Question

数学问题 急急急

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) (一)求sinC的值 (二)若a^2+b^2＝4(a+b)-8 求边c的值

Answer 1

(1) sinC+cosC=1-sinC/2
sinC=1-sinC/2-cosC
2sinC/2 cosC/2=1-sinC/2-1+2sin^2C/2
2sinC/2 cosC/2=sinC/2 (2sinC/2-1)
2(sinC/2-cosC/2)=1
(sinC/2-cosC/2)²=1/4=1-2sinC/2cosC/2 得2sinC/2cosC/2=3/4=sinC
即sinC=3/4
(2) (a²-4a+4)+(b²-4b+4)=0 即(a-2)²+(b-2)²=0
得a=b=2
由余弦定理c²=a²+b²-2abcosC=4+4-2X4X√7/4=8-2√7
c=√(8-2√7) =√7-1