Trigonometry

— Mathematics · AS & A Level Pure Mathematics 1 (9709) · Chapter 5 —

★ Exam Tip: This chapter is a toolkit: exact values, quadrant signs, graphs, inverse functions, equations and identities. The end-of-chapter questions are all 9709 Paper 11 (P1) past-paper style — solving equations and proving identities are what the exam loves. It all returns in P2 & P3 with sec, cosec, cot.

✓ By the end I must be able to…

sketch & use graphs of sin, cos, tan (any angle size, degrees or radians)
use exact values of sin, cos, tan of 30°, 45°, 60° and related angles
use sin⁻¹x, cos⁻¹x, tan⁻¹x for the principal values of the inverse functions
use the identities tan θ = sin θ / cos θ and sin²θ + cos²θ = 1
find all solutions of simple trig equations in a given interval

0. Warm-up — what I should already know

Right-angled triangles: Pythagoras + SOH CAH TOA.

Check: right angle at A, AC = 1, AB = r, angle θ at C. Then:

BC = √1 + r²sin θ = r√1 + r²cos θ = 1√1 + r²tan θ = r

opp to θ = r, adj to θ = 1

Degrees ↔ radiansπ rad = 180° ⇒ deg → rad: × π180 | rad → deg: × 180π

degrees	30°	45°	60°	90°	180°	270°	360°
radians	π/6	π/4	π/3	π/2	π	3π/2	2π

e.g. 45° = π/4 | 720° = 4π | 150° = 5π/6 and π/6 = 30° | 7π/2 = 630° | 13π/12 = 195°

Quadratics (needed for equations later) — factorise:

x² − 5x = 0x(x − 5) = 0 x = 0 or 5

2x² + 7x − 15 = 0(2x − 3)(x + 5) = 0x = 3/2 or −5

Why study trig? Oscillations & waves are everywhere — sound, light, water waves, electricity, vibrating aircraft wings, microwaves. Engineers model them with trig functions. The graphs of y = sin x and y = cos x are waves.

Calculator: check the mode! Degrees (D) for questions in °, Radians (R) for questions in π. Wrong mode = wrong answers.

1. Angles between 0° and 90°

sin θ = oppositehypotenuse = yr
cos θ = adjacenthypotenuse = xr
tan θ = oppositeadjacent = yx

Note: cos²θ means (cos θ)² — same for sin²θ, tan²θ.

Pythagoras: x² + y² = r²

Worked Ex 5.1exact values from one ratio

Given cos θ = √53, 0° ≤ θ ≤ 90°. a Find exact cos²θ, sin θ, tan θ. b Show that 1 − tan²θcos θ + sin θ = 3√5 − 65.

a i cos²θ = (cos θ)²= (√53)²5/9ii x = √3² − (√5)² = 2sin θ = 2/3iii tan θ = 2√5

draw the triangle ⇒ Pythagoras for the missing side

WE 5.1 bshow that (rationalise)

1 − tan²θ = 1 − 45 = 15cos θ + sin θ = √5 + 23LHS = 1/5 (√5+2)/3 = 35(√5 + 2)multiply top & bottom by (√5 − 2): (√5+2)(√5−2) = 5 − 2√5 + 2√5 − 4 = 1= 3(√5 − 2)5(√5+2)(√5−2)= 3√5 − 65 ✓ shown

2. Exact values of sin, cos, tan for 30°, 45°, 60°

Two triangles give all the exact values (π/6, π/4, π/3):

Triangle 1: right-angled isosceles, sides 1 ⇒ hyp = √(1²+1²) = √2

Triangle 2: equilateral, side 2, cut in half by the perpendicular bisector ⇒ height = √(2²−1²) = √3

	sin θ	cos θ	tan θ
θ = 30° = π/6	12	√32	1√3
θ = 45° = π/4	1√2	1√2	1
θ = 60° = π/3	√32	12	√3

Tips: 1/√2 can be written √2/2 and 1/√3 as √3/3 (rationalised). Memory: sin goes ½, √2/2, √3/2 (up) as angle grows; cos is the same list backwards; tan = sin ÷ cos.

