三角形ABC中,求证cosA+cosB+cosC＞1
1个回答
三角和差公式:
( cosA + cosB )
= 2 * cos[(A+B)/2] * cos[(A-B)/2]
( cosA - cosB )
= -2 * sin[(A+B)/2] * sin[(A-B)/2]
倍角公式:
cosC = cos(pi-A-B) = -cos(A+B)
= -2 * {cos[(A+B)/2]}^2 + 1
cosA + cosB + cosC
= (2 * cos[(A+B)/2] * cos[(A-B)/2]) + (-2 * {cos[(A+B)/2]}^2 + 1)
= 2 * cos[(A+B)/2] * ( cos[(A-B)/2] - cos[(A+B)/2] ) + 1
= 2 * sin(C/2) * [ 2 * sin(A/2) * sin(B/2) ] + 1
= 4 * sin(A/2) * sin(B/2) * sin(C/2) + 1
因为是锐角三角形,所以0 < A、B、C < π/2
因此sin(A/2) 、 sin(B/2) 、 sin(C/2) 均大于0
即 cosA + cosB + cosC > 1
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