高中数学必修 第一册（648题）

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已知α，β为锐角，\(\cos {α=\frac {\sqrt {10}} {10}}\)，\(\sin {\left ( {α+β} \right )}=\frac {\sqrt {5}} {5}\)，求\(\cos {β}\)

知识点：第五章 三角函数

参考答案：\(\because \sin {\left ( {α+β} \right )}=\frac {\sqrt {5}} {5}\)，
\(\therefore \cos {\left ( {α+β} \right )}=\pm \frac {2\sqrt {5}} {5}\)，
\(\because α，β\)为锐角，
\(\therefore π>α+β>α>0\)，
\(\therefore \sin {α=\frac {3\sqrt {10}} {10}}\)，
又\(y=\cos {x}\)在\(\left [ {0，π} \right ]\)上单调递减，
\(\therefore \cos {\left ( {α+β} \right )}<\cos {α=\frac {\sqrt {10}} {10}}\)，
\(\therefore \cos {\left ( {α+β} \right )=-\frac {2\sqrt {5}} {5}}\)，
\(\therefore \cos {β=}\cos {\left [ {\left ( {α+β} \right )-α} \right ]}=\cos {\left ( {α+β} \right )}\cos {α}+\sin {\left ( {α+β} \right )}\sin {α=\frac {\sqrt {10}} {10}\left ( {-\frac {2\sqrt {5}} {5}} \right )}+\frac {3\sqrt {10}} {10}\cdot \frac {\sqrt {5}} {5}=\frac {\sqrt {2}} {10}\)．