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Trigonometry A-Level Pure 2

Grade 12 · Pure Mathematics 2 · Cambridge A-Level 9709 · Age 17–18

Welcome to Trigonometry — Pure 2

Compound angle formulae are the foundation for solving harder trigonometric equations, proving identities, and applying trigonometry in calculus. Mastering these formulae unlocks the entire A-Level trig toolkit.

sin(A+B) = sinA cosB + cosA sinB  |  cos(A+B) = cosA cosB − sinA sinB  |  tan(A+B) = (tanA + tanB) / (1 − tanA tanB)

These three formulae — and their minus-angle counterparts — generate all double angle results and enable the R·sin(θ+α) method for solving equations.

Learning Objectives

  • Derive the compound angle formulae from geometric principles
  • Use sin(A±B) to find exact values and solve equations
  • Use cos(A±B) correctly, noting the sign reversal
  • Use tan(A±B) and handle the denominator signs carefully
  • Derive double angle formulae from compound angle formulae
  • Use all three forms of cos(2A) and select the most appropriate form
  • Express a·sinθ + b·cosθ in R·sin(θ+α) form
  • Find maximum and minimum values of a·sinθ + b·cosθ
  • Solve trigonometric equations using compound and double angle formulae
  • Prove trigonometric identities using compound angle results

Topic Overview

Compound Angles

sin/cos/tan of A±B — the core formulae for all advanced trig

Double Angle Formulae

sin(2A), three forms of cos(2A), and tan(2A)

R·sin(θ+α) Form

Combine a·sinθ + b·cosθ into a single sinusoidal wave

Solving Equations

Use identities to reduce equations to simpler forms

Proving Identities

Work from one side using compound angle results

Applications

Calculus, wave analysis, and exam-style problems

Learn 1 — Compound Angle Formulae (sin and cos)

These four formulae are the starting point for everything in this topic. Learn to recognise both the + and − versions, noting how the signs differ between sin and cos.

Sine Formulae

sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
Memory aid: For sin, the signs inside and outside always match — sin(A+B) uses +, sin(A−B) uses −.

Cosine Formulae

cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
Key point: For cos, the sign flips — cos(A+B) uses −, cos(A−B) uses +. This catches many students out!

Deriving Exact Values

These formulae let us find exact trig values for angles not in the standard table by splitting them into 30°, 45°, 60° combinations.

Example: Find the exact value of sin(75°)

Write 75° = 45° + 30°
sin(75°) = sin(45° + 30°) = sin45°cos30° + cos45°sin30°
= (√2/2)(√3/2) + (√2/2)(1/2)
= √6/4 + √2/4 = (√6 + √2)/4

Example: Find the exact value of cos(15°)

Write 15° = 45° − 30°
cos(15°) = cos(45° − 30°) = cos45°cos30° + sin45°sin30°
= (√2/2)(√3/2) + (√2/2)(1/2)
= √6/4 + √2/4 = (√6 + √2)/4
Note: sin(75°) = cos(15°) because they are complementary angles. This is a useful check!

Example: Find sin(A+B) given sinA = 3/5, cosA = 4/5, sinB = 5/13, cosB = 12/13

sin(A+B) = sinA cosB + cosA sinB
= (3/5)(12/13) + (4/5)(5/13)
= 36/65 + 20/65 = 56/65

Example: Find cos(A−B) given cosA = 3/5, sinA = 4/5, cosB = 12/13, sinB = 5/13

cos(A−B) = cosA cosB + sinA sinB
= (3/5)(12/13) + (4/5)(5/13)
= 36/65 + 20/65 = 56/65

Learn 2 — Tan Compound Formulae and Derivation

Tangent Compound Formulae

tan(A + B) = (tan A + tan B) / (1 − tan A · tan B)
tan(A − B) = (tan A − tan B) / (1 + tan A · tan B)
Sign trap: For tan(A+B) the denominator has minus; for tan(A−B) the denominator has plus. Opposite to what you might expect!

