The unit circle
One circle of radius 1 replaces every exact value you were told to memorize. This page builds it from two triangles, shows how to read any angle off it, and works through the questions it is used for — reference angles, signs, exact values, and solving trigonometric equations.
- 1. What it actually is
- 2. Radians
- 3. Build it from two triangles
- 4. Reading any angle: reference angle + CAST
- 5. The whole circle
- 6. The other three ratios
- 7. Solving equations with it
- 8. Draw it in 20 seconds
What it actually is
The idea: a circle of radius 1 centred at the origin. For an angle θ in standard position, the point where the terminal arm crosses the circle is (cos θ, sin θ). That is the whole trick — cosine is the x-coordinate and sine is the y-coordinate.
In a right triangle, cos θ = adjacent/hypotenuse and sin θ = opposite/hypotenuse. Put that triangle inside a circle of radius 1 and the hypotenuse is 1, so the fractions collapse:
- cos θ = x/1 = x
- sin θ = y/1 = y
- tan θ = y/x = sin θ / cos θ
Standard position means the angle starts at the positive x-axis and opens counterclockwise. A negative angle opens clockwise. Because x and y on a circle of radius 1 are never bigger than 1, sine and cosine are always between −1 and 1 — an answer of sin θ = 1.4 is a signal that something went wrong, not a number to keep working with.
If the radius is not 1
When a question gives a point on the terminal arm instead, such as P(−8, 15), find r with the Pythagorean theorem and use the full definitions: r = √((−8)² + 15²) = 17, so sin θ = y/r = 15/17, cos θ = x/r = −8/17 and tan θ = y/x = −15/8. The unit circle is the special case where r = 1.
Radians
The idea: one radian is the angle that wraps an arc as long as the radius. Since the circumference is 2πr, a full turn is 2π radians — and on the unit circle, the angle is the arc length.
| Turn | Radians | Degrees |
|---|---|---|
| Full circle | 2π | 360° |
| Half | π | 180° |
| Quarter | π/2 | 90° |
| Sixth | π/3 | 60° |
| Eighth | π/4 | 45° |
| Twelfth | π/6 | 30° |
- Degrees to radians: multiply by π/180. So 150° × π/180 = 5π/6.
- Radians to degrees: multiply by 180/π. So 7π/4 × 180/π = 315°.
- Reading a radian angle quickly: the denominator says how many slices a half-turn was cut into, and the numerator counts them. 5π/6 is five sixths of the way from 0 to π — just past 2π/3, not far from π.
- Arc length: a = θr, with θ in radians. A 2.5 rad angle in a circle of radius 12 cm subtends 30 cm of arc.
Watch out: your calculator has to be in the right mode. If a question is written with π in it, work in radians. A sine of 30 that comes out as −0.988 means the calculator is in radians while you are thinking in degrees.
Build it from two triangles
The idea: you do not memorize sixteen points. You memorize two triangles, and everything in the first quadrant falls out of them.
Scale each triangle so the hypotenuse is 1, and read off the legs:
| Angle | From the triangle | cos θ (x) | sin θ (y) |
|---|---|---|---|
| 30° = π/6 | 30-60-90, short side opposite | √3/2 | 1/2 |
| 45° = π/4 | 45-45-90, legs equal | √2/2 | √2/2 |
| 60° = π/3 | 30-60-90, flipped | 1/2 | √3/2 |
The √ pattern, if you prefer a pattern
Write the first-quadrant sines in order: √0/2, √1/2, √2/2, √3/2, √4/2 — which is 0, 1/2, √2/2, √3/2, 1 for 0, π/6, π/4, π/3, π/2. The cosines are the same list backwards. Use it as a check, not as the reason: if you understand the triangles, you can rebuild the list even when nerves erase it.
Why 30° has the small sine: a 30° angle barely lifts off the x-axis, so its height (sine) is small — 1/2 — and its horizontal reach (cosine) is large — √3/2 ≈ 0.87. If you ever mix up which is which, sketch the angle and ask whether the point is high or wide.
