把sinA=a/2R,sinB=b/2R,sinC=c/2R,代入
sin²A+sin²C-sinA×sinC=sin²B
约去(2R)²得：a²+c²-ac=b²,
cosB=(a²+c²-b²)/2ac=ac/2ac=1/2,
B=60°,∴A+C=120°,A-C=2A-120°.
2(cosA)²+cos(A-C)
=1+cos2A+cos(2A-120°)——［cos α+cos β=2cos[(α+β)/2]·cos[(α-β)/2]］
=1+2cos(2A-60°)*1/2
=1+cos(2A-60°)
∵-60°