Deriving tan(A+B)

Derivation

tan(A+B) = sin(A+B) / cos(A+B)
= (sinA cosB + cosA sinB) / (cosA cosB − sinA sinB)
Divide every term by cosA cosB:
= (tanA + tanB) / (1 − tanA tanB) ✓

Geometric Basis of sin(A+B)

Unit circle method: Place a point on the unit circle at angle A+B. Using the coordinates of the point (cos(A+B), sin(A+B)) and applying rotation matrix logic, we derive:
sin(A+B) = sinA cosB + cosA sinB

Alternatively, use a right-angled triangle diagram: draw a triangle with angle A, then add angle B at the apex. The vertical projection decomposes into sinA cosB (from the part along the cosB direction) plus cosA sinB (from the perpendicular sinB component).

Examples Using tan Compound

Example: Find the exact value of tan(75°)

75° = 45° + 30°, so tan75° = tan(45°+30°)
= (tan45° + tan30°) / (1 − tan45° · tan30°)
= (1 + 1/√3) / (1 − 1·1/√3)
= (1 + 1/√3) / (1 − 1/√3) × (√3/√3) = (√3 + 1)/(√3 − 1)
Rationalise: × (√3+1)/(√3+1) = (√3+1)² / (3−1) = (4+2√3)/2 = 2 + √3

Example: Given tanA = 2, tanB = 1/3, find tan(A+B)

tan(A+B) = (2 + 1/3)/(1 − 2·1/3) = (7/3)/(1/3) = 7
Since tan(A+B) = 7 = tan(arctan7), and also tan90° is undefined — but note that A+B = 45° + something special here. Actually tan(A+B) = 7 in this case.

Example: Solve tan(x + 30°) = 2tanx for 0° ≤ x ≤ 180°

Expand: (tanx + tan30°)/(1 − tanx·tan30°) = 2tanx
Let t = tanx. (t + 1/√3)/(1 − t/√3) = 2t
t + 1/√3 = 2t(1 − t/√3) = 2t − 2t²/√3
2t²/√3 − t + 1/√3 = 0 → 2t² − √3·t + 1 = 0
t = (√3 ± √(3−8))/4 — discriminant negative, so no real solutions in this range with these exact values. Check for specific cases where tanx = 0: x = 0° works trivially.

Learn 3 — Double Angle Formulae

Setting B = A in the compound angle formulae gives the double angle formulae. These are among the most important results at A-Level.

Double Angle for Sine

sin(2A) = 2 sin A cos A

Double Angle for Cosine — THREE Forms

cos(2A) = cos²A − sin²A
cos(2A) = 2cos²A − 1   (using sin²A = 1 − cos²A)
cos(2A) = 1 − 2sin²A   (using cos²A = 1 − sin²A)

Double Angle for Tangent

tan(2A) = 2tan A / (1 − tan²A)

Which Form of cos(2A) to Use?

Choose based on what you want to eliminate:
• To remove sin²A: use cos(2A) = 2cos²A − 1, so cos²A = (1 + cos2A)/2
• To remove cos²A: use cos(2A) = 1 − 2sin²A, so sin²A = (1 − cos2A)/2
• In integration and identity proofs, selecting the right form is crucial.

Example: Verify sin(60°) using double angle for sin(30°)

sin(2×30°) = 2sin30°cos30° = 2 × (1/2) × (√3/2) = √3/2 ✓

Example: Given sinA = 3/5 (A acute), find sin(2A) and cos(2A)

cosA = 4/5 (Pythagoras)
sin(2A) = 2sinA cosA = 2 × (3/5) × (4/5) = 24/25
cos(2A) = cos²A − sin²A = 16/25 − 9/25 = 7/25

Example: Express cos(2x) in terms of sinx only, then solve cos(2x) = 1 − 2sin²x = 0

cos(2x) = 1 − 2sin²x = 0
2sin²x = 1 → sin²x = 1/2 → sinx = ±1/√2
x = 45°, 135°, 225°, 315° for 0° ≤ x ≤ 360°

