Coordinates of a Point on a Circle
P = (r cosθ, r sinθ) · on unit circle: P = (cosθ, sinθ)
📌 Key Formula
From Topic 3.2: cosθ = xr and sinθ = yr. Multiply both sides by r:
x = r cosθ · y = r sinθ → P = (r cosθ, r sinθ)
On the unit circle (r = 1): P = (cosθ, sinθ). The x-coordinate IS cosθ and the y-coordinate IS sinθ.
📌 Example 1 — Circle of radius 4, equilateral triangle PQO inscribed. P is at (4, 0). Find the coordinates of Q.
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Key geometry fact: Three vertices equally spaced on a circle divide the full 360° into 3 equal parts. Each arc = 360°3 = 120° = 2π3. So the central angle at O between any two adjacent vertices is 2π3 — not π3. (π3 = 60° is the interior angle of the triangle at each vertex, a different thing.)1
Central angle at O = 2π3. P, Q, R are equally spaced, so each arc is 2π3. Starting from P at angle 0:P at 0
Q at 0 + 2π3 = 2π3 (120°)
R at 2π3 + 2π3 = 4π3 (240°)
2
Q is one arc-step from P. The angle to Q = 2π3. The blue arc in the diagram shows this central angle at O. The pink dashed arc shows the same 2π3 sweep along the circle from P to Q.3
Apply the formula P = (r cosθ, r sinθ) with r = 4 and θ = 2π3:Q = (4 cos2π3, 4 sin2π3)
The shortcut thinking
The three vertices divide the full circle (2π) into 3 equal parts, so each arc = 2π3. P is at 0, so Q is one arc away = 2π3, and R is two arcs away = 4π3. This is faster than counting "2 rotations of π3".
Answer: (C) Q = (4cos2π3, 4sin2π3)
📌 Example 2 — Circle of radius r, line y = −x. P = (6√2, −6√2). Find r.
x² + y² = r²
1
Substitute: (6√2)² + (−6√2)² = r²2
72 + 72 = 144 = r² → r = 12r = 12
Axis Values — Example 3
0, π2, π, 3π2 — read directly from the four axis points of the unit circle
| Angle θ | Point | cos θ = x | sin θ = y |
|---|---|---|---|
| 0 | P = (1, 0) | 1 | 0 |
| π2 | Q = (0, 1) | 0 | 1 |
| π | R = (−1, 0) | −1 | 0 |
| 3π2 | S = (0, −1) | 0 | −1 |
| 2π | P = (1, 0) again | 1 | 0 |
Deriving the Three Key Q1 Values
Use inscribed triangles + Pythagorean theorem to find exact coordinates
📐 Deriving cos(π6) and sin(π6) — equilateral triangle inscribed in unit circle
An equilateral triangle inscribed in a unit circle: the y-coordinate of P at θ = π6 is 12 (the midpoint of the vertical side).
Use Pythagorean theorem to find x: x² + (12)² = 1² → x² + 14 = 1 → x² = 34 → x = √32 (positive in Q1).
cos(π6) = √32 · sin(π6) = 12
📐 Deriving cos(π4) and sin(π4) — isosceles right triangle in unit circle
For an isosceles right triangle: x = y (45° means equal legs). Use Pythagorean theorem: x² + x² = 1 → 2x² = 1 → x² = 12 → x = 1√2 = √22.
cos(π4) = √22 · sin(π4) = √22
📐 Deriving cos(π3) and sin(π3) — equilateral triangle, different orientation
For θ = π3: the x-coordinate is 12 (by symmetry with π6 — cos and sin swap for complementary angles). y = √32.
cos(π3) = 12 · sin(π3) = √32
⭐ The 4 Q1 Values — memorize these!
0
(1, 0)
cos=1
sin=0
sin=0
π6 (30°)
(√32, 12)
cos=√32
sin=12
sin=12
π4 (45°)
(√22, √22)
cos=√22
sin=√22
sin=√22
π3 (60°)
(12, √32)
cos=12
sin=√32
sin=√32
The Symmetry Trick — cossin swap for complementary angles
Notice: cos(π6) = sin(π3) = √32 and sin(π6) = cos(π3) = 12.
The pair (π6, π3) have each other's sin and cos values swapped! This is because π6 + π3 = π2 — they are complementary angles. This same pattern works for all complementary angle pairs.
All Four Quadrants by Symmetry — Examples 4, 5, 6
Use Q1 values · flip the signs based on which quadrant P lands in
Sign Rules by Quadrant
Q1 (0 to π2): cos + · sin +
Q2 (π2 to π): cos − · sin +
Q3 (π to 3π2): cos − · sin −
Q4 (3π2 to 2π): cos + · sin −
Memory aid: "All Students Take Calculus" — All positive in Q1, Sine positive in Q2, Tangent positive in Q3, Cosine positive in Q4.
Example 4 — π6 Family (equilateral triangles QRO and PSO)
| Point | P (Q1) | Q (Q2) | R (Q3) | S (Q4) |
|---|---|---|---|---|
| Angle | π6 | π − π6 = 5π6 | π + π6 = 7π6 | 2π − π6 = 11π6 |
| Coordinates | (√32, 12) | (−√32, 12) | (−√32, −12) | (√32, −12) |
| cos | √32 | −√32 | −√32 | √32 |
| sin | 12 | 12 | −12 | −12 |
Example 5 — π4 Family (isosceles right triangles)
| Point | P (Q1) | Q (Q2) | R (Q3) | S (Q4) |
|---|---|---|---|---|
| Angle | π4 | 3π4 | 5π4 | 7π4 |
| Coordinates | (√22, √22) | (−√22, √22) | (−√22, −√22) | (√22, −√22) |
Example 6 — π3 Family (equilateral triangles PQO and SRO)
| Point | P (Q1) | Q (Q2) | R (Q3) | S (Q4) |
|---|---|---|---|---|
| Angle | π3 | 2π3 | 4π3 | 5π3 |
| Coordinates | (12, √32) | (−12, √32) | (−12, −√32) | (12, −√32) |
The Complete Unit Circle
All 16 standard angles with exact coordinates — memorize Q1, use symmetry for the rest
Complete Unit Circle — memorize Q1, use symmetry for Q2Q3/Q4