设△ABC三内角满足的方程(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0有两个相等的
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设△ABC三内角满足的方程(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0有两个相等的实根.
(1)求证:角B不大于π/3
(2)当角B取最大值时,判断△ABC的形状
过程详细~~
(1)求证:角B不大于π/3
(2)当角B取最大值时,判断△ABC的形状
过程详细~~
![设△ABC三内角满足的方程(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0有两个相等的](/uploads/image/z/14905089-9-9.jpg?t=%E8%AE%BE%E2%96%B3ABC%E4%B8%89%E5%86%85%E8%A7%92%E6%BB%A1%E8%B6%B3%E7%9A%84%E6%96%B9%E7%A8%8B%EF%BC%88sinB-sinA%29x%5E2%2B%28sinA-sinC%29x%2B%28sinC-sinB%29%3D0%E6%9C%89%E4%B8%A4%E4%B8%AA%E7%9B%B8%E7%AD%89%E7%9A%84)
(1)
(sinA-sinC)²-4(sinB-sinA)(sinC-sinB)
=sin²A-2sinAsinC+sin²C-4(sinBsinC-sinAsinC-sin²B+sinAsinB)
=(sinA+sinC)²-4sinB(sinA+sinC)+4sin²B=(sinA+sinC-2sinB)²
令上式=0得:2sinB=sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2]
===>2sinB/2cosB/2=cosB/2cos[(A-C)/2]===>sinB/2=cos[(A-C)/2]/2
∵0≤(A-C)/2<90º; ∴0<cos[(A-C)/2]≤1
∴0<sinB/2≤1/2,∴0º<B/2≤30º ∴0º<B≤60º
(2)
∠B=60º:
cos[(A-C)/2]/2=1/2===>A-C=0º===>A=C
∴ΔABC为等边三角形
(sinA-sinC)²-4(sinB-sinA)(sinC-sinB)
=sin²A-2sinAsinC+sin²C-4(sinBsinC-sinAsinC-sin²B+sinAsinB)
=(sinA+sinC)²-4sinB(sinA+sinC)+4sin²B=(sinA+sinC-2sinB)²
令上式=0得:2sinB=sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2]
===>2sinB/2cosB/2=cosB/2cos[(A-C)/2]===>sinB/2=cos[(A-C)/2]/2
∵0≤(A-C)/2<90º; ∴0<cos[(A-C)/2]≤1
∴0<sinB/2≤1/2,∴0º<B/2≤30º ∴0º<B≤60º
(2)
∠B=60º:
cos[(A-C)/2]/2=1/2===>A-C=0º===>A=C
∴ΔABC为等边三角形
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