IB DP Mathematics: Analysis and Approaches – Circular functions, trigonometric identities and equations Revision Notes

For full explanations and longer worked examples, use the circular functions study guide. These notes are for the final weeks of revision.

They cover IB Diploma Programme Mathematics: Analysis and Approaches, aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 3.5–3.11. Sections 3.5–3.8 are SL and HL; sections 3.9–3.11 are HL only (AHL). They follow the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so they apply to the May and November 2026, 2027 and 2028 sessions.

Test yourself afterwards with the practice questions. The IB DP Maths AA course hub and the printable syllabus checklist list the rest of the course.

Definitions

• Unit circle: the point at angle θ, measured anticlockwise from the positive x-axis, is (cos θ, sin θ).
• tan θ = sin θ / cos θ. The line through the origin at angle θ to the positive x-axis is y = x tan θ.
• Amplitude: half the distance from minimum to maximum, |a| in a sin(b(x + c)) + d.
• Period: the length of one complete cycle.
• Principal axis: the horizontal line y = d midway between maximum and minimum.
• Reciprocal ratios (HL only): sec θ = 1/cos θ, cosec θ = 1/sin θ, cot θ = 1/tan θ.
• Inverse functions (HL only): arcsin, arccos and arctan return one angle from a restricted range.

Radians are assumed on exam papers unless degrees are indicated.

Exact values

θ	0	π/6	π/4	π/3	π/2
sin θ	0	1/2	√2/2	√3/2	1
cos θ	1	√3/2	√2/2	1/2	0
tan θ	0	√3/3	1	√3	undefined

Signs by quadrant: first, all positive; second, sin only; third, tan only; fourth, cos only.

Formulas and identities

Result	Level
cos²θ + sin²θ = 1	SL and HL
sin 2θ = 2 sin θ cos θ	SL and HL
cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ	SL and HL
cos(−x) = cos x, sin(−x) = −sin x, sin(π + x) = −sin x	SL and HL
Period of a sin(bx) and a cos(bx) is 2π/b; of tan(bx) is π/b	SL and HL
1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ	HL only
sin(A ± B) = sin A cos B ± cos A sin B	HL only
cos(A ± B) = cos A cos B ∓ sin A sin B	HL only
tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)	HL only
tan 2θ = 2 tan θ / (1 − tan²θ)	HL only
sin(π − θ) = sin θ, cos(π − θ) = −cos θ, tan(π − θ) = −tan θ	HL only

Inverse functions (HL only)

Function	Domain	Range
arcsin x	[−1, 1]	[−π/2, π/2]
arccos x	[−1, 1]	[0, π]
arctan x	ℝ	]−π/2, π/2[

Method in steps

Find one ratio from another (3.6)

1. Use cos²θ + sin²θ = 1 (or, HL, 1 + tan²θ = sec²θ) to find the square of the missing ratio.
2. Take the square root and pick the sign from the quadrant.
3. If no quadrant is given, keep both signs.

Ambiguous case of the sine rule (3.5)

1. Use the sine rule to find sin of the unknown angle.
2. Take θ = arcsin(value) and also 180° − θ.
3. Keep 180° − θ only if it plus the given angle is under 180°.

Read a circular function (3.7) for f(x) = a sin(b(x + c)) + d:

1. Amplitude |a|, maximum d + |a|, minimum d − |a|.
2. Period 2π/b.
3. Horizontal shift −c: factorise b out of the bracket first.

Solve an equation in an interval (3.8)

1. Isolate the trig function: sin(bx + c) = k.
2. Change the interval to match bx + c.
3. List every angle in that interval with the right sign.
4. Solve for x and check each value is in the original interval.

Quadratic in one ratio (3.8)

1. Use cos²x = 1 − sin²x, or cos 2x = 1 − 2sin²x or 2cos²x − 1, so only one ratio remains.
2. Rearrange to = 0 and factorise.
3. Reject any root outside [−1, 1] for sin or cos.
4. Solve each remaining equation in the interval.

Worked reminders

Double angle. sin x = 1/3, find cos 2x: cos 2x = 1 − 2(1/9) = 7/9.

Transformations. f(x) = 2 sin(3(x − π/4)) + 5 is y = sin x stretched by scale factor 2 vertically and 1/3 horizontally, translated π/4 right and 5 up. Its period is 2π/3 and its range is 3 ≤ f(x) ≤ 7.

Factorise, don’t divide. 2sin²x = sin x on 0 ≤ x ≤ 2π. Write sin x(2 sin x − 1) = 0. sin x = 0 gives 0, π, 2π; sin x = 1/2 gives π/6, 5π/6.

