在△ABC中,cosA=4/5,sinB=5/13,求cosC
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cosA=4/5,sinB=5/13
sinA=3/5,cosB=12/13或cosB=-12/13
cosC
=cos(180-A-B)
=-cos(A+B)
=-[cosAcosB-sinAsinB]
=-[4/5*12/13-3/5*5/13]
=-[48/65-15/65]
=-33/65
cosC
=cos(180-A-B)
=-cos(A+B)
=-[cosAcosB-sinAsinB]
=-[4/5*(-12/13)-3/5*5/13]
=-[-48/65-15/65]
=63/65
sinA=3/5,cosB=12/13或cosB=-12/13
cosC
=cos(180-A-B)
=-cos(A+B)
=-[cosAcosB-sinAsinB]
=-[4/5*12/13-3/5*5/13]
=-[48/65-15/65]
=-33/65
cosC
=cos(180-A-B)
=-cos(A+B)
=-[cosAcosB-sinAsinB]
=-[4/5*(-12/13)-3/5*5/13]
=-[-48/65-15/65]
=63/65
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