cos is x, sin is y: On a circle of radius 1, the point at angle θ (measured anticlockwise from the positive x-axis) has coordinates (cos θ, sin θ). So cos θ is the x-coordinate and sin θ is the y-coordinate, and tan θ = sin θ / cos θ.
Why it matters: The unit circle lets sin, cos, tan work for any angle (including obtuse and beyond), which right-angled triangles can't do.
The special angles: You must know the exact sin, cos and tan of 0°, 30°, 45°, 60°, 90° (and their radian forms) — they appear on the non-calculator Paper 1.
sin
- 0 at 0°, 1 at 90°
cos
- 1 at 0°, 0 at 90°
tan
- = sin ÷ cos
sin and cos mirror each other: Notice sin and cos swap across 45°: sin 30° = cos 60° = ½. So you really only need one column plus the symmetry.
Practice with real exam questions
Answer exam-style questions and get AI feedback that shows you exactly what examiners want to see in a full-marks response.
Which ratios are positive where: Going round the circle, CAST tells you which ratio is positive in each quadrant: All (Q1), Sin (Q2), Tan (Q3), Cos (Q4). Outside its quadrant, a ratio is negative.
Positive ratios
- Q1 (0–90°): all positive
- Q2 (90–180°): only sin
- Q3 (180–270°): only tan
- Q4 (270–360°): only cos
So, for example
- cos 120° is negative (Q2)
- sin 200° is negative (Q3)
- tan 300° is negative (Q4)
Use it for signs, the acute angle for the value: Find the value from the matching acute angle, then attach the sign from CAST.
sin(180° − θ) = sin θ: Angles around the circle share values: sin(180° − θ) = sin θ (same sine, supplementary angle); cos(180° − θ) = −cos θ; cos(−θ) = cos θ. These let you find a second angle with the same sine — central to the ambiguous case.
IB-style question — sin from cos
Given cos θ = 2/3 with θ acute, find the exact value of sin θ.
Step by step
- Use sin²θ + cos²θ = 1.
- θ acute ⇒ sin positive.
Final answer
sin θ = √5 / 3 (exactly the audited exam step).
An angle on a straight line: If an angle sits on a straight line with θ (so it's 180° − θ), its sine is the same — handy in geometry questions.