锐角三角形ABC中,证明sinA+sinB+sinC>cosA+cosB+cosC
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/05/15 18:11:48
锐角三角形ABC中,证明sinA+sinB+sinC>cosA+cosB+cosC
∵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).
∴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).
锐角三角形ABC中,证明sinA+sinB+sinC>cosA+cosB+cosC
求证:在锐角三角形中,sinA+sinB+sinC>cosA+cosB+cosC
已知锐角三角形ABC,证明sinA+sinB+sinB>cosA+cosB+cosC
在锐角三角形ABC中,求证:sinA+sinB+sinC>cosA+cosB+cosC
三角函数、不等式锐角三角形ABC,求证:sinA+sinB+sinC > cosA+cosB+cosC
证明在锐角三角形ABC中sinA+sinB>cosA+cosB
sinA+sinB+sinC>=cosA+cosB+cosC
在三角形ABC中,sinA:sinB:sinC=2:3:4,则cosA:cosB:cosC=?
在三角形ABC中 sinA:sinB:sinC=4:5:6 则cosA:cosB:cosC
在锐角△ABC中,求证:sinA+sinB+sinC>cosA+cosB+cosC.
在三角形ABC中,求证sinA+sinB+sinC=4cosA/2cosB/2cosC/2.
在锐角三角形,比较sinA+sinB+sinC与cosA+cosB+cosC的大小