Question

求解一道高中数学题

已知在三角形ABC三个内角A B C 满足A+C=2B 1/cosA +1/cosC= - √2/cosB 求cos（A-C）/2 的值

Answer 1

A+C=2B
A+B+C=180
3B=180
B=60

A+C=120

1/cosA +1/cosC= - √2/cosB
1/cosA +1/cosC=-√2/(1/2)
(cosA+cosC)/cosAcosC=-√2/(1/2)
(cosC+cosA)/cosCcosA=-2√2
cosC+cosA=-2√2cosCcosA
2[cos(A+C)/2][cos(A-C)/2]=-2√2cosCcosA
2cos60[cos(A-C)/2]=-2√2cosCcosA
cos(A-C)/2=-2√2cosCcosA
cos(A-C)/2=-√2[cos(A+C)+cos(A-C)]
cos(A-C)/2=-√2[-1/2+cos(A-C)]
cos(A-C)/2=√2/2-√2cos(A-C)
cos(A-C)/2=√2/2-√2{2[(cos(A-C)/2]^2-1}
cos(A-C)/2=√2/2-2√2[(cos(A-C)/2]^2+√2
cos(A-C)/2=3√2/2-2√2[(cos(A-C)/2]^2

令n=cos(A-C)/2
n=3√2/2-2√2*n^2
2√2*n^2+n-3√2/2=0
4√2*n^2+2n-3√2=0
8n^2+2√2n-6=0
(√2n+3)(2√2n-2)=0
n=-3√2/2（舍去） 或 n=√2/2

所以n=√2/2
即cos(A-C)/2=√2/2

Answer 2

B=60°，再利用和差化积公式一套就出来了