r = √(x^2+y^2)
tanθ = y/x
so, (-√6,-√2) = (√(6+2),arctan(-√2/-√6)) = (2√2,7π/6)
To go the other way, just recall that
x = rcosθ
y = rsinθ
what is (sqrt 2, -pi/4) in rectangular coordinates?
what is (-sqrt 6, -sqrt 2) in polar coordinates?
1 answer
1 answer