Question

已知A、B、C是三角形ABC的三个内角，且满足2sinB=sinA+sinC，设B的最大值为B0

1，求B0；2，当B=3B0/4时，求cosA-cosC的值

Answer 1

1、∵2sinB=sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2]
=2sin[(π-B)/2]cos[(A-C)/2]=2cos(B/2)cos[(A-C)/2]
∴4sin(B/2)cos(B/2)=2cos(B/2)cos[(A-C)/2]
∴sin(B/2)=(1/2)cos[(A-C)/2]≤1/2
∴B/2≤π/6
∴B≤π/3
∴B0=π/3
2、B=3B0/4=π/4
则2sinB=sinA+sinC=√2……………………①
设cosA-cosC=x……………………②
①²+②²得
(sinA+sinC)²+(cosA-cosC)=2+x²
sin²A+sin²C+2sinAsinC+cos²A+cos²C-2cosAcosC=2+x²
2-2(cosAcosC-sinAsinC)=2+x²
2-2cos(A+C)=2+x²
所以x²=-2cos(A+C)=2cosB=√2
所以cosA-cosC=x=±(2的4次方根)