Reading any angle: reference angle + CAST
The idea: every angle has a twin in the first quadrant, its reference angle — the acute angle to the nearest part of the x-axis. The value comes from the twin; the sign comes from the quadrant.
| Quadrant | Reference angle | Positive there |
|---|---|---|
| I (0 to π/2) | θ itself | All of them |
| II (π/2 to π) | π − θ | Sine (and csc) |
| III (π to 3π/2) | θ − π | Tangent (and cot) |
| IV (3π/2 to 2π) | 2π − θ | Cosine (and sec) |
CAST is read counterclockwise starting in quadrant IV: Cosine, All, Sine, Tangent. You can skip memorizing it entirely if you remember that cosine is x and sine is y: in quadrant II, x is negative and y is positive, so cosine is negative and sine is positive. The letters are a shortcut for the coordinate signs, nothing more.
Worked: the exact value of cos(5π/6)
- Where is it? 5π/6 is between π/2 and π — quadrant II.
- Reference angle: π − 5π/6 = π/6.
- Value from the twin: cos(π/6) = √3/2.
- Sign from the quadrant: x is negative in quadrant II, so cosine is negative.
- Answer: cos(5π/6) = −√3/2.
Worked: sin θ = 3/5 with θ in quadrant II — find cos θ and tan θ
Here 3/5 is not a special angle, so use the definitions instead. sin θ = y/r gives y = 3 and r = 5. Then x² + y² = r²: x² = 25 − 9 = 16, so x = ±4, and in quadrant II, x = −4.
So cos θ = x/r = −4/5 and tan θ = y/x = −3/4. The same triangle then feeds the double-angle identities: sin 2θ = 2 sin θ cos θ = 2(3/5)(−4/5) = −24/25.
Watch out: a reference angle is always positive and always acute. If you get π − 5π/6 = −5π/6 + π and end up with something bigger than π/2, you subtracted in the wrong direction.
The whole circle
The idea: sixteen angles, four coordinate patterns, and the signs flipping quadrant by quadrant.
| Radians | Degrees | Point (x, y) | cos θ | sin θ |
|---|---|---|---|---|
| 0 | 0° | (1, 0) | 1 | 0 |
| π/6 | 30° | (√3/2, 1/2) | √3/2 | 1/2 |
| π/4 | 45° | (√2/2, √2/2) | √2/2 | √2/2 |
| π/3 | 60° | (1/2, √3/2) | 1/2 | √3/2 |
| π/2 | 90° | (0, 1) | 0 | 1 |
| 2π/3 | 120° | (−1/2, √3/2) | −1/2 | √3/2 |
| 3π/4 | 135° | (−√2/2, √2/2) | −√2/2 | √2/2 |
| 5π/6 | 150° | (−√3/2, 1/2) | −√3/2 | 1/2 |
| π | 180° | (−1, 0) | −1 | 0 |
| 7π/6 | 210° | (−√3/2, −1/2) | −√3/2 | −1/2 |
| 5π/4 | 225° | (−√2/2, −√2/2) | −√2/2 | −√2/2 |
| 4π/3 | 240° | (−1/2, −√3/2) | −1/2 | −√3/2 |
| 3π/2 | 270° | (0, −1) | 0 | −1 |
| 5π/3 | 300° | (1/2, −√3/2) | 1/2 | −√3/2 |
| 7π/4 | 315° | (√2/2, −√2/2) | √2/2 | −√2/2 |
| 11π/6 | 330° | (√3/2, −1/2) | √3/2 | −1/2 |
Notice how little there is to learn: the numbers 1/2, √2/2, √3/2, 0 and 1 are the only values that appear, and the pattern of signs is just which half of each axis you are on.
