} {4}π\)，\(\therefore -\frac {π} {2}<\frac {π} {4}-α<0\)，\(\therefore \sin {\left ( {\frac {\pi } {4}-α} \right )}=-\frac {4} {5}\)，\(\frac {π} {2}<β<\frac {3} {4}π\)，\(\because \frac {5π} {4}<\frac {3} {4}\pi +β<\frac {3} {2}π\)，\(\therefore \cos {\left ( {\frac {3} {4}π+β} \right )}=-\frac {5} {13}\)，\(\therefore \sin {\left ( {α+β} \right )}=\frac {5} {13}\times \frac {3} {5}-\left ( {-\frac {12} {13}} \right )\times \left ( {-\frac {4} {5}} \right )=\frac {15} {65}-\frac {48} {65}=-\frac {33} {65}\)