How do you convert #(-1/2, -sqrt3/2)# into polar form?

To convert from #color(blue)"cartesian to polar form"#

That is #(x,y)to(r,theta)#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(r=sqrt(x^2+y^2))color(white)(2/2)|)))#

and #color(red)(bar(ul(|color(white)(2/2)color(black)(theta=tan^-1(y/x))color(white)(2/2)|)))#

Here #x=-1/2" and " y=-sqrt3/2#

#rArrr=sqrt((-1/2)^2+(-sqrt3/2)^2)=sqrt(1/4+3/4)=1#

now #(-1/2,-sqrt3/2)# is in the 3rd quadrant so we must ensure that #theta# is in the 3rd quadrant.

#rArrtheta=tan^-1((-sqrt3/2)/(-1/2))#

#=tan^-1(sqrt3)=pi/3larr" reference angle"#

#rArrtheta=(pi+pi/3)=(4pi)/3larr" 3rd quadrant"#

#rArr(-1/2,-sqrt3/2)to(1,(4pi)/3)#

The polar coordinate system is an ordered pair, #(r, theta)#.

To compute r from from Cartesian coordinates, #(x,y)#, use the equation:

#r = sqrt(x^2 + y^2)#

#r = sqrt((-1/2)^2 + (-sqrt(3)/2)^2)#

#r = sqrt((1/4 + 3/4)#

#r = 1#

To compute #theta# from Cartesian coordinates, #(x,y)#, use the appropriate one of the following equations:

1. If #x > 0 and y>=0#m then use: #theta = tan^-1(y/x)#
2. If #x = 0 and y > 0#, then use: #theta = pi/2#
3. If #x = 0 and y < 0#, then use: #theta = (3pi)/2#
4. If #x < 0#, then use: #theta = pi + tan^-1(y/x)#
5. If #x > 0 and y < 0#, then use: #theta = 2pi + tan^-1(y/x)#

The appropriate one is equation 4:

#theta = pi + tan^-1((-sqrt(3)/2)/(-1/2))#

#theta = pi + tan^-1(sqrt(3))#

#theta = pi + pi/3#

#theta = (4pi)/3#

The polar point is #(1, (4pi)/3)#