Worked Ex 5.2exact value of expressions

Find the exact value of a sin30° cos45° b sin²(π/3) c 2 cos(π/4) sin(π/6)cos²(π/3) + sin²(π/3)

a sin30° cos45°= 12 × 1√2= 12√2rationalise the denominator= 1 × √22√2 × √2= √24

b sin²(π/3) = sin(π/3) × sin(π/3)sin²(π/3) means (sin π/3)²= √32 × √32= 3/4

c numerator: 2 × 1√2 × 12 = 1√2denominator: (12)² + (√32)² = 14 + 34 = 1= 1√2 = √2√2 √2= √22

3. The general definition of an angle

We need sin, cos, tan for any angle, so we define an angle generally:

Definition: an angle is a measure of rotation of a line segment OP about a fixed point O, measured from the positive x-direction.

anticlockwise = positive (+)
clockwise = negative (−)
The plane splits into four quadrants; θ is “in” the quadrant where OP lies.

Basic (reference) angle: the acute angle OP makes with the x-axis.

angle measured from +x, anticlockwise positive

Worked Ex 5.3quadrant + basic angle

Draw OP for each angle; find the acute angle OP makes with the x-axis.

a 120° anticlockwise
2nd quadrant, basic = 180−120 = 60°

b 430° = 360° + 70°
1 full turn + 70° ⇒ basic = 70°

c 3π/4 anticlockwise
basic = π − 3π/4 = π/4

d −2π/3 is clockwise
3rd quadrant, basic = π − 2π/3 = π/3

Quadrant	θ (degrees)	θ (radians)	basic angle b	given b ⇒ θ
1st	0°–90°	0–π/2	θ	b
2nd	90°–180°	π/2–π	180° − θ (π − θ)	180° − b (π − b)
3rd	180°–270°	π–3π/2	θ − 180° (θ − π)	180° + b (π + b)
4th	270°–360°	3π/2–2π	360° − θ (2π − θ)	360° − b (2π − b)

Trick: angle outside 0°–360°? Add or subtract full turns (360° / 2π) first to find where OP really is — e.g. 430° ≡ 70°, −500° ≡ −500 + 720 = 220° (3rd quadrant). For “given b, quadrant, range” questions find θ in 0–360° then add/subtract 360° (2π) until it is in the range asked.

4. Trigonometric ratios of general angles

For any angle θ in any quadrant, P(x, y) on the rotating line, r = length OP:

sin θ = yr cos θ = xr tan θ = yx
(tan only when x ≠ 0)

Where: r = √x² + y² (always positive); x, y are the coordinates of P.

Key Point 5.1 — signs of x and y decide the signs of the ratios

Quadrant	x	y	sin θ = y/r	cos θ = x/r	tan θ = y/x
1st	+	+	+	+	+
2nd	−	+	+	−	−
3rd	−	−	−	−	+
4th	+	−	−	+	−

which ratios are positive in each quadrant

Mnemonic: “All Students Trust Cambridge” — go anticlockwise from the 1st quadrant: A, S, T, C.

Only the sign depends on the quadrant; the size comes from the basic angle.

Method (any angle): ① find the quadrant ② find the basic angle ③ sign from ASTC ④ value of the ratio of the basic angle.

Worked Ex 5.4express with acute angles

a sin140°: 2nd quad (sin +), basic 40°
sin 140° = sin 40°

b cos(−130°): 3rd quad (cos −), basic 50°
cos(−130°) = −cos 50°

Worked Ex 5.5find sin & tan from cos

Given cos θ = −3/5 and 180° ≤ θ ≤ 270°, find sin θ and tan θ.

θ in 3rd quadrant: sin −, tan +x = −3, r = 5y² + (−3)² = 5²y² = 25 − 9 = 16y < 0 ⇒ y = −4sin θ = −4/5tan θ = −4−3 = 4/3

x = −3, r = 5, y = −4 (negative, below the axis)

Worked Ex 5.6exact values, no calculator

a sin120°120° in 2nd quad ⇒ sin +basic = 180° − 120° = 60°sin120° = sin60° = √32

b cos(7π/6)7π/6 in 3rd quad ⇒ cos −basic = 7π/6 − π = π/6cos(7π/6) = −cos(π/6) = −√32

