請簡化下列三角函數方程式:?

1)
Br=sin(wt-120°)*(-j)
Bs=sin(wt)*(-√3i+j)/2
Bt=sin(wt-240°)*(√3i+j)/2
Beq=Br+Bs+Bt
=-jsin(wt-120°)-√3/2i sin(wt)+j/2 sin(wt) + √3/2i sin(wt-240°)+ j/2 sin(wt-240°)
=√3/2i ( sin(wt-240°) - sin(wt) )+ j/2 ( sin(wt-240°)+j/2 sin(wt) ) - jsin(wt-120°)
=√3/2i ( 2 cos (1/2) ( wt-240°+wt)* sin (1/2) ( wt-240°- wt) + j/2 ( sin(wt) + sin(wt-240°)) - j sin(wt-120°)
=√3/2i ( 2 cos (wt-120°)*sin ( -120° ) + j/2 ( 2sin(1/2)(wt+wt-240°)*cos (1/2))(wt-wt+240°) - j sin(wt-120°)
=√3/2i ( 2 cos (wt-120°)* ( -√3/2 ) ) + j/2 ( 2sin( wt-120°)*cos ( 120°) ) - j sin(wt-120°)
= -3/2i *cos (wt-120°) + j/2 ( 2sin( wt-120°)*( -1/2) ) - j sin(wt-120°)
= -1.5i cos (wt-120°) - j/2 sin( wt-120° ) - j sin(wt-120°)
= -1.5i cos (wt-120°) - 1.5 j sin( wt-120° )

2)
Br=sin(wt-240°)*(-j)
Bs=sin(wt-120°)*(-√3i+j)/2
Bt=sin(wt)*(√3i+j)/2
Beq=Br+Bs+Bt
=-j sin (wt-240°)-√3i/2 sin(wt-120°)+ j/2 sin(wt-120°)+ √3i/2 sin(wt) + j/2 sin(wt)
=√3i/2 [ sin(wt) - sin(wt-120°) ] + j/2[ sin(wt) + sin(wt-120°) ] - j sin (wt-240°)
=√3i/2 [ 2 cos (1/2) ( wt + wt -120°)*sin(1/2)(wt- (wt-120°)) ] + j/2[2 sin(1/2)(wt+wt-120°)*cos(1/2)(wt-wt +120°) ] - j sin (wt-240°)
=√3i/2 [ 2 cos ( wt -60°)*sin60° ] + j/2[2 sin(wt-60°)*cos(60°) ] - j sin (wt-240°)
=√3i/2 [ 2 cos ( wt -60°)*√3/2 ] + j/2[ sin(wt-60°) ] - j sin (wt-240°)
=3i/2 [ cos ( wt -60°) ] + j/2[ sin(wt-60°) ] - j sin (wt-60°-180°)
=1.5 i [ cos ( wt -60°) ] + j/2[ sin(wt-60°) ] - j [sin (wt-60°)*cos180° - cos (wt-60°)sin180°]
=1.5 i [ cos ( wt -60°) ] + j/2[ sin(wt-60°) ] - j [sin (wt-60°)*(-1) ]
=1.5 i [ cos ( wt -60°) ] + j/2[ sin(wt-60°) ] + j [sin (wt-60°) ]
=1.5 i cos ( wt -60°) + 1.5 j sin(wt-60°)