三角形ABC,求cosA+cosB+cosC的值域
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cosA+cosB+cosC
=2cos[(A+B)/2]*cos[(A-B)/2]-cos(A+B)
=2cos[(A+B)/2]*cos[(A-B)/2]-2{cos[(A+B)/2]}^2+1
=2cos[(A+B)/2]*{cos[(A-B)/2]-cos[(A+B)/2]}+1
≤2sin(C/2)[1-sin(C/2)]+1
=-2[sin(C/2)]^2-sin(C/2)+1/4-1/4]+1
=-2[sin(C/2)-1/2]^2+3/2
≤3/2
又∵cosA+cosB+cosC
=2cos[(A+B)/2]*{cos[(A-B)/2]-cos[(A+B)/2]}+1
=4sin(A/2)sin(B/2)sin(C/2)+1>1
∴值域是(1,3/2]
=2cos[(A+B)/2]*cos[(A-B)/2]-cos(A+B)
=2cos[(A+B)/2]*cos[(A-B)/2]-2{cos[(A+B)/2]}^2+1
=2cos[(A+B)/2]*{cos[(A-B)/2]-cos[(A+B)/2]}+1
≤2sin(C/2)[1-sin(C/2)]+1
=-2[sin(C/2)]^2-sin(C/2)+1/4-1/4]+1
=-2[sin(C/2)-1/2]^2+3/2
≤3/2
又∵cosA+cosB+cosC
=2cos[(A+B)/2]*{cos[(A-B)/2]-cos[(A+B)/2]}+1
=4sin(A/2)sin(B/2)sin(C/2)+1>1
∴值域是(1,3/2]
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