Worked Ex 5.7in terms of b, given sin 50° = b

a sin230°: 3rd quad (sin −)basic 230° − 180° = 50°sin230° = −sin50° = −bb triangle: hyp 1, opp badj = √1 − b²cos50° = √1 − b²

draw the triangle with angle 50° (then 40° is the other angle)

WE 5.7 c, dtan in terms of b

c tan40° = oppadj = √1 − b²bd tan140°: 2nd quad (tan −), basic 40°tan140° = −tan40° = −√1 − b²b

5. Graphs of trigonometric functions

Explore 5.2 (Ferris wheel, radius 50 m): entering level with the centre, one turn anticlockwise: vertical displacement = 50 sin(angle), horizontal = 50 cos(angle). Two turns ⇒ the graphs just repeat.

On the unit circle (r = 1) P moves round: P = (cos x, sin x).

height of P: 0 → 1 → 0 → −1 → 0 ⇒ sin x
horizontal displacement: 1 → 0 → −1 → 0 → 1 ⇒ cos x

P goes round the unit circle

y = sin x, 0° ≤ x ≤ 360°

y = cos x, 0° ≤ x ≤ 360°

Both can be continued beyond 0°–360° (and to negative x):

sin: passes through O, odd symmetry

cos: starts at 1, symmetric about the y-axis

Periodic functions: sin and cos repeat over and over. Period = length of one cycle = 360° (2π rad). Amplitude = distance between a max (or min) point and the principal axis (middle line) = 1 for y = sin x and y = cos x.

Symmetry of y = sin x ⇒ important relationshipssin(−x) = −sin x | sin(180° − x) = sin x | sin(180° + x) = −sin x
sin(360° − x) = −sin x | sin(360° + x) = sin x

Explore 5.3 / 5.4 — same idea from the cos and tan curves:

angle	sin	cos	tan
(−x)	−sin x	cos x	−tan x
(180° − x)	sin x	−cos x	−tan x
(180° + x)	−sin x	−cos x	tan x
(360° − x)	−sin x	cos x	−tan x
(360° + x)	sin x	cos x	tan x

Link to ASTC: these rows are just the quadrant-sign rule — e.g. 180° − x is in the 2nd quadrant so only sin stays positive. Use whichever you remember best.

The graph of y = tan x

red dashed lines = asymptotes x = ±90°, 270°, 450°… (branches get closer, never touch)

tan behaves very differently from sin and cos.
Period = 180° (π rad) — repeats every 180°.
Asymptotes at x = ±90°, 270°, 450°… (x = π/2 + nπ).
tan has no amplitude; range is all real numbers.

	period	amplitude	range
sin x	360° (2π)	1	−1 ≤ y ≤ 1
cos x	360° (2π)	1	−1 ≤ y ≤ 1
tan x	180° (π)	none	all real y

6. Transformations of trigonometric graphs

Rewind (Sec 2.6): the rules for transforming y = f(x) — y = af(x), y = f(ax), y = f(x) + a, y = f(x + a) — work on trig graphs too.

Four basic transformations of y = sin x

y = a sin x: stretch SF a ∥ y-axis, amp a

y = sin(ax): stretch SF 1/a ∥ x-axis, period 360°/a

y = a + sin x: translation (0, a)

y = sin(x + a): translation (−a, 0)

Equation	Transformation of y = sin x	amplitude	period
y = a sin x	stretch SF a, parallel to y-axis	a	360°
y = sin(ax)	stretch SF 1/a, parallel to x-axis	1	360°/a
y = a + sin x	translation by vector (0, a)	1	360°
y = sin(x + a)	translation by vector (−a, 0)	1	360°

Careful: y = sin(x + 90) moves the graph left by 90 (vector (−90, 0)); y = sin(x − 90) moves it right (vector (90, 0)). Period: sin bx, cos bx → 360°/b (2π/b); tan bx → 180°/b (π/b).

