Question

△ABC角中,A,B,C的对边分别是abc,已知sinC+cosC=1-sinC/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