gbu94mz95e的部落格

標題:

a-maths~

發問:

a=ksinA,b=ksinB,c=ksinC,其中∠A+∠B+∠C=180°且k>0。利用複角公式証明a^2=b^2+c^2-2bc cosA

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最佳解答:

a/sinA = b/sinB = c/sinC = ka = ksinA,b = ksinB,c = ksinC a^2 = b^2 + c^2 - 2bc cosA (b^2 + c^2 - a^2)/2bc = cosA LHS= (b2+c2-a2)/2bc= [(ksinB)^2+(ksinC)^2-(ksinA)^2]/2(ksinB)(ksinC)=[(sinB)^2+(sinC)^2-(sinA)^2]/2(sinB)(sinC)= [(sinB)^2+(sinC)^2-sin(180-(B+C))^2]/2(sinB)(sinC)= [(sinB)^2+(sinC)^2-sin(B+C)^2]/2(sinB)(sinC)= [(sinB)^2+(sinC)^2-(sinBcosC+cosBsinC)^2]/2(sinB)(sinC)= [(sinB)^2+(sinC)^2-(sinBcosC)^2-(cosBsinC)^2-2sinBsinCcosBcosC]/2(sinB)(sinC)= [2(sinB)^2(sinC)^2 - 2sinBsinCcosBcosC]/2(sinB)(sinC)= sinBsinC - cosBcosC = -cos(B + C)= cos(180 - (B + C))= cosA= RHS Hope it helps !!!

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