Reading a, b, c from a graph of y = a + b sin cx (or a + b cos cx)a = (max + min)/2 = middle line | b = (max − min)/2 = amplitude | c = 360° ÷ period (or 2π ÷ period)

Building up a sketch: y = 2 cos(x + 90) + 1, 0° ≤ x ≤ 360°

Step 1: y = cos x period 360°, amp 1

Step 2: y = cos(x + 90): translate (−90, 0)

Step 3: y = 2cos(x + 90): stretch SF 2 ∥ y, amp 2

Step 4: + 1: translate (0, 1), period 360°, amp 2

Worked Ex 5.8sketch y = sin(x − 90)

Sketch y = sin x and y = sin(x − 90) on the same grid, 0° ≤ x ≤ 360°.

translation of y = sin x by the vector (90, 0) — shift 90° right

Worked Ex 5.9f(x) = 3 cos 2x, 0° ≤ x ≤ 360°

a period & amplitude b max/min points c sketch y = f(x) d sketch y = 1 + 3cos 2x

a period = 360°2 = 180° amplitude = 3b y = cos x has max/min at (0°,1), (180°,−1), (360°,1), (540°,−1), (720°,1)halve the x-values, multiply y by 3:f(x) max/min at (0°,3) (90°,−3) (180°,3) (270°,−3) (360°,3)

c y = 3cos 2x: two full cycles in 360°

d y = 1 + 3cos 2x = translation by (0, 1) (dashed = original)

Worked Ex 5.10number of solutions from graphs

a sketch y = sin 2x and y = 1 + 3cos 2x for 0° ≤ x ≤ 360°. b number of solutions of sin 2x = 1 + 3cos 2x?

the graphs cross at 4 points (green dots) ⇒ 4 solutions

b: the graphs intersect 4 times ⇒ the equation has 4 solutions in the interval.

Idea: solutions of f(x) = g(x) are the x-coordinates of the intersection points of y = f(x) and y = g(x).

7. Inverse trigonometric functions

Rewind (Ch 2): only one-one functions have an inverse; the graph of f⁻¹ is the reflection of f in the line y = x.

sin x, cos x, tan x (x ∈ ℝ) are many-one. Restrict the domain suitably ⇒ one-one ⇒ inverse exists. The notations are sin⁻¹x, cos⁻¹x, tan⁻¹x.

Common mistake: sin⁻¹x does not mean 1/sin x — it means “the angle whose sine is x”.

sin x on −π/2 ≤ x ≤ π/2 & its reflection

cos x on 0 ≤ x ≤ π & its reflection

tan x on −π/2 < x < π/2 (blue, with asymptotes x = ±π/2) & tan⁻¹x (red): the dashed black line is y = x

Function	Domain	Range
y = sin x	−π/2 ≤ x ≤ π/2	−1 ≤ sin x ≤ 1
y = sin⁻¹x	−1 ≤ x ≤ 1	−π/2 ≤ sin⁻¹x ≤ π/2
y = cos x	0 ≤ x ≤ π	−1 ≤ cos x ≤ 1
y = cos⁻¹x	−1 ≤ x ≤ 1	0 ≤ cos⁻¹x ≤ π
y = tan x	−π/2 < x < π/2	tan x ∈ ℝ
y = tan⁻¹x	x ∈ ℝ	−π/2 < tan⁻¹x < π/2

Remember: domain of f⁻¹ = range of f, and range of f⁻¹ = domain of f. In degrees: sin⁻¹, tan⁻¹ give −90° to 90°; cos⁻¹ gives 0° to 180°.

Principal angle (principal value): the angle your calculator gives is the one in the range of the inverse function. e.g. sin x = 0.5, 0 ≤ x ≤ π: calculator x = sin⁻¹0.5 = π/6 — but there is a second angle 5π/6 with sin x = 0.5 (use the symmetry of the curve or the quadrant rules).

Worked Ex 5.11no calculator, answer in degrees

The output of sin⁻¹, cos⁻¹, tan⁻¹ can be given in degrees if needed.

a sin⁻¹0 = angle whose sine is 0, in −90° ≤ angle ≤ 90° ⇒ 0°b cos⁻¹(√32) = angle whose cosine is √3/2, in 0° ≤ angle ≤ 180° ⇒ 30°c tan⁻¹(−1) = angle whose tangent is −1, in −90° < angle < 90° ⇒ −45°

Worked Ex 5.12inverse of a trig function

f(x) = 3 sin(x/2) − 1, domain −π ≤ x ≤ π. a sketch, explain why f⁻¹ exists b range c find f⁻¹(x) and its domain.