One ratio from another. cos θ = −1/3 with θ in the second quadrant: sin²θ = 1 − 1/9 = 8/9, and sin is positive there, so sin θ = 2√2/3 and tan θ = −2√2.

Model from a context. A point on a wheel of radius 6 m, with its centre 8 m above the ground, turns once every 30 seconds and starts at the bottom. Then d = 8, the amplitude is 6, b = 2π/30 = π/15 and h(t) = 8 − 6 cos(πt/15). The minimum is 2 m and the maximum is 14 m.

Ambiguous case check. Given angle A = 40°, BC = 9 and AC = 12: sin B = 12 sin 40°/9 = 0.857, so B = 59.0° or 121°. Since 40° + 121° < 180°, both triangles exist.

Graphical solving (Paper 2). Graph both sides of the equation in the correct angle mode and read the intersections inside the interval only. Write down the equation you graphed, then give each answer to 3 significant figures.

Double angle for tan (HL only). Put B = A in tan(A + B): tan 2A = 2 tan A/(1 − tan²A). With tan A = 3, tan 2A = 6/(1 − 9) = −3/4.

Symmetry (HL only). If sin x = 0.4 has solution α in [0, π/2], the other solution in [0, π] is π − α, because sin(π − θ) = sin θ.

Compound angle (HL only). cos 15° = cos(45° − 30°) = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.

Inverse (HL only). arccos(−1/2) = 2π/3, not −π/3, because the range of arccos is [0, π].

Must-know distinctions

• Period vs frequency factor: b is not the period; the period is 2π/b.
• a vs |a|: a negative a reflects the graph in the principal axis; the amplitude is still |a|.
• sin 2x vs 2 sin x: sin 2x = 2 sin x cos x; they are not equal.
• sin²x vs sin x²: sin²x means (sin x)².
• arcsin x vs cosec x (HL only): arcsin x is an angle; cosec x = 1/sin x is a ratio.
• Finite interval vs general solution: the guide only requires solutions in a stated interval.
• Degrees vs radians: match the interval. An interval like 0 ≤ x ≤ 2π means radians.
• Given angle obtuse: if the given angle is 90° or more, the ambiguous case cannot give two triangles.

Quick self-test

1. Find the exact values of sin(4π/3), cos(−5π/4) and tan 150°.
2. Given sin x = 2/3 and x is obtuse, find the exact value of tan x.
3. Given sin x = 1/3, find cos 2x.
4. State the amplitude, period and range of f(x) = 5 cos 4x − 2.
5. State the period of y = tan(x/2).
6. Solve 2 cos x = −1 for 0 ≤ x ≤ 2π.
7. Solve tan²x = 3 for 0 ≤ x ≤ π.
8. Solve 2sin²x = sin x for 0 ≤ x ≤ 2π.
9. Find the equation of the line through the origin making an angle of 120° with the positive x-axis.
10. (HL only) Given tan θ = 2, find sec²θ.
11. (HL only) Find arctan √3, arccos 0 and arcsin(1/2).
12. (HL only) Find the exact value of cos 15°.

Answers

1. −√3/2, −√2/2, −√3/3
2. cos x = −√5/3, so tan x = −2/√5 = −2√5/5
3. 7/9
4. Amplitude 5, period π/2, range −7 ≤ f(x) ≤ 3
5. 2π
6. x = 2π/3, 4π/3
7. tan x = ±√3, so x = π/3, 2π/3
8. x = 0, π/6, 5π/6, π, 2π
9. y = −√3 x
10. sec²θ = 1 + 4 = 5
11. π/3, π/2, π/6
12. (√6 + √2)/4

Where marks are usually lost

• Using the period b instead of 2π/b when writing or reading a model.
• Not changing the interval for sin(2x) or cos(3x), so solutions are missing.
• Dividing by sin x or cos x and losing the solutions where it equals zero.
• Keeping cos x = 2 or sin x = −3/2 from a quadratic instead of rejecting it.
• Missing the second triangle in the ambiguous case, or keeping one whose angles add to more than 180°.
• Wrong sign when finding cos θ from sin θ because the quadrant was ignored.
• Reading the shift in sin(2x − π/2) as π/2 instead of π/4.
• Decimal answers on Paper 1 where exact values are expected.
• (HL only) Giving arccos of a negative number as a negative angle.
• (HL only) Sign errors in cos(A + B): the sign in the middle is a minus.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).