已知锐角△ABC中,sin(A+B)=3/5,sin(A-B)=1/5求tanB

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已知锐角△ABC中,sin(A+B)=3/5,sin(A-B)=1/5求tanB

sin(A+B)=3/5,sin(A-B)=1/5 sin(a+b)=sinAcosB+sinBcosA=3/5 sin(a-b)=sinAcosB-sinBcosA=1/5 两式相加相减后可得:sinAcosB=2/5 sinBcosA=1/5 将两式相除,可得tanA=2tanB tan(B)=sinB/cosB=sinBcosA/cosAcosB cos(A...