Question

在△ABC,sinC+cosC=1-sinC/2,求sinC；若a^2+b^2=4(a+b)-8,求边C的值

Answer 1

你好：
(1)原方程变形为2sinC/2cosC/2+cos²C/2-sin²C/2=sin²C/2+cos²C/2-sinC/2
即sinC/2-cosC/2=1/2
(sinC/2-cosC/2)²=1/4=1-2sinC/2cosC/2 得2sinC/2cosC/2=3/4=sinC
即sinC=3/4
(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

Answer 2

1
sinC+cosC-1+sin(C/2)=0
sinC-2(sin(C/2))^2+sin(C/2)=0
2cos(C/2)-2sin(C/2)+1=0
1/2=√2sin(C/2-45)
sin(C/2-45)=1/(2√2)
1-2(sin(C/2-45))^2=cos(C-90)=sinC=1-1/4=3/4
cosC=√7/4
2
a^2+b^2=4(a+b)-8
(a-2)^2+(b-2)^2=0
a=2,b=2
c^2=a^2+b^2-2abcosC
=8-2√7
c=√7-1

Answer 3

sinC+cosC=sinC+cosC-sinC/2 化简最后C=90°所以sinC=1.

Answer 4

△ABC