A point is named by a distance and an angle.
The triangle is the same one from the unit circle, scaled up by r. That is the whole derivation of both conversion formulas.
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- The Polar Coordinate System
The Polar Coordinate System
How far, and in which direction — instead of how far across and how far up.
Convert to rectangular form
(4, π3)
x = r cos θ = 4 cos(π3) = 4(12) = 2, and y = r sin θ = 4 sin(π3) = 4(√32) = 2√3. So the point is (2, 2√3). B swaps them, using sine for x — cosine always gives the x-coordinate, exactly as on the unit circle.
Polar to rectangular: multiply r by the cosine and the sine.
With r and θ known, x = r cos θ and y = r sin θ give the point directly. The signs come out of the cosine and sine by themselves.
Convert (8, 5π6) to rectangular form.
x = 8 cos(5π6) = 8(−√32) = −4√3
cosine is negative in quadrant II
y = 8 sin(5π6) = 8(12) = 4
sine is positive there
(−4√3, 4)
the rectangular point
check: (−4√3)² + 4² = 48 + 16 = 64 = 8² ✓
x² + y² must equal r²
Swapping sine and cosine
Using x = r sin θ gives (4, −4√3), a point in quadrant IV, nowhere near where 5π6 points. Cosine belongs to x, exactly as on the unit circle.
Going back needs the quadrant, not just a calculator.
r = √x² + y² = √9 + 27
= √36 = 6
tan θ = yx = 3√3−3 = −√3
reference angle π3
(−3, 3√3) is in quadrant II
x negative, y positive
θ = π − π3 = 2π3
(6, 2π3)
arctan cannot tell quadrant II from quadrant IV
Its range is only [−π2, π2], so it returns −π3 for this point — an angle in quadrant IV, pointing the opposite way. Always check the signs of x and y and place the angle yourself.
One point, infinitely many names.
(5, π6)
the usual name
r positive, θ in [0, 2π)
(5, 13π6)
same point
one extra full turn
(−5, 7π6)
same point again
negative r reverses the direction
In rectangular coordinates every point has exactly one name. In polar coordinates it has infinitely many, which is a feature rather than a flaw — but it means "the" polar form usually needs a stated convention.
Simple polar equations describe shapes rectangular ones struggle with.
r = 4
a circle of radius 4
vs x² + y² = 16
θ = π3
a straight line through the origin
vs y = √3x
r = 2cos θ
a circle, off-centre
much uglier in rectangular form
Anything organised around a centre — circles, spirals, petal curves, radar sweeps — is simpler in polar form. Anything organised around a grid is simpler in rectangular.
Variations: a negative r, a point on an axis, and a polar equation turned rectangular.
The same substitutions, where a sign, a zero or an equation makes them less automatic.
A negative r: convert (−2, π3) to rectangular form
x = −2 cos(π3) = −2(12) = −1
the formulas work with negative r
y = −2 sin(π3) = −2(√32) = −√3
no special rule needed
(−1, −√3)
quadrant III, opposite to π3
A negative r walks backwards from the direction θ, landing in the opposite quadrant.
A point on an axis: convert (0, −5) to polar form
r = √0 + 25 = 5
distance from the origin
yx = −50 is undefined
arctan cannot be used at all
straight down the negative y-axis: θ = 3π2
read it off the picture
(5, 3π2)
the polar point
On an axis, the angle is 0, π2, π or 3π2; look, do not divide.
An equation: convert r = 6 sin θ to rectangular form
r² = 6r sin θ
multiply both sides by r
x² + y² = 6y
r² = x² + y² and r sin θ = y
x² + (y − 3)² = 9
complete the square
a circle, centre (0, 3), radius 3
it touches the origin
Multiplying by r creates the r² and r sin θ that have rectangular partners.
Converting an equation uses the same two substitutions.
x² + y² = 49
a circle, radius 7
x² + y² = r²
the key substitution
r² = 49, so r = 7
two symbols instead of six
x² + y² = r² is worth memorising on its own. It follows straight from x = r cos θ and y = r sin θ together with sin²θ + cos²θ = 1, and it turns most circle conversions into one line.
On the exam
Tests ask to convert a point either way, to name what a simple polar equation such as r = 5 or θ = π6 draws, or to rewrite x² + y² = k in polar form. The fastest route: x = r cos θ, y = r sin θ one way; r = √x² + y² and the quadrant of (x, y) for θ the other. A 10-second check: x² + y² must equal r², and the point must sit in the quadrant θ points to.
Now you try
Practice problem
Convert to rectangular form
(6, π6)
x = 6 cos(π6) = 6(√32) = 3√3, and y = 6 sin(π6) = 6(12) = 3, giving (3√3, 3). B swaps sine and cosine. A useful check: π6 is a shallow angle, so the point should be far across and not very high — and 3√3 ≈ 5.2 against 3 fits that.
Hint: cosine gives x and sine gives y, the same way round as on the unit circle. Sketching the angle tells you which coordinate should be larger.