設(shè)AB=c、BC=a、AC=b AB=c=2
1.
2sin(2C)cosC-sin(3C)=√3(1-cosC)
2sin(2C)cosC-sin(2C)cosC-cos(2C)sinC=√3(1-cosC)
sin(2C)cosC-cos(2C)sinC)=√3(1-cosC)
sinC=√3(1-cosC)
sinC+√3cosC=1
(1/2)sinC+(√3/2)cosC=1/2
sin(C+π/3)=1/2
C為三角形內(nèi)角,0sinC=√3(1-cosC)
sinC+√3cosC=1
錯了哦，還真是，重新寫一下：

1.
2sin(2C)cosC-sin(3C)=√3(1-cosC)
2sin(2C)cosC-sin(2C)cosC-cos(2C)sinC=√3(1-cosC)
sin(2C)cosC-cos(2C)sinC)=√3(1-cosC)
sinC=√3(1-cosC)
sinC+√3cosC=√3
(1/2)sinC+(√3/2)cosC=√3/2
sin(C+π/3)=√3/2
C為三角形內(nèi)角，0只有C+π/3=2π/3
C=π/3
2.
sinC+sin(B-A)=2sin(2A)
sin(A+B)+sin(B-A)=2sin(2A)
sinAcosB+cosAsinB-sinAcosB+cosAsinB=4sinAcosA
2cosAsinB=4sinAcosA
sinB/sinA=2
由正弦定理得b=2a
由余弦定理得
c²=a²+b²-2abcosC
a²+(2a)²-2a(2a)cos(π/3)=2²
a²+(2a)²-2a(2a)(1/2)=4
整理，得
3a²=4
a²=4/3
S△ABC=(1/2)absinC
=(1/2)a(2a)sin(π/3)
=a²sin(π/3)
=(4/3)(√3/2)
=2√3/32cosAsinB=4sinAcosA
缺cosA=o的情況？？？