Question

若在三角形ABC中，a、b、c三边成等差数列，则：（cosA+cosC）/（1+cosAcosC）=？（大题做法）

Answer 1

a,b,c等差数列,所以a+c=2b
正弦定理,sinA+sinC=2sinB=2sin(A+C)=2sin[2*(A+C)/2]=4sin[(A+C)/2]cos[(A+C)/2]
和差化积：2sin[(A+C)/2]cos[(A-C)/2]=4sin[(A+C)/2]cos[(A+C)/2]
两边同除以2sin[(A+C)/2]得cos[(A-C)/2]=2cos[(A+C)/2] ……（1）
两边平方并用二倍角余弦公式得1+cos(A-C)=4+4cos(A+C)
展开，整理得5sinAsinC=3(1+cosAcosC)
再利用和差化积和（1）式
(cosA+cosC)/(1+cosAcosC)
=3(cosA+cosC)/(5sinAsinC)
=6{cos[(A+C)/2]cos[(A-C)/2]}/(5sinAsinC)
=3cos^2[(A-C)/2]/(5sinAsinC)
=3[1+cos(A-C)]/(10sinAsinC)
=3(1+cosAcosC+sinAsinC)/(10sinAsinC)
=(5sinAsinC+3sinAsinC)/(10sinAsinC)
=8/10=4/5

Answer 2

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