Question

三角形ABC,求cosA+cosB+cosC的值域。

Answer 1

1）逐步调整法由和差化积公式得
cosA+cosB+cosC+cos(π/3)
=2cos[(A+B)/2]cos[(A-B)/2]+2cos[(C+π/3)/2]cos[(C-π/3)/2]
<=2{cos[(A+B)/2]+cos[(C+π/3)/2]}
=4cos[(A+B+C+π/3)/4]cos[(A+B-C-π/3)/4]
<=4cos[(A+B+C+π/3)/4]
=4cos[(π+π/3)/4]
=4cos(π/3),
所以 cosA+cosB+cosC<=3cos(π/3)=3/2.
2) 一元化方法
cosA+cosB+cosC=cosA+2cos[(B+C)/2]cos[(B-C)/2]
<=cosA+2cos[(B+C)/2]
=1-2[sin(A/2)]^2+2sin(A/2)
=-2[(sin(A/2)-1/2]^2+3/2
<=3/2

最小值
cosA+cosB+cosC
=cosA+cosB-cos(A+B)
=2cos[(A+B)/2]cos[(A-B)/2]-{[cos((A+B)/2)]^2-1}
=1+2cos[(A+B)/2]{cos[(A-B)/2]-cos[(A+B)/2]}
=1+2sin(C/2)[-2sin(A/2)sin(-B/2)]
=1+4sin(A/2)sin(B/2)sin(C/2)
因为△ABC中sin(A/2)>0,sin(B/2)>0,sin(C/2)>0,所以
1+sin(A/2)sin(B/2)sin(C/2)>1
图形分析:当钝角达到接近180度时，设C接近180度
cosA+cosB+cosC = 1+1-1=1