IB DP Maths AA · Topic 3 Geometry and trigonometry · SL 3.5–3.8 · AHL 3.9–3.11 · ~9 min read

Trig identities, equations and graphs

The unit circle and exact values, the Pythagorean and double angle identities, graphs of sin, cos and tan with transformations, modelling with trig functions, solving trig equations in an interval, and for HL: sec, cosec, cot, inverse trig functions and compound angle identities.

Unit circleExact valuessin² + cos² = 1Double angle identitiesTrig graphsAmplitude & periodTrig modelsTrig equations

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No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

On the unit circle, the point at angle θ is (cos θ, sin θ). That one picture defines sin, cos and tan for every angle and gives their signs (CAST).
Key identities: sin²θ + cos²θ = 1, tan θ = sin θ/cos θ, sin 2θ = 2 sin θ cos θ, cos 2θ = cos²θ − sin²θ.
y = a sin(b(x − c)) + d: amplitude |a|, period 2π/b, horizontal shift c, principal axis y = d.
Trig equations: find one solution, then use symmetry and the period to find every solution in the given interval.
HL: sec, cosec and cot, 1 + tan²θ = sec²θ, compound angle formulas, and the inverse functions arcsin, arccos and arctan with their restricted ranges.

By the end you can

Use the unit circle to find exact values and the signs of trig ratios in each quadrant.
Use sin²θ + cos²θ = 1, tan θ = sin θ/cos θ and the double angle identities.
Sketch and transform the graphs of sin, cos and tan, and model periodic situations.
Solve trig equations in a finite interval, including quadratic types.
Use sec, cosec and cot, and the identities 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ. HL
Use the inverse trig functions and their domains and ranges. HL
Use compound angle identities and prove trig identities. HL

1The unit circle and exact values

The x-coordinate is cos θ, the y-coordinate is sin θ, and tan θ is the gradient of the radius.

θ	0	π/6 (30°)	π/4 (45°)	π/3 (60°)	π/2 (90°)
sin θ	0	½	√2/2	√3/2	1
cos θ	1	√3/2	√2/2	½	0
tan θ	0	√3/3	1	√3	undefined

Memory trickDown the sin row the numbers are √0/2, √1/2, √2/2, √3/2, √4/2. The cos row is the same list backwards.

Symmetry rules: sin(π − θ) = sin θ, cos(π − θ) = −cos θ, cos(−θ) = cos θ, sin(−θ) = −sin θ, and tan has period π.

Worked example

Find the exact value of cos(4π/3).

4π/3 is in the third quadrant, π/3 past π M1
cos is negative there R1
cos(4π/3) = −½ A1

2Identities

cos²θ + sin²θ = 1, tan θ = sin θcos θ FORMULA BOOKLET

Pythagorean identity

sin 2θ = 2 sin θ cos θ FORMULA BOOKLET

double angle

cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ FORMULA BOOKLET

pick the version that leaves one ratio

Worked example

θ is obtuse and sin θ = 35. Find cos θ and sin 2θ.

cos²θ = 1 − 925 = 1625 M1
θ is obtuse, so cos θ = −45 A1R1
sin 2θ = 2(35)(−45) = −2425 A1

Common mistakeForgetting the sign. A square root gives ±; the quadrant decides which. "Obtuse" means the second quadrant, where cos is negative.

3Graphs and models

sin and cos are the same wave shifted by π/2; tan repeats every π and has vertical asymptotes.

y = a sin(b(x − c)) + d: amplitude |a|; period 2πb; shifted c to the right; principal axis y = d. The same applies to cos. For tan, the period is π/b.
Reading a model from data: amplitude = (max − min)/2; axis d = (max + min)/2; period = time between two maxima, so b = 2π/period.

A tide model: maximum at t = 0, so a cosine curve fits without a shift.

Worked example

The depth of water in a harbour is d = 5 + 3cos(πt6) metres, where t is hours after midnight, 0 ≤ t ≤ 12. Find when the depth is 6.5 m.

cos(πt6) = 0.5 M1
πt6 = π3 or 5π3 A1
t = 2 h or t = 10 h A1

4Solving trig equations

Method: rearrange to sin x = k (or cos, tan); find the principal value; use the unit circle or the graph to find the second solution; then add or subtract periods until you leave the interval.
Change the interval for compound angles: for sin 2x on 0 ≤ x ≤ 2π, solve for 2x on 0 ≤ 2x ≤ 4π, then halve.

