Question

已知角A,B,C为△ABC的三个内角，m=(cosB-sinB,cosC)，n=(sinC,coB)

已知角A,B,C为△ABC的三个内角，m=(cosB-sinB,cosC)，n=(sinC,coB+sinB)，m﹡n=7/5
求：tanA值

Answer 1

解答过程如下：
m*n=(cosB-sinB)*sinC+cosC*(cosB+sinB)
=cosB*sinC-sinB*sinC+cosC*cosB+cosC*sinB
=cosB*sinC+sinB*cosC+cosC*cosB-sinB*sinC
=sin(B+C)+cos(B+C)
=sin(180°-A)+cos(180°-A)
=sinA-cosA=7/5
又sinA*sinA+cosA*cosA=1,解得sinA=4/5,cosA=-3/5或sinA=3/5,cosA=-4/5
所以tanA=sinA/cosA=-4/3或-3/4

Answer 2

m*n=(cosB-sinB)*sinC+cosC*(cosB+sinB)
=cosB*sinC-sinB*sinC+cosC*cosB+cosC*sinB
=cosB*sinC+sinB*cosC+cosC*cosB-sinB*sinC
=sin(B+C)+cos(B+C)
=sin(180°-A)+cos(180°-A)
=sinA-cosA=7/5
又sinA*sinA+cosA*cosA=1,解得sinA=4/5,cosA=-3/5或sinA=3/5,cosA=-4/5
所以tanA=sinA/cosA=-4/3或-3/4