Example: Given tanA = 2 (A acute), find tan(2A)

tan(2A) = 2tanA/(1 − tan²A) = 2(2)/(1 − 4) = 4/(−3) = −4/3

Example: Find cos(2A) when cosA = −2/3 (A obtuse)

cos(2A) = 2cos²A − 1 = 2(4/9) − 1 = 8/9 − 1 = −1/9

Learn 4 — R·sin(θ + α) Form

Any expression of the form a·sinθ + b·cosθ can be written as a single sinusoidal wave R·sin(θ + α). This is essential for finding maximum/minimum values and solving equations.

The General Rule

a·sinθ + b·cosθ = R·sin(θ + α)
where R = √(a² + b²) and tan α = b/a, with R > 0 and 0 < α < 90°

How to Find R and α

Step 1: Expand R·sin(θ + α) = R sinθ cosα + R cosθ sinα
Step 2: Compare with a·sinθ + b·cosθ: R cosα = a and R sinα = b
Step 3: R = √(a² + b²) from squaring and adding
Step 4: tanα = b/a (from dividing the two equations)
Step 5: α = arctan(b/a) — ensure α is in the correct quadrant

Also: R·cos(θ − α) Form

a·cosθ + b·sinθ = R·cos(θ − α)
where R = √(a² + b²) and tan α = b/a

Maximum and Minimum Values

Since −1 ≤ sin(θ + α) ≤ 1, we have −R ≤ R·sin(θ+α) ≤ R
Maximum value = R, occurs when sin(θ+α) = 1, i.e. θ+α = 90°
Minimum value = −R, occurs when sin(θ+α) = −1, i.e. θ+α = 270°

Example: Write 3sinθ + 4cosθ in R·sin(θ+α) form

R = √(3² + 4²) = √(9 + 16) = √25 = 5
tanα = 4/3 → α = arctan(4/3) ≈ 53.13°
3sinθ + 4cosθ = 5sin(θ + 53.13°)
Maximum value = 5, when θ + 53.13° = 90°, so θ ≈ 36.87°
Minimum value = −5, when θ ≈ 216.87°

Example: Solve 3sinθ + 4cosθ = 2 for 0° ≤ θ ≤ 360°

From above: 5sin(θ + 53.13°) = 2
sin(θ + 53.13°) = 2/5 = 0.4
θ + 53.13° = arcsin(0.4) = 23.58° or 156.42°
But 23.58° gives θ = −29.55° which is out of range
Also consider: 360° + 23.58° = 383.58° → θ = 330.45°
And: 156.42° → θ = 103.29°
Solutions: θ ≈ 103.3° and θ ≈ 330.5°

Learn 5 — Solving Equations and Proving Identities

Solving Equations with Double/Compound Angles

The key strategy is to use identities to express an equation in terms of a single trig ratio, then solve as usual.

Example: Solve sin(2x) = sinx for 0° ≤ x ≤ 360°

Use sin(2x) = 2sinx cosx
2sinx cosx = sinx
2sinx cosx − sinx = 0
sinx(2cosx − 1) = 0
sinx = 0 → x = 0°, 180°, 360°
2cosx − 1 = 0 → cosx = 1/2 → x = 60°, 300°
Solutions: x = 0°, 60°, 180°, 300°, 360°

Example: Solve cos(2x) − cosx = 0 for 0° ≤ x ≤ 360°

Use cos(2x) = 2cos²x − 1: 2cos²x − 1 − cosx = 0
2cos²x − cosx − 1 = 0 → (2cosx + 1)(cosx − 1) = 0
cosx = −1/2 → x = 120°, 240°
cosx = 1 → x = 0°, 360°
Solutions: x = 0°, 120°, 240°, 360°

Proving Identities — Strategy

Rule 1: Work on ONE side only (usually the more complex side).
Rule 2: Apply compound/double angle formulae to expand.
Rule 3: Use Pythagorean identities (sin²+cos²=1) to simplify.
Rule 4: Look for common factors and cancel.
Rule 5: Never cross-multiply or move terms across the equals sign.

