a b c成等差数列 cosA+cosC/1+cosAcosC

题目

a b c成等差数列 cosA+cosC/1+cosAcosC
三角形中 边长a b c成等差数列 且 c

答案

a,b,c等差数列,a+c=2b
正弦定理,sinA+sinC=2sinB
和差化积,2sin(A+C)/2cos(A-C)/2=4sin(A+C)/2cos(A+C)/2
cos(A-C)/2=2cos(A+C)/2 .1
两边平方,1+cos(A-C)=4+4cos(A+C)
展开,整理,5sinAsinC=3(1+cosAcosC)
再利用和差化积和1式
(cosA+cosC)/(1+cosAcosC)
=3(cosA+cosC)/(5sinAsinC)
=6cos(A+C)/2cos(A-C)/2/(5sinAsinC)
=3cos^2(A-C)/(5sinAsinC)
=3(1+cos(A-C))/(10sinAsinC)
=3(1+cosAcosC+sinAsinC)/(10sinAsinC)
=(5sinAsinC+3sinAsinC)/(10sinAsinC)
=8/10=4/5
得证