因?yàn)?A+B+C=180,A+C=2B,
3B=180,B=60度
1/cosA+1/cosC=-√2/cosB,
(cosC+cosA)/cosA*cosC=-√2/cosB,
利用公式:
cosA+cosC=2*cos(A+C)/2*cos(A-C)/2,(和差化積)
cosA*cosC=1/2*[cos(A+C)+cos(A-C)],(積化和差公式)
就可得到:
[2*cos(A+C)/2*cos(A-C)/2]/{1/2*[cos(A+C)+cos(A-C)]=-√2/cosB,
[2cos60*cos(A-C)/2]*cosB=-√2*1/2*[cos120+cos(A-C)],
化簡后,得
cos(A-C)/2=-√2*[cos(A-C)-1/2],
而,cos(A-C)=2cos^2(A-C)/2-1,(倍角公式),則有
2√2cos^2(A-C)/2+cos(A-C)/2-3√2/2=0,
令,cos(A-C)/2=t,|t|≤1,則有
2√2t^2+t-3√2/2=0,
4√2t^2+2t-3√2=0,
(2√2t+3)(2t-√2)=0,
t1=-3/(2√2),(不合,舍去,|t|≤1,)
t2=√2/2.
即,cos(A-C)/2=√2/2.
(注:三角函數(shù)主要是公式間的轉(zhuǎn)化)