Question

在三角形abc中，若acos^2C/2+ccos^2A/2=3b/2，求证：a+c=2b

Answer 1

a(1+cosC)/2+c(1+cosA)/2=3b/2
a+a(a^2+b^2-c^2)/2ab+c+c(b^2+c^2-a^2)/2bc=3b
a+(a^2+b^2-c^2)/2b+c+(b^2+c^2-a^2)/2b=3b
2ab+a^2+b^2-c^2+2bc+b^2+c^2-a^2=6b^2
2ab+2bc=4b^2
两边除以2b
a+c=2b

Answer 2

a*cos^2(C/2)+c*cos^2(A/2)=3/2*b
a(cosC+1)/2+c(cosA+1)/2=1.5b
acosC+ccosA=3b-a-c
a*(a^2+b^2-c^2/2ab+c*（c^2+b^2-a^2）/2bc
=3b-a-c
2b^2/2b=3b-a-c
b=3b-a-c
a+c=2b
[cos(C/2)]^2=(cosc+1)/2 所以a*cosC+a+c*cosA+c=3b a*cosC+c*cosA=b 化简为a+c=2b,