asinx±bcosx=â(a^2+b^2)* sin(x±γ) 0<γ<Ï/2
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1 â3sina-cosa=2sin(a-Ï/6). å 为tanγ=1/â3,æ以γ=Ï/6
2 -sinB+â3cosB=-(sinB-â3cosB)= -2sin(B-Ï/3) å 为tanγ=â3ï¼ æ以γ=Ï/3
3 âsinx-cosx=-(sinx+cosx)= - â2sin(x+Ï/4) å 为tanγ=1ï¼æ以γ=Ï/4
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