已知△ABC的三个内角为A,B,C,所对的三边分别为a,b,c,若△ABC的面积为S=a^2-(b-c)^2,则tanA/2等于
问题描述：
已知△ABC的三个内角为A,B,C,所对的三边分别为a,b,c,若△ABC的面积为S=a^2-(b-c)^2,则tanA/2等于
A.1/2 B.1/4 C.1/8 D.1
答
S = a^2-(b-c)^2 = a^2-b^2-c^2 + 2bc = -2bc*cosA + 2bc = 4bc*[sin(A/2)]^2
S = bc*(sinA)/2
==> 4bc*[sin(A/2)]^2 = bc*(sinA)/2 = bc*sin(A/2)cos(A/2)
==> tan(A/2) = 1/4
正余弦定理的应用