(a+b)/c = cos[1/2(α-β)] / sin(γ/2)在任何三角形中适用,求证.a,b,c是三角形三边,α,β,γ是三角形三个角字母对应，是每个角的对边。
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（a+b)/c
由正弦定理= (sinA + sinB )/sinC
其中由和差化积公式
sinA + sinB = 2sin[(A+B)/2]cos[(A-B)/2]
由二倍角公式
sinC = 2sin(c/2)cos(c/2)
又因为 A + B + C = 180
A/2 + B/2 + C/2 = 90
所以sin[(A+B)/2] = cos(c/2)
所以原式= cos[(A-B)/2]/sin(c/2)
A B C 是对应的三个角
======以下答案可供参考======
供参考答案1:
α，β，γ还是用a，b，c表示吧。
(a+b)/c
=a/c+b/c
=sina/sinc+sinb/sinc
=(sina+sinb)/sinc
=(sin[1/2(a+b)+1/2(a-b)]+sin[1/2(a+b)-1/2(a-b)]/sinc
=2sin[1/2(a+b)]cos[1/2(a-b)]/2sin(c/2)cos(c/2)
=cos(c/2)cos[1/2(a-b)]/sin(c/2)cos(c/2)
=cos[1/2(a-b)]/sin(c/2)
供参考答案2:
Cos[1/2 (A - B)]/Sin[(C)/2]
= Cos[(A - B)/2]/Sin[(Pi - A - B)/2]
= Cos[(A - B)/2]/Cos[(A + B)/2]
= (Cos[(A - B)/2]*2 Sin[(A + B)/2])/(2 Sin[(A + B)/2]*Cos[(A + B)/2])
= (Cos[(A - B)/2]*2 Sin[(A + B)/2])/Sin[(A + B)]
= (Sin[A] + Sin[B])/Sin[(A + B)]
= (Sin[A] + Sin[B])/Sin[Pi - (A + B)]
= (Sin[A] + Sin[B])/Sin[C], (正弦定理得)
= (a + b)/c