Question

sin(x+π/3)+2sin(x-π/3)-√3cos(2π/3 - x)

化简、、、

Answer 1

原式=sinxcosπ/3+cosxsinπ/3+2sinxcosπ/3-2cosxsinπ/3-√3cosxcos2π/3-√3sinxsin2π/3
=1/2*sinx+√3/2*cosx+sinx-√3cosx+√3/2*cosx-3/2*sinx
=0

Answer 2

展开得到
原式
=sinx*cosπ/3
+cosx
*sinπ/3
-
√3
*cos2π/3
*cosx
-√3
*sin2π/3*sinx
+2sinx*cosπ/3
-2cosx
*sinπ/3
=sinx*
1/2
+cosx
*√3/2
-√3
*(-1/2)
*cosx
-√3
*√3/2
*sinx
+2sinx
*1/2
-2cosx
*√3/2
=(1/2
-3/2+1)*sinx
+(√3/2
+√3/2
-√3)*cosx
=0

Answer 3

sin(x+π/3)+2sin(x-π/3)-√3cos(2π/3 - x)
=sin(x+π/3)-√3cos(2π/3 - x)+2sin(x-π/3)
=sin(x+π/3)-√3cos(π-(π/3 + x))+2sin(x-π/3)
=sin(x+π/3)+√3cos(π/3 + x)+2sin(x-π/3)
=2(cos(π/3)sin(x+π/3)+sin(π/3)cos(π/3 + x))+2sin(x-π/3)
=2sin(x+2π/3)+2sin(x-π/3)
=2sin(π-(π/3-x))+2sin(x-π/3)
=2sin(π/3-x)+2sin(x-π/3)
=0