Question

三角形ABC中,cosA+cosB+cosC的最大值是多少?求确切过程哦~

Answer 1

证明一
（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.
注：仿上可证：sinA+sinB+sinC<=3√3/2
证明二
(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
证明三
(配方法)
cosA+cosB+cosC=<3/2
(1-cosA-cosB)^2+(sinA-sinB)^2>=0