已知SIN(α-β)=3/5,sin(α+β)=-3/5且(α-β)∈(2/π,π) ,(α+β)∈(3/2π,2π),求COS2β.
人气:398 ℃ 时间:2020-06-12 14:59:47
解答
cos2β = cos[(α+β)-(α-β)]
=cos(α+β)cos(α-β) + sin(α+β)sin(α-β)
因(α-β)∈(π/2,π) ,(α+β)∈(3π/2,2π)
则cos(α-β)=-4/5,cos(α+β)=4/5
代入上式得
cos2β = cos(α+β)cos(α-β) + sin(α+β)sin(α-β)
=-4/5 * 4/5 + (-3/5) * 3/5
=-1
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