f is one-one on this domain ⇒ inverse exists

a f is a one-one function with this domain ⇒ has an inverseb range: −4 ≤ f(x) ≤ 2sin(x/2) runs −1→1 so 3sin(x/2) − 1 runs −4→2

WE 5.12 c3-step inverse method

Step 1: write as y = : y = 3 sin(x2) − 1Step 2: swap x and y: x = 3 sin(y2) − 1Step 3: make y the subject:x + 1 = 3 sin(y2)sin(y2) = x + 13take sin⁻¹y2 = sin⁻¹(x + 13)y = 2 sin⁻¹(x + 13)f⁻¹(x) = 2 sin⁻¹(x + 13) for −4 ≤ x ≤ 2domain of f⁻¹ = range of f

8. Trigonometric equations

Solve sin x = 0.5 for −360° ≤ x ≤ 360°. Calculator: x = sin⁻¹(0.5) = π/6 (30°) — but there are more solutions. Sketch y = sin x and the line y = 0.5:

4 intersections between −360° and 360° ⇒ use the calculator value 30° + symmetry for the rest

Solution: sin x = 0.5 for −360° ≤ x ≤ 360° ⇒ x = −330°, −210°, 30°, 150°

Finding all solutions in one go (x₁ = calculator value)sin x = k: x = x₁ and x = 180° − x₁ (π − x₁)
cos x = k: x = x₁ and x = 360° − x₁ (2π − x₁), or −x₁
tan x = k: x = x₁ and x = x₁ + 180° (π)
…then add/subtract the period (360° / 2π for sin & cos; 180° / π for tan) until you have every value in the interval.

method for every equation of this type

Worked Ex 5.13cos x = −0.7, 0° ≤ x ≤ 360°

cos x = −0.7calculator: cos⁻¹(−0.7) ⇒ x = 134.4°sketch: two solutions in 0°–360°symmetry: 360° − 134.4° = 225.6°x = 134.4° or 225.6°(1 d.p.)

cos x = −0.7 cuts twice

Worked Ex 5.14tan 2A = −2.1, 0° ≤ A ≤ 180°

Let 2A = x. Then 0° ≤ A ≤ 180° means 0° ≤ x ≤ 360° (the interval doubles!).

tan x = −2.1calculator: tan⁻¹(−2.1) ⇒ x = −64.54°tan repeats every 180° ⇒ add 180°:x = −64.54° + 180° = 115.46°x = 115.46° + 180° = 295.46°(−64.54° is outside 0°–360°, so it is dropped)2A = 115.46° ⇒ A = 57.7°2A = 295.46° ⇒ A = 147.7°A = 57.7° or 147.7°

dashed grey = asymptotes; solutions in 0–360°: 115.46°, 295.46°

Worked Ex 5.15sin(2A + π/6) = 0.6, 0 ≤ A ≤ π (radians)

Let 2A + π/6 = x. Interval: 0 ≤ A ≤ π ⇒ π/6 ≤ x ≤ 13π/6 (so up to ≈ 6.81).

sin x = 0.6calculator (radians mode!): x = 0.6435symmetry: x = π − 0.6435 = 2.498(next ones, 6.927 & 8.78, are above 13π/6 ≈ 6.81)2A + π6 = 0.6435A = ½(0.6435 − π/6) = 0.06002A + π6 = 2.498A = ½(2.498 − π/6) = 0.987(3 s.f.) A = 0.0600 or 0.987 radians

y = 0.6 meets sin x at 0.6435 and 2.498

Common mistakes: ① forgetting to change the interval when the angle is 2A, 3x, x + 30… ② calculator in the wrong mode ③ giving only the calculator value and missing the other solutions ④ not rounding as asked (1 d.p. for degrees, 3 s.f. for radians).

9. Two key identities (and equations that use them)

Take a right-angled triangle with sides x, y, hypotenuse r and angle θ:

the triangle

Rule 1: tan θ = yxdivide top & bottom by r= y/rx/r = sin θcos θRule 2: x² + y² = r²divide both sides by r²(xr)² + (yr)² = 1cos²θ + sin²θ = 1

★ Key Point 5.2 & 5.3 — true for ALL θ (valid values)tan θ = sin θcos θ (cos θ ≠ 0) | cos²θ + sin²θ = 1

Forms I’ll use all the time: sin²x = 1 − cos²x | cos²x = 1 − sin²x | 1 − sin²x = (1 + sin x)(1 − sin x) | sin x = tan x cos x.

