How do you graph the point $D(-2,(13pi)/6)$ ?

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Answer 1 · 2017-01-22T08:05:00

See the small circle representing the point of zero radius, at D.
graph{(x+sqrt3)^2+(y+1)^2-.01=0 [-10, 10, -5, 5]}

In cartesian form, it is

$D( -2cos(13/6pi), -2sin(13/6pi))=D(-2cos(pi/6), -2sin(pi/6))$

$= D(-sqrt3, -1)$

Using Socratic utility, mark the point by a circle of small radius, with

center at D

How do you graph the point D(-2,(13pi)/6)D(-2,(13pi)/6)?