在△ABC中,a,b,c,分别是角A,B,C的对边,S为△ABC的面积,若a+b=2,且2S=c^2-(a-b)^2

显示全部楼层 · 2012-6-24 17:50:37

（1）s=1/2absinCcosC=(a^2+b^2-c^2)/2ab于是ab.sinC=2ab-2ab.cosC解得 sinC/(1-cosC)=2 （1）(2)由S\sin^2 C+cos^ 2C=1(2)联立求得 5cosC^2-8cosC+3=0求得 cos C=1
(C角度为0度舍去）或者cosC=0.6于是sinC=0.8于是S=1/2ab*sinC=0.4ab因为ab≤1/4*(a+b)^2=1/4*2^2=1于是S的最大值为0.4（等号成立条件 a=b=1)...

千问 · 2012-6-24 17:50:37

S=sqrt(s(s-a)(s-b)(s-c)), s = (a+b+c)/2 = 1+c/2, c= 2(s-1)2S = c^2 - (a-b)^2 = (a+c-b)(c-a+b)a = c sinA/sinC = 2(s-1) sinA/sinCb = c sinB/sinC = 2(s-1) sinA/sinC代入即可，我现在没时间...