These come from the unit-circle definition, so they hold for any angle. Use them to solve harder equations.

Worked Ex 5.163cos²x − sin x cos x = 0, 0° ≤ x ≤ 360°

3cos²x − sin x cos x = 0cos x (3cos x − sin x) = 0factorise — do NOT cancel cos x!cos x = 0 ⇒ x = 90°, 270°or 3cos x − sin x = 0sin x = 3cos xtan x = 3x = 71.6° or 180° + 71.6° = 71.6°, 251.6°

tan x = 3: 71.6° and 251.6°

Solution: x = 71.6°, 90°, 251.6° or 270°

Common mistake: dividing both sides by cos x would lose the solutions 90° and 270° (where cos x = 0). Factorise instead. Use sin x = 3 cos x ⇒ tan x = 3 only for the second bracket.

Worked Ex 5.172sin²x + 3cos x − 3 = 0, 0 ≤ x ≤ 2π

2sin²x + 3cos x − 3 = 0replace sin²x with 1 − cos²x2(1 − cos²x) + 3cos x − 3 = 02cos²x − 3cos x + 1 = 0(expand & collect)(2cos x − 1)(cos x − 1) = 0cos x = ½ ⇒ x = π/3 or 2π − π/3 = 5π/3cos x = 1 ⇒ x = 0 or 2π

cos x = ½ (red) and cos x = 1 (green)

Solution: x = 0, π/3, 5π/3 or 2π

10. Trigonometric identities — simplify & prove

x + x = 2x is an identity — true for all x. We often write ≡ instead of = to emphasise this. Two commonly used trig identities:

sin²x + cos²x ≡ 1 and tan x ≡ sin xcos x

Proving an identity: start with the more complicated side and show it simplifies to the other side. LHS = left-hand side, RHS = right-hand side. Never move terms across the ≡ sign.

Toolbox: ① tan = sin/cos ② sin² + cos² = 1 (to swap sin² ↔ cos²) ③ combine fractions (common denominator) ④ factorise & cancel ⑤ difference of two squares, 1 − sin² = (1 + sin)(1 − sin).

Worked Ex 5.18express in terms of cos x

4cos²x − 3sin²x ≡ 4cos²x − 3(1 − cos²x)replace sin²x with 1 − cos²x≡ 4cos²x − 3 + 3cos²x≡ 7cos²x − 3

Worked Ex 5.19prove 1 + sin xcos x + cos x1 + sin x ≡ 2cos x

LHS ≡ 1 + sin xcos x + cos x1 + sin xadd the two fractions≡ (1 + sin x)² + cos²xcos x (1 + sin x)expand the brackets≡ 1 + 2sin x + sin²x + cos²xcos x (1 + sin x)use sin²x + cos²x ≡ 1≡ 2 + 2sin xcos x (1 + sin x)factorise the numerator≡ 2(1 + sin x)cos x (1 + sin x)divide top & bottom by (1 + sin x)≡ 2cos x ≡ RHS ✓

Worked Ex 5.20prove 1 + sin x1 − sin x ≡ (tan x + 1cos x)²

RHS ≡ (tan x + 1cos x)²use tan x = sin x / cos x≡ (sin xcos x + 1cos x)²add the fractions≡ (sin x + 1cos x)² ≡ (1 + sin x)²cos²xreplace cos²x with 1 − sin²x≡ (1 + sin x)²1 − sin²x1 − sin²x = (1 + sin x)(1 − sin x)≡ (1 + sin x)²(1 + sin x)(1 − sin x)divide top & bottom by (1 + sin x)≡ 1 + sin x1 − sin x ≡ LHS ✓

Explore 5.5many expressions that simplify to cos x

Each of these equals cos x — a good test of the toolbox:

1 − sin²xcos x = cos²xcos xsin xtan x = cos x1sin x tan x + cos x = 1(sin²x + cos²x)/cos xsin²xtan²x cos x = sin²xsin²x/cos xtan²x cos³xsin²x = cos xcos³x + cos x sin²x = cos x(cos²x + sin²x)cos³x1 − sin²x = cos³xcos²xsin²x cos x(1 − cos x)(1 + cos x) = sin²x cos xsin²x

Task: make up your own expressions that simplify to sin x (at least two different ratios) and compare with classmates.

