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cosα=1/3,cos(α+β)=-1/3,且α,β∈(0,π/2),求cos(α-β)
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cosα=1/3,cos(α+β)=-1/3,且α,β∈(0,π/2),求cos(α-β)
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答案和解析
∵(sinα)^2 + (cosα)^2=1
∴sinα=±√[1 - (cosα)^2] = ±2√2/3
∵α∈(0,π/2)
∴sinα=2√2/3
同理：sin(α+β)=2√2/3
则sin2α=2sinα*cosα=4√2/9 ,cos2α=2(cosα)^2 - 1 = -7/9
cos(α-β)=cos[2α-(α+β)] =cos2α*cos(α+β) + sin2α*sin(α+β) =23/27
∴sinα=±√[1 - (cosα)^2] = ±2√2/3
∵α∈(0,π/2)
∴sinα=2√2/3
同理：sin(α+β)=2√2/3
则sin2α=2sinα*cosα=4√2/9 ,cos2α=2(cosα)^2 - 1 = -7/9
cos(α-β)=cos[2α-(α+β)] =cos2α*cos(α+β) + sin2α*sin(α+β) =23/27
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