把Z=根号3SINa+COSa-2iCOS(a+π/6）化为三角形式

z=√3*sina+cosa-2i*cos(a+π/6)z=2(√3/2*sina+1/2*cosa)-2i*cos(a+π/6)z=2(sina*cosπ/6+cosa*sinπ/6)-2i*cos(a+π/6)z=2sin(a+π/6)-2i*cos(a+π/6)z=2[sin(a+π/6)-i*cos(a+π/6)]z=2[cos(a+π/6-π/2)+i*sin(...