The other three ratios
The idea: cosecant, secant and cotangent are just reciprocals. Find the sine, cosine or tangent first, then flip it.
| Ratio | Definition | On the unit circle | Undefined when |
|---|---|---|---|
| csc θ | 1/sin θ | 1/y | y = 0, so θ = 0, π, 2π |
| sec θ | 1/cos θ | 1/x | x = 0, so θ = π/2, 3π/2 |
| cot θ | cos θ/sin θ | x/y | y = 0, so θ = 0, π, 2π |
For example, sec(2π/3): first cos(2π/3) = −1/2, so sec(2π/3) = 1 ÷ (−1/2) = −2. And tan(π/2) is undefined because it would be 1/0 — the terminal arm is vertical and never crosses a vertical line at a finite height.
Watch out: csc pairs with sine even though it starts with a c, and sec pairs with cosine. Match the third letter if it helps: csc ↔ sine, sec ↔ cosine.
Solving equations with it
The idea: isolate the ratio, find the reference angle from the positive version, then place one answer in each quadrant where that ratio has the right sign.
Worked: 2 sin θ + 1 = 0 on 0 ≤ θ < 2π
- Isolate: sin θ = −1/2.
- Reference angle from the positive value: sin⁻¹(1/2) = π/6.
- Which quadrants? Sine is negative where y < 0 — quadrants III and IV.
- Build them: π + π/6 = 7π/6, and 2π − π/6 = 11π/6.
- Check the domain: both are in [0, 2π), so both stay.
The general solution, if asked: θ = 7π/6 + 2πn and θ = 11π/6 + 2πn, n ∈ I.
Worked: sin 2θ = −1/2 on 0 ≤ θ < 2π
When the angle is 2θ, solve for 2θ first and widen the domain to match: if 0 ≤ θ < 2π then 0 ≤ 2θ < 4π — two full turns.
From above, 2θ = 7π/6, 11π/6, and then the same two plus 2π: 19π/6 and 23π/6. Divide every one by 2: θ = 7π/12, 11π/12, 19π/12, 23π/12 — four solutions, which is what the doubled domain predicted.
- Second-degree equations factor first: 2sin²θ − sin θ = 0 becomes sin θ(2 sin θ − 1) = 0, so sin θ = 0 or sin θ = 1/2 — then solve each on the unit circle. Never divide both sides by sin θ, or you lose the sin θ = 0 answers.
- Mixed ratios need an identity first: 2cos²θ + sin θ = 1 becomes 2(1 − sin²θ) + sin θ = 1, a quadratic in sin θ.
- Count the answers before you finish. A first-degree equation usually has 2 solutions per turn; sin 2θ has 4 in one turn; sin 3θ has 6.
Draw it in 20 seconds
The idea: the exam gives you a formula sheet, but the unit circle is not on it. Redraw it on your scrap paper before you start.
- Draw a circle and the two axes. Label the four axis points: (1, 0), (0, 1), (−1, 0), (0, −1).
- In quadrant I, mark three dots: low, middle, high. They are π/6, π/4, π/3.
- Write their coordinates: (√3/2, 1/2), (√2/2, √2/2), (1/2, √3/2). The low dot is wide, the high dot is tall.
- Mirror those three dots into the other three quadrants — same numbers, signs by quadrant.
- Label the mirrored angles by counting sixths and quarters of π, or in degrees if that is faster for you.
Check yourself: cos(4π/3)
Quadrant III, reference angle 4π/3 − π = π/3, cos(π/3) = 1/2, and x is negative in quadrant III, so −1/2.
Check yourself: tan(3π/4)
Quadrant II, reference angle π/4, tan(π/4) = 1, and tangent is negative in quadrant II (y positive over x negative), so −1.
Check yourself: csc(7π/6)
sin(7π/6) is quadrant III with reference angle π/6, so sin = −1/2. Its reciprocal is −2.
Check yourself: the angle in [0, 2π) with cos θ = √2/2 and sin θ < 0
Cosine positive and sine negative is quadrant IV; reference angle π/4 gives θ = 2π − π/4 = 7π/4.