Question

在三角形abc中,角A,B,C的对边是abc,已知sinC+cosC=1-sin2/C(1)求sinC的值

Answer 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
根据辅助角公式得sin(C/2-π/4)=√2 /4
平方后根据被叫公式得cos(C-π/2)=3/4
∴sinC=3/4
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Answer 2

(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 3

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)
（sinC/2-cosC/2)=1/2 两边
同时平方cos^2C/2+sin^C/2-2cosC/2sinC/2=1/4即1+sinC=1/4,所以解得sinC=3/4