题目

锐角三角形ABC中,证明sinA+sinB+sinC>cosA+cosB+cosC

提问时间：2020-10-27

答案

∵0 < C < π/2,
∴0 < C/2 < π/4,
∴cos(C/2) > sin(C/2).
又∵0 < A,B < π/2,
∴-π < A-B < π,
∴-π/2 < (A-B)/2 < π/2,
∴cos((A-B)/2) > 0,
∴sin(A)+sin(B) = 2sin((A+B)/2)cos((A-B)/2)
= 2sin((π-C)/2)cos((A-B)/2)
= 2cos(C/2)cos((A-B)/2)
> 2sin(C/2)cos((A-B)/2) (∵cos(C/2) > sin(C/2),cos((A-B)/2) > 0)
= sin((C-A+B)/2)+sin((C+A-B)/2)
= sin((π-2A)/2)+sin((π-2B)/2)
= cos(A)+cos(B).
同理,可证sin(B)+sin(C) > cos(B)+cos(C),sin(C)+sin(A) > cos(C)+cos(A),
三式相加除以2即得sin(A)+sin(B)+sin(C) > cos(A)+cos(B)+cos(C).