Question

sinC(acosB-1/2b)=(sinA-sinB)(a+ b)求A

Answer 1

解：sinC=sin(π-A-B)=sin(A+B)
由正弦定理a/sinA=b/sinB=2R得
a=2RsinA
b=2RsinB
将sinC和a、b代入，并约掉2R，得
sin(A+B)(sinAcosB-1/2*sinB)=(sinA-sinB)(sinA+sinB) ①
将sin(A+B)=sinAcosB+cosAsinB代入上式，得
(sinAcosB+cosAsinB)(sinAcosB-1/2*sinB)=(sinA-sinB)(sinA+sinB)
sin²Acos²B+sinAcosAsinBcosB-1/2*sinAsinBcosB-1/2*cosAsin²B=sin²A-sin²B
sin²A-sin²Asin²B+sinAcosAsinBcosB-1/2*sinAsinBcosB-1/2*cosAsin²B=sin²A-sin²B
-sin²AsinB+sinAcosAcosB-1/2*sinAcosB-1/2*cosAsinB=-sinB
cos²AsinB+sinAcosAcosB-1/2*sinAcosB-1/2*cosAsinB=0
cosAsinB(cosA-1/2)+sinAcosB(cosA-1/2)=0
sin(A-B)(cosA-1/2)=0
得A=B或A=π/3
若A=B，代入式①得：
sin(2A)(sinAcosA-1/2*sinA)=0
sin(2A)sinA(cosA-1/2)=0
得A=π/2或A=π/3