Example: Prove (1 − cos2x) / sin2x = tanx

LHS: (1 − cos2x)/sin2x
= (1 − (1 − 2sin²x))/(2sinx cosx) [using cos2x = 1−2sin²x and sin2x = 2sinx cosx]
= 2sin²x / (2sinx cosx)
= sinx/cosx = tanx = RHS ✓

Example: Prove sin(A+B)/sin(A−B) = (tanA + tanB)/(tanA − tanB)

LHS: (sinA cosB + cosA sinB)/(sinA cosB − cosA sinB)
Divide numerator and denominator by cosA cosB:
= (tanA + tanB)/(tanA − tanB) = RHS ✓

Worked Examples

Example 1 — Find the exact value of sin(105°)

105° = 60° + 45°
sin(105°) = sin60°cos45° + cos60°sin45°
= (√3/2)(√2/2) + (1/2)(√2/2)
= √6/4 + √2/4 = (√6 + √2)/4
M1 A1

Example 2 — Solve sin(x + 30°) = cosx for 0° ≤ x ≤ 360°

Expand: sinx cos30° + cosx sin30° = cosx
(√3/2)sinx + (1/2)cosx = cosx
(√3/2)sinx = cosx − (1/2)cosx = (1/2)cosx
tanx = (1/2)/(√3/2) = 1/√3
x = 30°, 210° A1 A1

Example 3 — Write 5sinθ − 12cosθ in R·sin(θ − α) form

Expand R·sin(θ−α) = R sinθ cosα − R cosθ sinα
Compare: R cosα = 5 and R sinα = 12
R = √(25 + 144) = √169 = 13
tanα = 12/5 → α = arctan(12/5) ≈ 67.38°
5sinθ − 12cosθ = 13sin(θ − 67.38°) M1 A1 A1

Example 4 — Maximum value of 5sinθ − 12cosθ

From Example 3: f(θ) = 13sin(θ − 67.38°)
Maximum = 13, occurs when sin(θ − 67.38°) = 1
θ − 67.38° = 90° → θ = 157.38°
Minimum = −13, occurs when θ − 67.38° = 270° → θ = 337.38°

Example 5 — Prove sin2A / (1 + cos2A) = tanA

LHS = 2sinA cosA / (1 + 2cos²A − 1)
= 2sinA cosA / 2cos²A
= sinA/cosA = tanA = RHS ✓ M1 A1

Example 6 — Solve cos(2x) + 3cosx + 2 = 0 for 0° ≤ x ≤ 360°

Use cos(2x) = 2cos²x − 1:
2cos²x − 1 + 3cosx + 2 = 0
2cos²x + 3cosx + 1 = 0
(2cosx + 1)(cosx + 1) = 0
cosx = −1/2 → x = 120°, 240°
cosx = −1 → x = 180°
Solutions: x = 120°, 180°, 240° M1 M1 A1 A1 A1

Example 7 — Show sin(3A) = 3sinA − 4sin³A

sin(3A) = sin(2A + A) = sin2A cosA + cos2A sinA
= (2sinA cosA)cosA + (1 − 2sin²A)sinA
= 2sinA cos²A + sinA − 2sin³A
= 2sinA(1 − sin²A) + sinA − 2sin³A
= 2sinA − 2sin³A + sinA − 2sin³A
= 3sinA − 4sin³AM1 M1 A1

Example 8 — Solve 3sinx + 4cosx = 3 for 0 ≤ x ≤ 2π

Write as 5sin(x + α), tanα = 4/3 → α ≈ 0.9273 rad
5sin(x + 0.9273) = 3 → sin(x + 0.9273) = 0.6
x + 0.9273 = arcsin(0.6) = 0.6435 rad or π − 0.6435 = 2.4981 rad
Case 1: x = 0.6435 − 0.9273 = −0.284 (out of range)
Add 2π: x = −0.284 + 2π ≈ 5.999 ≈ 6.00 rad
Case 2: x = 2.4981 − 0.9273 = 1.571 rad (≈ π/2)
Solutions: x ≈ 1.57 rad and x ≈ 6.00 rad

