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 = x/r and sinθ = y/r. 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. P is at (4,0). Find the coordinates of Q.
1
Three equilateral triangles fill half a circle (π radians). Each central angle θ = π/3.2
P is at angle 0. Q requires 2 rotations of θ = π/3, so Q is at angle 2π/3.3
Use the formula: Q = (r cos(2π/3), r sin(2π/3)) = (4cos(2π/3), 4sin(2π/3))Answer: (C) Q = (4cos(2π/3), 4sin(2π/3))
📌 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, π/2, π, 3π/2 — 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 |
| π/2 | Q = (0, 1) | 0 | 1 |
| π | R = (−1, 0) | −1 | 0 |
| 3π/2 | 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(π/6) and sin(π/6) — equilateral triangle inscribed in unit circle
An equilateral triangle inscribed in a unit circle: the y-coordinate of P at θ = π/6 is 1/2 (the midpoint of the vertical side).
Use Pythagorean theorem to find x: x² + (1/2)² = 1² → x² + 1/4 = 1 → x² = 3/4 → x = √3/2 (positive in Q1).
cos(π/6) = √3/2 · sin(π/6) = 1/2
📐 Deriving cos(π/4) and sin(π/4) — 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 = 1/2 → x = 1/√2 = √2/2.
cos(π/4) = √2/2 · sin(π/4) = √2/2
📐 Deriving cos(π/3) and sin(π/3) — equilateral triangle, different orientation
For θ = π/3: the x-coordinate is 1/2 (by symmetry with π/6 — cos and sin swap for complementary angles). y = √3/2.
cos(π/3) = 1/2 · sin(π/3) = √3/2
⭐ The 4 Q1 Values — memorize these!
0
(1, 0)
cos=1
sin=0
sin=0
π/6 (30°)
(√3/2, 1/2)
cos=√3/2
sin=1/2
sin=1/2
π/4 (45°)
(√2/2, √2/2)
cos=√2/2
sin=√2/2
sin=√2/2
π/3 (60°)
(1/2, √3/2)
cos=1/2
sin=√3/2
sin=√3/2
The Symmetry Trick — cos/sin swap for complementary angles
Notice: cos(π/6) = sin(π/3) = √3/2 and sin(π/6) = cos(π/3) = 1/2.
The pair (π/6, π/3) have each other's sin and cos values swapped! This is because π/6 + π/3 = π/2 — 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 π/2): cos + · sin +
Q2 (π/2 to π): cos − · sin +
Q3 (π to 3π/2): cos − · sin −
Q4 (3π/2 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 — π/6 Family (equilateral triangles QRO and PSO)
| Point | P (Q1) | Q (Q2) | R (Q3) | S (Q4) |
|---|---|---|---|---|
| Angle | π/6 | π − π/6 = 5π/6 | π + π/6 = 7π/6 | 2π − π/6 = 11π/6 |
| Coordinates | (√3/2, 1/2) | (−√3/2, 1/2) | (−√3/2, −1/2) | (√3/2, −1/2) |
| cos | √3/2 | −√3/2 | −√3/2 | √3/2 |
| sin | 1/2 | 1/2 | −1/2 | −1/2 |
Example 5 — π/4 Family (isosceles right triangles)
| Point | P (Q1) | Q (Q2) | R (Q3) | S (Q4) |
|---|---|---|---|---|
| Angle | π/4 | 3π/4 | 5π/4 | 7π/4 |
| Coordinates | (√2/2, √2/2) | (−√2/2, √2/2) | (−√2/2, −√2/2) | (√2/2, −√2/2) |
Example 6 — π/3 Family (equilateral triangles PQO and SRO)
| Point | P (Q1) | Q (Q2) | R (Q3) | S (Q4) |
|---|---|---|---|---|
| Angle | π/3 | 2π/3 | 4π/3 | 5π/3 |
| Coordinates | (1/2, √3/2) | (−1/2, √3/2) | (−1/2, −√3/2) | (1/2, −√3/2) |
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 Q2/Q3/Q4