Question

在三角形ABC中,角A,B,C的对边分别是a,b,c,已知sinC+cosC=1-sin(C/2).

Answer 1

解：
∵sinC=2sin0.5C
×cos0.5C,
cosC=cos0.5C×cos0.5C-sin0.5C×sin0.5C
∴2sin0.5C
×cos0.5C+cos0.5C×cos0.5C-sin0.5C×sin0.5C+sin0.5C=1
∵∠C<180°
∴sin0.5C,cos0.5C>0,
设sin0.5C=x
则有：2x×（1-x^2）^(1/2)+1-2×x^2+x=1
解得：x=[7^(1/2)-1]/4
∴cosC=cos0.5C×cos0.5C-sin0.5C×sin0.5C=1-2sin0.5C×sin0.5C
=1-2×x^2=7^(1/2)/4
又∵a^2+b^2=4(a+b)-8
化简得到：（a-2）^2+(b-2)^2=0
∴a=b=2
根据余弦定理：c^2
=
a^2
+
b^2
-
2·a·b·cosC
=4+4-8×7^(1/2)/4
∴c=7^(1/2)-1