11. Further trigonometric equations

Use identities first to turn a harder equation into a simple one.

Worked Ex 5.21 aprove 1 − tan²θ1 + tan²θ ≡ 2cos²θ − 1

LHS ≡ 1 − tan²θ1 + tan²θtan θ = sin θ / cos θ≡ 1 − (sin θ/cos θ)²1 + (sin θ/cos θ)²multiply top & bottom by cos²θ≡ cos²θ − sin²θcos²θ + sin²θsin²θ + cos²θ = 1≡ cos²θ − sin²θsin²θ = 1 − cos²θ≡ cos²θ − (1 − cos²θ)≡ 2cos²θ − 1 ≡ RHS ✓

WE 5.21 bhence solve 1 − tan²θ1 + tan²θ = 5cos θ − 3, 0° ≤ θ ≤ 360°

2cos²θ − 1 = 5cos θ − 3use the result from part a, then rearrange2cos²θ − 5cos θ + 2 = 0(2cos θ − 1)(cos θ − 2) = 0cos θ = ½ or cos θ = 2cos θ = 2 has no solutions (−1 ≤ cos θ ≤ 1)θ = cos⁻¹(½) = 60° or 360° − 60°θ = 60° or 300°

★ Exam Tip — which tool? ① sin and cos both, same angle, no squares ⇒ divide to make tan (or factorise). ② sin² and cos² or cos only ⇒ swap with sin² + cos² = 1 to get a quadratic in one ratio ⇒ factorise. ③ common factor ⇒ factorise, don’t cancel. ④ If a ratio comes out outside −1…1 (e.g. cos θ = 2) ⇒ reject it.

★ Quick Recap — Full Chapter Cheat-Sheet

	sin θ	cos θ	tan θ
30° = π/6	1/2	√3/2	1/√3
45° = π/4	1/√2	1/√2	1
60° = π/3	√3/2	1/2	√3

✓ Angles & signs

anticlockwise +, clockwise −; measure from +x
sin θ = y/r, cos θ = x/r, tan θ = y/x (r = √(x²+y²))
All Students Trust Cambridge: 1st all, 2nd sin, 3rd tan, 4th cos
basic angle: 2nd 180°−θ, 3rd θ−180°, 4th 360°−θ (π, 2π in radians)

✓ Graphs

sin, cos: period 360° (2π), amplitude 1
tan: period 180° (π), asymptotes x = ±90°, 270°…, no amplitude
y = a sin x: amp a | y = sin(ax): period 360°/a
y = a + sin x: up a | y = sin(x + a): left a

✓ Symmetry (sin; cos; tan)

sin(−x) = −sin x; cos(−x) = cos x; tan(−x) = −tan x
sin(180°−x) = sin x; cos(180°−x) = −cos x; tan(180°−x) = −tan x
sin(180°+x) = −sin x; cos(180°+x) = −cos x; tan(180°+x) = tan x
360° ± x: sin(360°−x) = −sin x, cos(360°−x) = cos x

✓ Inverse functions

sin⁻¹: [−1,1] → [−π/2, π/2]
cos⁻¹: [−1,1] → [0, π]
tan⁻¹: ℝ → (−π/2, π/2)
f⁻¹: swap x, y; domain f⁻¹ = range f

Identitiestan x ≡ sin xcos x | sin²x + cos²x ≡ 1 | sin²x = 1 − cos²x | cos²x = 1 − sin²x

✓ Solving equations

sub X = (…), change interval, calculator ⇒ principal value, sketch ⇒ symmetry ⇒ all values, undo substitution
sin: x₁, 180°−x₁ | cos: x₁, 360°−x₁ | tan: x₁, x₁+180° (then ± periods)
factorise (don’t cancel); quadratic in sin/cos via sin²+cos²=1; reject |sin|,|cos| > 1

Where to practise: Ex 5A exact values · 5B basic angles & quadrants · 5C ratios of general angles · 5D graphs & transformations · 5E inverse functions · 5F equations · 5G proving identities · 5H further equations · End-of-chapter review (9709 P11 past papers).