#cos 210 = cos (180+30) = -cos 30 = -sqrt(3)/2#.

#sin 210 = sin (180+30) = -sin 30 = -1/2#.

#3(cos 210+i sin 210) = 3(-sqrt(3)/2)+3i(-1/2)#

#-(3/2)sqrt(3)-(3/2)i#

My favourite way of seeing that #sin 30 = 1/2# and #cos 30 = sqrt(3)/2# is to picture an equilateral triangle with sides of length 1. This will have internal angles all 60 degrees (so the three angles add up to 180 degrees. Then cut the triangle in two to make two right-angled triangles. These will have internal angles of 30, 60 and 90 degrees. The shortest side (opposite the 30 degree angle) will have length #1/2# and the other side at right angles, will have length #sqrt(1^2-(1/2)^2) = sqrt(1-1/4) = sqrt(3/4) = sqrt(3)/2#.