Question

在三角形ABC中,角A.B.C的对边分别为a.b.c且cos((A+B)/2)=1-cosC

若1+tanA/tanB=2c/b且c=4求三角形面积。
求过程谢谢w

Answer 1

cos((A+B)/2)=cos((180-C)/2)=cos(90-C/2)=sinC/2=根号((1-cosC)/2)
(1-cosC)/2=(1-cosC)²
因为1-cosC≠0
所以1-cosC=1/2
cosC=1/2
C=60°

1+tanA/tanB=1+sinAcosB/sinBcosA=(sinBcosA+sinAcosB)/sinBcosA
=sin(A+B)/sinBcosA
=sinC/sinBcosA
=(c/b)/cosA ..........(正弦定理)
所以(c/b)/cosA=2c/b
cosA=1/2
A=60°

△ABC为等边三角形
S=1/2*4*4*sin60°=4√3

Answer 2

解：cos(A+B)=2cos²((A+B)/2)-1=2(1-cosC)²-1=1+2cos²C-4cosC,
A+B=π-C,
=> c0s(A+B)=c0s(π-C)=-cosC=1+2cos²C-4cosC,
=> 2cos²C-4cosC+1=-cosC
=> 2cos²C-3osC+1=0,
=> cosC=1/2或1
cosC=1舍去，cosC=1/2
=> C=π/3, A+B=2π/3;
作CD垂直AB于点D，设AD=x，则
1+tanA/tanB=1+(CD/x)/[CD/(c-x)]=1+(c-x)/x=c/x=2c/b,
=> x=b/2,
CD垂直于AB,
=> A=π/3,
A+B=2π/3,
=> B=π/3,
为等边三角形，
三角形面积SΔABC=bcsinA/2=4*4*sin(π/3)/2=4√3