Question

三角形ABC中，角A,B,C的对边为a，b，c，已知sinC+cosC=1-sin1/2C 1，求sinC值

Answer 1

sinC+cosC=1-sinC/2，移项得
sinC-sinC/2
=
1-cosC
由二倍角公式得
2sinC/2
cosC/2-sinC/2
=
2(sinC/2)^2
因为sinC/2≠0，所以两边消去sinC/2得
2cosC/2-1
=
2sinC/2
整理得
sinC/2-cosC/2=1/2
根据辅助角公式得sin(C/2-π/4)=√2
/4
再由二倍角公式得cos(C-π/2)=1-2sin(C/2-π/4)^2=3/4
∴sinC=cos(C-π/2)=3/4
请点击“采纳为答案”

Answer 2

1-(sinC+cosC)=sin(C/2)>0
1
-
2
(sinC
+
cosC)
+(sinC)^2+(cosC)^2
+
2 sinC
*
cosC
=
[sin
(C/2)]^2
=
(1-cosC)/2
2(1-sinC)(1-cosC)
= (1-cosC)/2
讨论（1）cosC=0，
C=90
，不符题意(1-(sinC+cosC)=sin(C/2)>0)，舍去
讨论
（2）cosC不等于0，sin
C
=
3/4，解毕。