Picture a point walking around a circle of radius 1: Draw a circle of radius 1 centred at the origin. Start on the positive x-axis and rotate anticlockwise by θ: you land on (cos θ, sin θ).
Across and up: So cos θ is how far across (the x-coordinate) and sin θ is how far up (the y-coordinate).
This one picture explains everything
- Why sin and cos sit between −1 and 1 — you can't go further than the edge of the circle.
- Why they repeat every 360°.
- Why they go negative in some quadrants.
Signs by quadrant (anticlockwise from the right): Reading the sign off the picture beats memorising 'CAST' — just ask whether the point is left/right and up/down.
| Quadrant | Across / up | Signs |
|---|---|---|
| Q1 (0°–90°) | across +, up + | cos +, sin + |
| Q2 (90°–180°) | across −, up + | cos −, sin + |
| Q3 (180°–270°) | across −, up − | both − |
| Q4 (270°–360°) | across +, up − | cos +, sin − |
IB-style question — read coordinates off the circle
A wind-turbine blade tip moves on a circle of radius 1 (in suitable units). After rotating 120° anticlockwise from the positive x-axis, where is the tip?
Give its exact coordinates.
Step by step
- The tip is at (cos 120°, sin 120°). 120° lands in Quadrant 2 (across negative, up positive).
- 120° is 60° past the y-axis, so it has the same size ratios as 60° but cos is now negative.
- Write the point.
Final answer
The blade tip is at (−½, √3⁄2) — left of and above the centre, as Quadrant 2 demands.
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Five special angles worth knowing by heart: For the 'nice' angles the coordinates come out as clean exact numbers.
Why they matter: You meet these constantly in AI questions about oscillations and waves, and a GDC in degree mode confirms them.
cos θ from 0° to 90°
- Decreases 1 → 0
- The point moves left
sin θ from 0° to 90°
- Increases 0 → 1
- The point rises
The Pythagorean identity comes free from the circle: The point (cos θ, sin θ) sits on a circle of radius 1, so by Pythagoras its coordinates satisfy x² + y² = 1. That is the identity:
sin²θ + cos²θ = 1.
Use it to get one ratio from the other when you only know one of them (and the quadrant tells you the sign).
IB-style question — use the identity
A tidal-height model gives cos θ = 0.6 for an acute angle θ (so θ is in Quadrant 1).
Find the exact value of sin θ.
Step by step
- Start from the identity.
- Substitute cos θ = 0.6 = 3⁄5.
- Take the square root. θ is in Quadrant 1, so sin θ is positive.
Final answer
sin θ = 0.8. (The angle that fits a 3-4-5 triangle — a tidy check that sin²+cos² = 0.36 + 0.64 = 1.)
IB-style question — combine two sine waves
Two alternating voltages v₁ = 3 sin(t)° and v₂ = 4 cos(t)° are added.
Write 3 sin t + 4 cos t in the form R sin(t + α)°, and state the maximum voltage.
Step by step
- Expand R sin(t + α) = R cos α sin t + R sin α cos t and match the coefficients of sin t and cos t.
- R is the hypotenuse; the ratio gives α.
- So α = 53.1°. (A GDC sinusoidal regression on the combined values gives the same R and α.)
Final answer
3 sin t + 4 cos t = 5 sin(t + 53.1°); the maximum voltage is R = 5 (reached when the sine equals 1).