Question

在三角形ABC中，已知(sinA+sin+B+sinC)(sinA+sinB-sinC)=3,a<b,且acosA+bcosB=ccosC，求三角形各内角大小

详细过程

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, 且a<b
∴cosB=0
∴B=π/2
A=π/2-C
(sinA+sin+B+sinC)(sinA+sinB-sinC)\
=(1+SinA)^2-SinC ^2
=1+2SinA+SinA ^2 -CosA ^2
=1+2SinA - Cos 2A =3
2SinA-Cos2A=2 ?条件似乎缺点什么？