sin(2⼀3π-C)+sinC=√3,求角C

答：
sin(2π/3-C)+sinC=√3
sin2π/3cosC-cos2π/3sinC+sinC=√3
(√3/2)cosC+(1/2)sinC+sinC=√3
(3/2)sinC+(√3/2)cosC=√3
(√3/2)sinC+(1/2)cosC=1
sin(C+π/6)=1
因为：C是三角形内角
所以：C+π/6=π/2
所以：C=π/3

sin2Pai/3cosC-cos2Pai/3sinC+sinC=根号3

根号3/2cosC+1/2sinC+sinC=根号3
1/2cosC+根号3/2sinC=1
sin(C+Pai/6)=1
C+Pai/6=2kPai+Pai/2
C=2kPai+Pai/3.

利用sinθ+sinφ = 2 sin[(θ+φ)/2] cos[(θ-φ)/2]
sin(2/3π-C)+sinC=√3
2sin[(2/3π-C+C)/2]cos[(2/3π-C-C)/2]=√3
2sin(1/3π)cos(1/3π-C)=√3
cos(1/3π-C)=1
1/3π-C=0
c=1/3π

sin(2/3π-C)+sinC=√3,
√3/2cosC+3/2sinC=√3
1/2cosC+√3/2sinC=1
sin30°cosC+cos30°+sinC=1
sin(30°+C)=1
30°+C=90°
C=60°

sin(2/3π-C)+sinC=√3
sin2/3πcosC-cos2/3πsinC+sinC=√3
√3/2cosC+1/2sinC+sinC=√3
√3/2cosC+3/2sinC=√3
1/2cosC+√3/2sinC=1
cos(C-π/3)=1
C-π/3=0
C=π/3