Question

1/cosA+1/cosC=-√2/cosB,A+C=2B,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