Question

三角形ABC中,求证cosA+cosB+cosC＞1

Answer 1

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