Common Mistakes

Mistake 1 — Mixing up signs in cos(A±B)

✗ cos(A+B) = cosAcosB + sinAsinB
✓ cos(A+B) = cosAcosB − sinAsinB (the sin terms subtract)

For cos, the sign flips relative to the angle. cos(A+B) uses minus, cos(A−B) uses plus. Contrast with sin where the sign matches.

Mistake 2 — Not square rooting when finding R

✗ For 3sinθ + 4cosθ, R = 3² + 4² = 25
✓ R = √(3² + 4²) = √25 = 5

R is the square root of the sum of squares — always take the positive square root.

Mistake 3 — Using the wrong form of cos(2A)

✗ Using cos(2A) = 2cos²A−1 when the equation contains sinx terms
✓ If equation has sinx, use cos(2A) = 1−2sin²A to eliminate cos²A

Choose the form of cos(2A) that eliminates the trig ratio you don't want.

Mistake 4 — Sign error in tan(A−B) denominator

✗ tan(A−B) = (tanA − tanB)/(1 − tanAtanB)
✓ tan(A−B) = (tanA − tanB)/(1 + tanAtanB) — denominator has PLUS

The denominators of tan(A+B) and tan(A−B) have opposite signs to what you'd expect. Memorise this carefully.

Mistake 5 — Missing solutions when solving R·sin(θ+α) = k

✗ Only finding θ+α = arcsin(k/R) and stopping with one solution
✓ Also consider θ+α = 180° − arcsin(k/R), then subtract α from both, and check range

Sine equations always have two solutions per period. Use the principal value and the supplementary angle, then adjust for range.

Mistake 6 — Incorrect double angle for sine

✗ sin(2A) = 2sinA
✓ sin(2A) = 2sinA cosA — the cosA factor is essential!

Doubling the angle is not the same as doubling the ratio. The formula requires both sinA and cosA.

Mistake 7 — Not checking all solutions are in the given range

✗ Including x = −30° when the domain is 0° ≤ x ≤ 360°
✓ Add or subtract 360° (or 2π) to bring all solutions into the required interval

Always check every solution against the domain restriction at the end.

Key Formulas Reference

FormulaExpressionNotes
sin(A+B)sinA cosB + cosA sinBSigns match the angle
sin(A−B)sinA cosB − cosA sinBSigns match the angle
cos(A+B)cosA cosB − sinA sinBSign flips!
cos(A−B)cosA cosB + sinA sinBSign flips!
tan(A+B)(tanA + tanB)/(1 − tanA tanB)Denominator: minus
tan(A−B)(tanA − tanB)/(1 + tanA tanB)Denominator: plus
sin(2A)2sinA cosAFrom sin(A+A)
cos(2A) — Form 1cos²A − sin²ADirect substitution
cos(2A) — Form 22cos²A − 1Use when eliminating sinA
cos(2A) — Form 31 − 2sin²AUse when eliminating cosA
tan(2A)2tanA/(1 − tan²A)From tan(A+A)
R·sin(θ+α)R = √(a²+b²), tanα = b/aa sinθ + b cosθ form
R·cos(θ−α)R = √(a²+b²), tanα = b/aa cosθ + b sinθ form
Max of R·sin(θ+α)RWhen sin(θ+α) = 1
Min of R·sin(θ+α)−RWhen sin(θ+α) = −1
sin²A (rearranged)(1 − cos2A)/2Useful in integration
cos²A (rearranged)(1 + cos2A)/2Useful in integration

Proof Bank

Proof 1 — Geometric Derivation of sin(A+B)

Consider a right-angled triangle OPQ where angle at O is (A+B).
Draw a point M on OQ such that angle POM = A, so angle PMQ = B.
Drop perpendiculars from P to OQ (at N) and from P to OM extended (at L).

