题目内容
例题:已知

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【答案】分析:先根据α和β的范围利用同角三角函数基本关系求得sinα和cos(α+β),进而利用两角和公式根据cosβ=cos(α-α+β)求得答案.
解答:解:∵
∴≤α+β≤π
∴sinα=
=
,sin(α+β)=
=
∴cosβ=cos(α-α+β)=cosαcos(α+β)+sinαsin(α+β)=
故答案为:
点评:本题主要考查了同角三角函数基本关系的应用,两角和与差的余弦函数.要特别注意三角函数值的正负的判定.
解答:解:∵

∴≤α+β≤π
∴sinα=




∴cosβ=cos(α-α+β)=cosαcos(α+β)+sinαsin(α+β)=

故答案为:

点评:本题主要考查了同角三角函数基本关系的应用,两角和与差的余弦函数.要特别注意三角函数值的正负的判定.

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