在△ABC中,若(sinA+sinB+sinC)/(cosA+cosB+cosC)=根号3,求证:△ABC中至少有一个角
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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°
![在△ABC中,若(sinA+sinB+sinC)/(cosA+cosB+cosC)=根号3,求证:△ABC中至少有一个角](/uploads/image/z/1864846-46-6.jpg?t=%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%8B%A5%28sinA%2BsinB%2BsinC%29%2F%28cosA%2BcosB%2BcosC%29%3D%E6%A0%B9%E5%8F%B73%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E2%96%B3ABC%E4%B8%AD%E8%87%B3%E5%B0%91%E6%9C%89%E4%B8%80%E4%B8%AA%E8%A7%92)
(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
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
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