Question

已知三角形ABC，三内角满足A+B=2C，1/COSA+1/COSC=负根号2处以COSB，求COS（A-C)/2

已知三角形ABC，三内角满足A+B=2C，1/COSA+1/COSC=负根号2处以COSB，求COS（A-C)/2

Answer 1

因A+B+C=π,又A+C=2B
得B=π/3
1/cosA+1/cosC=-2√2
=>(cosA+cosC)=-2√2cosAcosC
=>2cos(A-C)/2cos(A+C)/2=-√2[cos(A+C)+cos(A-C)]
=>cos(A-C)/2=-√2[-1/2+cos(A-C)]
=>cos(A-C)/2=-√2[-1/2+2cos²(A-C)/2-1]
=>4cos²(A-C)/2+√2cos(A-C)/2-3=0(|A-C|/2<π/3)
cos(A-C)/2=[-√2+√(2+48)]/8(负根舍去)
=√2/2

Answer 2

因A+B+C=π,又A+C=2B
得B=π/3
1/cosA+1/cosC=-2√2
=>(cosA+cosC)=-2√2cosAcosC
=>2cos(A-C)/2cos(A+C)/2=-√2[cos(A+C)+cos(A-C)]
=>cos(A-C)/2=-√2[-1/2+cos(A-C)]
=>cos(A-C)/2=-√2[-1/2+2cos²(A-C)/2-1]
=>4cos²(A-C)/2+√2cos(A-C)/2-3=0(|A-C|/2<π/3)
cos(A-C)/2=[-√2+√(2+48)]/8(负根舍去)
=√2/2

是这样的，好好看