Question

在△ABC中，sinC=（sinA=sinB)/（cosA=cosB），求△ABC形状

Answer 1

在△ABC中,若(sinA+sinB)/(cosA+cosB)=sinC,试判定△ABC的形状
解法1 (sinA+sinB)/(cosA+cosB)=sinC
<==>sinA+sinB=sin(A+B)(cosA+cosB)
<==>sinA+sinB=(sinAcosB+cosAsinB)(cosA+cosB)
<==>sinA+sinB=sinAcosAcosB+sinA(cosB)^2+(cosA)^2sinB)+cosAcosBsinB
<==>sinA[1-(cosB)^2]+sinB[1-(cosA)^2]=(sinA+sinB)cosAcosB
<==>sinA(sinB)^2+sinB(sinA)^2=(sinA+sinB)cosAcosB
<==>sinAsinB(sinA+sinB)=(sinA+sinB)cosAcosB
<==>sinAsinB=cosAcosB
<==>cosAcosB-sinAsinB=0
<==>cos(A+B)=0
<==>cosC=0
<==>C=90º
所以△ABC是直角三角形.

解法2 (转化为边的关系)由正弦定理和余弦定理得
(sinA+sinB)/(cosA+cosB)=sinC
<==>sinA+sinB=sinC(cosA+cosB)
<==>a+b=c[(b^2+c^2-a^2)/(2bc)+(c^2+a^2-b^2)/(2ca)]
<==>2ab(a+b)=a(b^2+c^2-a^2)+b(c^2+a^2-b^2)
<==>ab(a+b)=(a+b)c^2-(a^3+b^3)
<==>ab(a+b)=(a+b)[c^2-(a^2-ab+b^2)]
<==>a^2+b^2=c^2
所以△ABC是直角三角形.

Answer 2

括号里的是sinA-sinB，cosA—cosB么？

Answer 3

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