(a+b)/c = cos[1/2(α-β)] / sin(γ/2)在任何三角形中适用,求证.a,b

(a+b)/c = cos[1/2(α-β)] / sin(γ/2)在任何三角形中适用,求证.a,b

（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 = 180A/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

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