Polar coordinates

A polar pair $(r, \theta)$ means: stand at the origin (the pole), face the direction θ measured counterclockwise from the positive x-axis, and walk out a distance r. The translation back to rectangular coordinates is pure right-triangle trigonometry:

1. $x = r\cos\theta$
2. $y = r\sin\theta$

Worked: (4, 2π/3) gives x = 4 cos(2π/3) = 4(−1/2) = −2 and y = 4 sin(2π/3) = 4(√3/2) = 2√3, so the point is (−2, 2√3), second quadrant, exactly where a 120° direction should land you. Keep the pairing straight: cosine builds x, sine builds y, and swapping them reflects every point across the line y = x.

What does $(-2, \pi/6)$ mean? Face the π/6 direction, then walk backward 2: a negative r puts the point on the ray pointing the opposite way. So (−2, π/6) is the same point as (2, π/6 + π) = (2, 7π/6), in the third quadrant, not the first. In rectangular terms: x = −2 cos(π/6) = −√3, y = −2 sin(π/6) = −1.

The general rule: $(-r, \theta) = (r, \theta + \pi)$. A negative r is not "invalid," and it is not a reflection across the x-axis (that would be changing the sign of θ). It is a 180° about-face. When plotting r = f(θ) curves later, the intervals where f goes negative are exactly the arcs that swing to the far side of the pole, so this convention is load-bearing, not a technicality.

Rectangular names are unique; polar names are not. Starting from $(3, \pi/4)$, all of these are the same point:

1. $(3, \pi/4 + 2\pi k)$ for any integer k: extra full turns change nothing.
2. $(-3, 5\pi/4)$: face the opposite direction and walk backward.

Every point except the pole has infinitely many polar names, two families of them. The pole itself is the extreme case: (0, θ) is the origin for every θ. This non-uniqueness is why "find the polar coordinates" problems usually pin down a window like r > 0 and 0 ≤ θ < 2π, and why checking whether two polar names agree means converting to rectangular, not comparing symbols.

From rectangular (x, y) back to polar:

1. $r^2 = x^2 + y^2$, so r = √(x² + y²) for the standard positive-r name.
2. $\tan\theta = y/x$, then check the quadrant.

The quadrant check is not optional. For (−3, 3), a calculator's arctan(3/(−3)) = arctan(−1) returns −π/4, a fourth-quadrant direction, but the point sits in quadrant II. Arctangent only reports angles from its own half-turn window and cannot see the signs of x and y separately; you can. Since (−3, 3) is in quadrant II, the correct angle is −π/4 + π = 3π/4, giving (3√2, 3π/4). Convert, look at where the point actually lives, and adjust by π when the calculator answers from the wrong side.