Question

在三角形abc中 角a b c的对边分别为abc，已知sinc+cosc=1-sinc/2

Answer 1

1、sinC+cosC=1-sin（C/2）2sin(C/2)cos(C/2)=2sin²(C/2)-sin（C/2)∵sin(C/2)≠0∴2cos(C/2)=2sin(C/2)-1sin(C/2)-cos(C/2)=1/2[sin(C/2)-cos(C/2)]^2=1/4 1-sinC=1/4, sinC=3/42、∵a^2+b^=4（a+b）-8∴(a-2)^2+(b-2)^2=0∴a=2,b=2∵sin(C/2)-cos(C/2)=1/2∴[sin(C/2)+cos(C/2)]^2=1+3/4=7/4∴sin(C/2)+cos(C/2)=√7/2∴sin(C/2)=(√7+1)/4∴cosC=1-sin(C/2)-sinC=1/4-sin(C/2)=-√7/4∴c^2=a^2+b^2-2abcosC=8+2√7∴c=1+√7