Question

在△ABC中,若(sinA+sinB+sinC)/(cosA+cosB+cosC)=根号3，求证：△ABC中至少有一个角为60°

Answer 1

(sinA+sinB+sinC)/(cosA+cosB+cosC) = √3
sinA+sinB+sinC = √3cosA+√3cosB+√3cosC
(sinA - √3cosA) + (sinB - √3cosB) + (sinC - √3cosC) = 0
2(sinAcosπ/3-cosAsinπ/3) + 2(sinBcosπ/3-cosBsinπ/3) + 2(sinCcosπ/3-cosCsinπ/3) = 0
2sin(A-π/3) + 2sin(B-π/3) + 2sin(C-π/3) = 0
sin(A-π/3) + sin(B-π/3) + sin(C-π/3) = 0
sin(A-π/3) + { sin(B-π/3) + sin(C-π/3) } = 0
sin(A-π/3) + 2sin{[(B-π/3)+(C-π/3)]/2} cos{[(B-π/3)-(C-π/3)]/2}= 0
sin(A-π/3) + 2sin{[(B+C)/2-π/3)} cos{(B-C)/2}= 0
sin(A-π/3) + 2sin(π/2-A/2-π/3) cos{(B-C)/2}= 0
sin(A-π/3) + 2sin(π/6-A/2) cos{(B-C)/2}= 0
-sin(π/3-A) + 2sin{π/6-A/2)} cos{(B-C)/2}= 0
-2sin(π/6-A/2)cos(π/6-A/2) + 2sin{π/6-A/2) cos{(B-C)/2}= 0
sin(π/6-A/2) {cos(π/6-A/2) - cos{(B-C)/2} } = 0
sin(π/6-A/2) * (-2) * sin{[(π/6-A/2)+(B-C)/2]/2 } sin{[(π/6-A/2)-(B-C)/2]/2 } = 0
sin(π/6-A/2) * sin{π/12-(A+C-B)/4} * sin{π/12-(A+B-C)/4 } = 0
sin(π/6-A/2) =0，或者sin{π/12-(A+C-B)/4} = 0，或者sin{π/12-(A+B-C)/4 }=0

假设sin(π/6-A/2) =0，则π/6-A/2=0，A=π/3

假设sin(π/6-A/2) ≠0，则必须sin{π/12-(A+C-B)/4} = 0或者sin{π/12-(A+B-C)/4 }=0
即：π/12-(A+C-B)/4 = 0，或π/12-(A+B-C)/4=0，即：
A+C-B=π/3 ...(1)
或A+B-C=π/3 (2)
又：A+B+C=π，
∴A+C=π-B，或A+B=π-C分别代入*1）、（2）：
π-B-B=π/3
或π-C-C=π/3
∴B=π/3，或C=π/3

综上：A=π/3，或B=π/3，或C=π/3