Worked example

Solve 2cos²x − cos x − 1 = 0 for 0 ≤ x ≤ 2π.

(2cos x + 1)(cos x − 1) = 0 M1
cos x = −½ or cos x = 1 A1
x = 0, 2π3, 4π3, 2π A1A1

Worked example

Solve sin 2x = cos x for 0 ≤ x ≤ 2π.

2 sin x cos x − cos x = 0 ⇒ cos x(2 sin x − 1) = 0 M1
cos x = 0 ⇒ x = π2, 3π2 A1
sin x = ½ ⇒ x = π6, 5π6 A1

Common mistakeDividing both sides by cos x. You lose the solutions where cos x = 0. Factorise instead.

GDC (calculator) methodPaper 2: graph both sides on the given interval and use intersect. Set the window to the interval exactly, and check the mode (radians or degrees) first.

5Reciprocal trig functions HL

sec θ = 1cos θ, cosec θ = 1sin θ, cot θ = cos θsin θ

learn these; they are not given

1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ FORMULA BOOKLET

divide sin² + cos² = 1 by cos²θ or sin²θ

Worked example

Solve sec²x = 3 tan x − 1 for 0 ≤ x ≤ π.

1 + tan²x = 3 tan x − 1 ⇒ tan²x − 3 tan x + 2 = 0 M1
tan x = 1 or 2 A1
x = π4 or x = 1.11 A1A1

6Inverse trig functions HL

Function	Domain	Range
arcsin x	−1 ≤ x ≤ 1	−π/2 ≤ y ≤ π/2
arccos x	−1 ≤ x ≤ 1	0 ≤ y ≤ π
arctan x	x ∈ ℝ	−π/2 < y < π/2

sin x is not one-to-one, so its domain is restricted to [−π/2, π/2] before it can have an inverse. The calculator always gives the value in the range above (the principal value).

Worked example

Find the exact value of arcsin(−√32) + arccos(−½).

arcsin(−√3/2) = −π3 A1
arccos(−½) = 2π3 A1
total π3 A1

7Compound angle identities HL

sin(A ± B) = sin A cos B ± cos A sin B FORMULA BOOKLET

same signs

cos(A ± B) = cos A cos B ∓ sin A sin B FORMULA BOOKLET

opposite signs

tan(A ± B) = tan A ± tan B1 ∓ tan A tan B FORMULA BOOKLET

watch the sign flip in the denominator

Worked example

Find the exact value of tan 75°.

tan(45° + 30°) = 1 + √3/31 − √3/3 M1
= 3 + √33 − √3 = (3 + √3)²6 M1
= 2 + √3 A1

Worked example

Prove that 1 − cos 2θsin 2θ ≡ tan θ.

LHS = 1 − (1 − 2sin²θ)2 sin θ cos θ M1
= 2sin²θ2 sin θ cos θ = sin θcos θ A1
= tan θ = RHS AG

Exam tipProving an identity: start from the messier side and work towards the other. Never move terms across an "≡" as if it were an equation.

Linking questionExact values come straight from the 30-60-90 and 45-45-90 triangles (Trigonometry: triangles). The derivative of sin x is cos x only in radians (Differentiation), and complex numbers in polar form turn compound angle identities into simple multiplication (Complex numbers, HL).

Idea for your IA / explorationModel daylight hours in your city over a year with y = a sin(b(t − c)) + d, using published sunrise and sunset times, and compare the fitted model with the data.

Test yourself — cover the answers first

Quick check

Find the exact value of sin(5π/6).
State the period of y = 3sin(2x).
State the amplitude and axis of y = 4 − 2cos x.
In which quadrant are sin negative and cos positive?
Write cos 2θ in terms of sin θ only.
HL Find sec(π/3).
HL State the range of arccos x.

Exam-style

Solve 2sin(x + π4) = √3 for 0 ≤ x ≤ 2π.
HL tan A = ½ and tan B = ⅓, with A and B acute. Show that A + B = π4.

Show answers

½
π
amplitude 2, axis y = 4
the fourth
1 − 2sin²θ
2
0 ≤ y ≤ π
sin(x + π/4) = √3/2
x + π/4 = π/3 or 2π/3 (within π/4 ≤ x + π/4 ≤ 9π/4)
x = π12 or 5π12
tan(A + B) = ½ + ⅓1 − ⅙ = 5/65/6 = 1
A + B is between 0 and π, so A + B = π4 AG

Practise this chapter

IB Maths AA practice papers

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