求证:在△ABC中,sinA*cosB*cosC+cosA*sinB*cosC+cosA*cosB*sinC=sinA*sinB*sinC
热心网友
证明:有A、B、C为△ABC内角∴A+B+C=π∴左边=cosC*(sinA*cosB+cosA*sinB)+cosA*cosB*sinC =cosC*sin(A+B)+cosA*cosB*sinC =sinC[-cos(A+B)+cosA*cosB] =sinC[-cosA*cosB+sinA*sinB+cosA*cosB] =sinA*sinB*sinC