A point described in polar coordinates also has rectangular coordinates. Converting between them is the same calculation as finding vector components and magnitudes.
Key idea
Polar to rectangular: $x = r\cos\theta$, $y = r\sin\theta$. Rectangular to polar: $r^2 = x^2 + y^2$ and $\tan\theta = \frac{y}{x}$, with $\theta$ chosen in the quadrant where $(x, y)$ actually is. For equations, substitute $x = r\cos\theta$, $y = r\sin\theta$, $x^2 + y^2 = r^2$.
Why it works
The point at distance $r$ along the ray at angle $\theta$ is the unit-circle point $(\cos\theta, \sin\theta)$ stretched by $r$. Going back, $r$ is the distance from the origin (the Pythagorean theorem), and $\frac{y}{x}$ is the slope of the ray, which is $\tan\theta$.
Worked examples
Worked example: Polar to rectangular
Convert $\left(4, \frac{2\pi}{3}\right)$ to rectangular coordinates.
- $x = 4\cos\frac{2\pi}{3} = 4 \cdot \left(-\frac{1}{2}\right) = -2$.
- $y = 4\sin\frac{2\pi}{3} = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}$.
Answer: $\left(-2, 2\sqrt{3}\right)$
Worked example: Rectangular to polar
Convert $\left(-1, -\sqrt{3}\right)$ to polar coordinates with $r > 0$ and $0 \le \theta < 2\pi$.
- $r = \sqrt{1 + 3} = 2$.
- $\tan\theta = \frac{-\sqrt{3}}{-1} = \sqrt{3}$, so the reference angle is $\frac{\pi}{3}$.
- The point is in Quadrant III, so $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$.
Answer: $\left(2, \frac{4\pi}{3}\right)$
Worked example: An equation
Write $r = 4\cos\theta$ in rectangular form.
- Multiply both sides by $r$: $r^2 = 4r\cos\theta$.
- Substitute: $x^2 + y^2 = 4x$.
- Complete the square: $(x - 2)^2 + y^2 = 4$, a circle of radius $2$ centered at $(2, 0)$.
Answer: $x^2 + y^2 = 4x$
Worked example: The other way
Write the line $x = 3$ as a polar equation solved for $r$.
- Replace $x$: $r\cos\theta = 3$.
- Solve: $r = \frac{3}{\cos\theta} = 3\sec\theta$.
Answer: $r = 3\sec\theta$
Common mistake
Using $\theta = \tan^{-1}\frac{y}{x}$ for a point in Quadrant II or III. For $\left(-1, -\sqrt{3}\right)$, $\tan^{-1}\sqrt{3} = \frac{\pi}{3}$ points into Quadrant I, the wrong side of the origin. Add $\pi$ when $x < 0$. Always check that your $(r, \theta)$ converts back to the original point.