在三角形ABC中,角A,B,C的对边分别是a,b,c.已知sinC+cosC=1-sin0.5C.若a^2+b^2=4(a+b)-8.求边c的值

题目

在三角形ABC中,角A,B,C的对边分别是a,b,c.已知sinC+cosC=1-sin0.5C.若a^2+b^2=4(a+b)-8.求边c的值

答案

∵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
∵∠C0, 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