From triangle OLP: OL = cosA (if OP = 1), PL = sinA
From triangle PLM: LM = sinA sinB, PM = sinA cosB
From triangle OLN: ON = cosA cosB, LN = sinA sinB... wait, more carefully:

Using the right-angled triangle approach (OP = 1, angle A+B at O):
Height = sin(A+B)
Decompose into two right triangles sharing a diagonal of length 1:
sin(A+B) = (height from A triangle) + (height from B triangle)
= sinA cosB + cosA sinB ✓

This geometric approach works because the two height contributions are additive and orthogonal projections are preserved under the decomposition.

Proof 2 — Derivation of sin(2A)

Start from sin(A+B) = sinA cosB + cosA sinB
Set B = A:
sin(A+A) = sinA cosA + cosA sinA
sin(2A) = 2sinA cosA ✓

This is a direct consequence of the compound angle formula with equal angles.

Proof 3 — Derivation of all three forms of cos(2A)

Start from cos(A+B) = cosA cosB − sinA sinB
Set B = A:
cos(2A) = cos²A − sin²A    ...(Form 1)

Form 2: Replace sin²A with (1 − cos²A):
cos(2A) = cos²A − (1 − cos²A) = 2cos²A − 1

Form 3: Replace cos²A with (1 − sin²A):
cos(2A) = (1 − sin²A) − sin²A = 1 − 2sin²A

All three are equivalent; choice depends on which squared term you wish to eliminate.

Visualiser — a·sinθ + b·cosθ vs R·sin(θ+α)

Adjust a and b to see how the combined wave changes. The blue line is a·sinθ + b·cosθ. The red dashed line is R·sin(θ+α). They are identical — this proves the R-form works!

3
4

Exercise 1 — Exact Values using Compound Angles

Exercise 2 — Double Angle Formulae

Exercise 3 — R·sin(θ+α) Form

Exercise 4 — Solving Trig Equations

Exercise 5 — Mixed Compound & Double Angle

Practice — 30 Questions

Challenge — 15 Hard Questions

Exam Style Questions

Question 1 [5 marks]

(a) Express 8sinθ + 6cosθ in the form R sin(θ + α), where R > 0 and 0° < α < 90°. State the exact value of R and give α to 2 decimal places. [3]

(b) Hence find the maximum value of 8sinθ + 6cosθ and the smallest positive θ at which it occurs. [2]

(a) R = √(64+36) = √100 = 10, tanα = 6/8 = 3/4, α = 36.87° [M1 A1 A1]
8sinθ + 6cosθ = 10sin(θ + 36.87°)
(b) Maximum = 10 [B1], occurs when sin(θ+36.87°) = 1 → θ+36.87° = 90° → θ = 53.13° [A1]

Question 2 [6 marks]

Solve the equation 5sinx − 12cosx = 6.5 for 0° ≤ x ≤ 360°, giving solutions to 1 decimal place. [6]

R = √(25+144) = 13, tanα = 12/5, α = 67.38° [M1 A1]
13sin(x − 67.38°) = 6.5 [M1]
sin(x − 67.38°) = 0.5 [A1]
x − 67.38° = 30° → x = 97.4° [A1]
x − 67.38° = 150° → x = 217.4° [A1]

Question 3 [5 marks]

(a) Show that cos(2θ) + 3cosθ + 2 = 0 can be written as (2cosθ + 1)(cosθ + 1) = 0. [2]

(b) Hence solve cos(2θ) + 3cosθ + 2 = 0 for 0° ≤ θ ≤ 360°. [3]

(a) Use cos2θ = 2cos²θ − 1 [M1]: 2cos²θ − 1 + 3cosθ + 2 = 2cos²θ + 3cosθ + 1 = (2cosθ+1)(cosθ+1) [A1]
(b) cosθ = −1/2 → θ = 120°, 240° [A1 A1]; cosθ = −1 → θ = 180° [A1]

Question 4 [4 marks]

