Question

在三角形ABC中，若aCosA+bCosB=cCosC,则三角形是什么形状？

Answer 1

∵acosA+bcosB=ccosC
∴sinAcosA+sinBcosB=sinCcosC
∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)
∴0=sin2A+sin2B+sin(2A+2B)
=sin2A+sin2B+sin2Acos2B+sin2Bcos2A
=sin2A(1+cos2B)+sin2B(1+cos2A)
=4sinAcosA(cosB)^2+4sinBcosB(cosA)^2
=4cosAcosBsin(A+B)
∵sin(A+B)=sin(π-C)=sinC>0
∴cosA=0或cosB=0
∴A=π/2或B=π/2
∴△ABC是以a或b为斜边的直角三角形

Answer 2

正弦定理a/sinA=b/sinB=c/sinC=2r(r为外接圆半径)
代入acosA+bcosB=ccosC得 sinAcosA+sinBcosB=sinCcosc
sin2A+sin2B=sin2C
2sin(A+B)cos(A-B)=2sinCcosC和化积
sin(Π-C)cos(A-B)=sinCcosC
sinCcos(A-B)=sinCcosC
cos(A-B)-cosC=0
-2sin(A-B+C)/2 *sin(A-B-C)/2 =0 差化积
sin(A+C-B)/2*sin(A-B-C)/2 =0
sin(Π-2B)/2* sin(2A-Π)/2=0
cosB*(-cosA)=0
所以A或B中有一个为π/2
所以为直角三角形