已知8sinα+10cosβ=5,8cosα+10sinβ=5√3,求证:sin(α+β)=-sin(π/3+α)
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已知8sinα+10cosβ=5,8cosα+10sinβ=5√3,求证:sin(α+β)=-sin(π/3+α)
8sinα+10cosβ=5=10cosπ/3.(1)8cosα+10sinβ=5√3=10sinπ/3.(2)
(1)^2+(2)^2得:64+100+160sin(α+β)=100,
64=-160sin(α+β).8=-20sin(α+β).
(1)*sinα+(2)*cosα得:8+10sin(α+β)=10sin(π/3+α),
-20sin(α+β)+10sin(α+β)=10sin(π/3+α),
sin(α+β)=-sin(π/3+α).
(1)^2+(2)^2得:64+100+160sin(α+β)=100,
64=-160sin(α+β).8=-20sin(α+β).
(1)*sinα+(2)*cosα得:8+10sin(α+β)=10sin(π/3+α),
-20sin(α+β)+10sin(α+β)=10sin(π/3+α),
sin(α+β)=-sin(π/3+α).
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