Prove the identity: (sin2A + sin A) / (cos2A + cosA + 1) = tanA [4]

Numerator: sin2A + sinA = 2sinA cosA + sinA = sinA(2cosA + 1) [M1]
Denominator: cos2A + cosA + 1 = (2cos²A−1) + cosA + 1 = 2cos²A + cosA = cosA(2cosA+1) [M1]
Ratio: sinA(2cosA+1) / [cosA(2cosA+1)] = sinA/cosA = tanA ✓ [A1 A1]

Question 5 [6 marks]

f(θ) = 3sinθ + 4cosθ

(a) Write f(θ) in the form R sin(θ + α) where R > 0 and 0 < α < π/2. Give R exactly and α to 3 significant figures. [3]

(b) Hence find all solutions of f(θ) = 2 for 0 ≤ θ ≤ 2π. Give answers to 3 significant figures. [3]

(a) R = 5, α = arctan(4/3) = 0.927 rad [M1 A1 A1]
(b) 5sin(θ+0.927) = 2 → sin(θ+0.927) = 0.4 [M1]
θ+0.927 = 0.4115 (out of range, add 2π: gives 6.694, θ=5.77) or θ+0.927 = π−0.4115 = 2.730, θ = 1.80 [A1 A1]
θ ≈ 1.80 rad and θ ≈ 5.77 rad

Question 6 [5 marks]

Given that sin(A+B) = 2sin(A−B), show that tanA = 3tanB. [5]

sinA cosB + cosA sinB = 2(sinA cosB − cosA sinB) [M1 expanding both sides A1]
sinA cosB + cosA sinB = 2sinA cosB − 2cosA sinB [A1]
3cosA sinB = sinA cosB [M1]
Divide by cosA cosB: 3tanB = tanA ✓ [A1]

Question 7 [6 marks]

(a) Find the exact value of sin(π/12). [3]

(b) Hence, or otherwise, find the exact value of sin²(π/12). [1]

(c) Find ∫₀^(π/12) sin²(x) dx, leaving your answer in exact form. [2]

(a) π/12 = 15° = 45°−30°; sin15° = sin45°cos30°−cos45°sin30° = (√6−√2)/4 [M1 A1 A1]
(b) sin²(π/12) = ((√6−√2)/4)² = (8−4√3)/16 = (2−√3)/4 [A1]
(c) ∫sin²x dx = ∫(1−cos2x)/2 dx = x/2 − sin2x/4 + c [M1]
[x/2 − sin2x/4]₀^(π/12) = π/24 − sin(π/6)/4 = π/24 − 1/8 [A1]

Question 8 [4 marks]

Solve the equation sin(2θ) = sinθ for 0 ≤ θ ≤ 2π. [4]

2sinθcosθ = sinθ [M1 using sin2θ = 2sinθcosθ]
sinθ(2cosθ − 1) = 0 [M1 factorising]
sinθ = 0: θ = 0, π, 2π [A1]
cosθ = 1/2: θ = π/3, 5π/3 [A1]

Past Paper Questions — Cambridge 9709

Past Paper 1 — Cambridge 9709/22 (adapted)

The function f is defined by f(x) = 5sinx − 12cosx for x ∈ ℝ.

(a) Express f(x) in the form R sin(x − α), where R > 0 and 0 < α < π/2. [3]

(b) State the range of f. [1]

(c) Solve f(x) = 6 for 0 ≤ x ≤ 2π. [3]

(a) R = √(25+144) = 13, Rcosα=5, Rsinα=12, α = arctan(12/5) ≈ 1.176 rad [M1 A1 A1]
f(x) = 13sin(x − 1.176)
(b) Range: −13 ≤ f(x) ≤ 13 [B1]
(c) 13sin(x−1.176) = 6 → sin(x−1.176) = 6/13 [M1]
x−1.176 = 0.4759 or π−0.4759 = 2.666 [A1]
x = 1.652 or x = 3.842 [A1]

Past Paper 2 — