IB Maths AI HL Trig Identities & Equations Paper 1 & 2 Unit circle & symmetries ~9 min read

The Unit Circle

The unit circle (radius 1, centred at the origin) is the engine behind every trig value beyond 90°. A point on it at angle θ has coordinates (cos θ, sin θ); tan θ is the gradient of the radius. The four-quadrant symmetries let you express trig values of any angle in terms of an acute one — and find every solution of sin/cos/tan equations within an interval.

📘 What you need to know

Unit circle: radius 1, centre (0,0). Angles measured from the positive x-axis, anticlockwise positive, clockwise negative.
Coordinates & trig: a point on the unit circle at angle θ is (cos θ, sin θ); tan θ = sin θ / cos θ (gradient of the radius).
Quadrant signs (ASTC): Q1 (0–90°) All +; Q2 (90–180°) Sine +; Q3 (180–270°) Tangent +; Q4 (270–360°) Cosine + — “All Students Take Calculus”.
Key symmetries: sin(180°−θ) = sin θ; sin(180°+θ) = −sin θ; sin(360°−θ) = −sin θ. Same pattern for cos and tan with their own signs (see table below).
Periodicity: sin and cos repeat every 360° (2π); tan repeats every 180° (π). So sin(θ+360°) = sin θ etc.
Negative angles: sin(−θ) = −sin θ; cos(−θ) = cos θ; tan(−θ) = −tan θ.

Coordinates, quadrants, and the ASTC rule

Every angle θ measured anticlockwise from the positive x-axis lands on a unique point of the unit circle. Its coordinates are (cos θ, sin θ); the gradient of the radius from origin to that point is tan θ. The four quadrants tell you which of sin / cos / tan are positive: in Q1 all three are positive; in Q2 only sin; in Q3 only tan; in Q4 only cos. The “ASTC” mnemonic (“All Students Take Calculus”) captures this, going round anticlockwise from Q1.

The unit circle defines sin and cos as the y– and x-coordinates of the rotating point. ASTC tells you which ratios are positive in each quadrant; the symmetry table converts any angle into an acute-angle equivalent.

Unit circle point at angle θ: (x, y) = (cos θ, sin θ) · tan θ = sin θcos θ = yx Pythagorean identity: x² + y² = 1, i.e. sin²θ + cos²θ = 1

Solving simple trig equations with the unit circle

Calculators give one solution (the principal value) when you press sin⁻¹, cos⁻¹, or tan⁻¹. The unit circle gives you all the others. For sin x = k, the second solution in 0–360° is 180°−principal (same y-coordinate); for cos x = k, the second is 360°−principal (same x-coordinate); for tan x = k, the second is principal+180° (same gradient). Then add or subtract multiples of 360° (or 180° for tan) until you’ve covered the whole interval.

How many solutions to expect: for sin/cos with −1 < k < 1, there are two per full revolution (360°). For tan, one per half-revolution (180°). Multiply by how many full revolutions fit your interval to get the count.

🧭 Recipe — unit circle problems

Read off the quadrant from the signs of the coordinates (or from where the angle lies).
Find the reference angle: use sin⁻¹(|y|) or cos⁻¹(|x|) to get the acute angle from the x-axis.
Convert to the actual angle using the quadrant: Q1 = ref; Q2 = 180°−ref; Q3 = 180°+ref; Q4 = 360°−ref.
For equations: get the principal value, find the second solution using the right symmetry, then add ±360° (or ±180° for tan) to cover the interval.
Check signs at the end using ASTC — one wrong sign usually means wrong quadrant.

Worked examples

WE 1

Find the angle from unit-circle coordinates

A point P on the unit circle has coordinates (0.342, 0.940), to 3 s.f. The radius OP makes angle θ with the positive x-axis (0° ≤ θ < 360°). Find θ, to the nearest degree.

match coordinates to (cos θ, sin θ) cos θ = 0.342, sin θ = 0.940 both positive → Q1; use either ratio θ = cos⁻¹(0.342) = 70.00…° check: sin⁻¹(0.940) = 70.05…° ✓ θ ≈ 70° (nearest degree)

WE 2

Negative-coordinate point — pick the right quadrant

A point Q on the unit circle has coordinates (−0.766, −0.643), to 3 s.f. Given that θ is the angle from the positive x-axis to OQ measured anticlockwise (0° ≤ θ < 360°), find θ, to the nearest degree.

both coordinates negative → Q3 reference angle from x-axis ref = cos⁻¹(0.766) = 40.00…° Q3: θ = 180° + ref θ = 180° + 40° = 220° θ ≈ 220° (nearest degree) check: cos(220°) ≈ −0.766, sin(220°) ≈ −0.643 ✓

WE 3

Express trig of large angles in terms of acute angles

Using the unit-circle symmetries, express each of the following in terms of a trig ratio of an acute angle:
(a) sin(165°), (b) cos(230°), (c) tan(305°).

(a) 165° = 180° − 15° (Q2) sin(180° − 15°) = +sin(15°) sin(165°) = sin(15°) (b) 230° = 180° + 50° (Q3) cos(180° + 50°) = −cos(50°) cos(230°) = −cos(50°) (c) 305° = 360° − 55° (Q4) tan(360° − 55°) = −tan(55°) tan(305°) = −tan(55°) use ASTC to check signs: sin + in Q2, cos − in Q3, tan − in Q4 ✓

WE 4

Solve sin x = 0.766 in [−360°, 360°]

One solution of sin x = 0.766 is x = 50° (to the nearest degree). Find all other solutions in the range −360° ≤ x ≤ 360°, to the nearest degree.

principal value x = 50° (Q1) second solution: 180° − 50° = 130° (Q2) sin(130°) = sin(50°) ✓ subtract 360° for negative range 50° − 360° = −310° 130° − 360° = −230° x = −310°, −230°, 50°, 130° interval width is 720° → expect 4 solutions for sin. ✓

WE 5

Solve cos θ = 0.4 in [−2π, 2π] (radians)

Given that one solution of cos θ = 0.4 is θ = 1.159 rad (to 3 d.p.), find all other solutions in the range −2π ≤ θ ≤ 2π. Give answers to 3 s.f.

principal value (Q1) θ = 1.159 rad cos symmetry: cos(−θ) = cos θ ⇒ θ = −1.159 also a solution add ±2π to each 1.159 − 2π = −5.124 −1.159 + 2π = 5.124 (1.159 + 2π and −1.159 − 2π are outside [−2π, 2π]) θ ≈ −5.12, −1.16, 1.16, 5.12 rad cos is even, so solutions are symmetric about 0 — they come in ±pairs.

WE 6

Find cos and the angle from a partial value

It is given that sin θ = −0.6, and that θ lies in the third quadrant (180° < θ < 270°). Find: (a) cos θ; (b) θ to the nearest degree.

(a) use x² + y² = 1 x² = 1 − (−0.6)² = 1 − 0.36 = 0.64 x = ±0.8 Q3: cos θ negative cos θ = −0.8 (b) reference angle ref = sin⁻¹(0.6) = 36.87° Q3 → θ = 180° + ref θ = 180° + 36.87° = 216.87° θ ≈ 217° (nearest degree) Q3 means both sin and cos are negative — sign of cos was fixed by the quadrant.

💡 Top tips

Always sketch the unit circle — even a rough one shows which quadrant your angle is in and prevents sign errors.
Reference angle first, then quadrant: get the acute angle from |sin| or |cos|, then shift it by 0°/180°/360° to land in the correct quadrant.
ASTC for sign checks: after finding an angle, verify the trig values have the right signs for that quadrant.
cos is even, sin is odd: cos(−θ) = cos θ but sin(−θ) = −sin θ — useful shortcut for negative angles.
Convert between degrees and radians carefully: 180° = π rad. Set calculator mode to match the question’s units.

⚠ Common mistakes

Only giving the principal value — the inverse trig button returns one solution; the unit circle gives the rest.
Sign errors — forgetting that sin is negative in Q3/Q4 or cos is negative in Q2/Q3.
Wrong quadrant reflection: for sin x = k use 180°−principal; for cos x = k use 360°−principal (or −principal); for tan use principal+180°.
Forgetting the periodicity step when the interval extends beyond one revolution — add multiples of 360° (or 180° for tan).
Calculator in the wrong mode — radians vs degrees mismatch turns answers into nonsense.

Next up — Simple Identities. The Pythagorean identity sin²θ + cos²θ = 1 and the tan identity tan θ = sin θ / cos θ let you transform trig equations into solvable forms.

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