Question

已知△ABC的三个内角A、B、C满足A+C=2B,且1/cosA+1/cosC=-根号2/cosB，求cos[(A-c)/2]的值。要过程。

Answer 1

解：
A+C=2B
3B=A+C+B=π
∴ B=π/3
1/cosA+1/cosC=-√2/(1/2)=-2√2
cosA+cosC=-2√2cosAcosC
2cos[(A+C)/2]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
2cos60°cos[(A-C)/2]=-√2[cos120°+cos(A-C)]
cos[(A-C)/2]=-√2{-1/2+2cos²[(A-C)/2]-1}
cos[(A-C)/2]=3√2/2-2√2cos²[(A-C)/2]
2√2cos²[(A-C)/2]+cos[(A-C)/2]-3√2/2=0
4cos²[(A-C)/2]+√2cos[(A-C)/2]-3=0
解方程，cos[(A-C)/2]=√2/2或cos[(A-C)/2]=-3√2/4（舍）
所以 cos[(A-C)/2]=√2/2

Answer 2

三角形ABC中，A+C=2B，
得：B=60°, A+C=120°
1/cosA+1/cosC=-√2/cosB =-2√2
cosA+cosC=-2√2cosAcosC （左边用和差化积，右边用积化和差）
2cos[(A+C)/2]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
cos[(A-C)/2]=-√2[-1/2+cos(A-C)]
=-√2{2[cos(A-C)/2]^2-3/2}
解得：cos[(A-C)/2]=√2/2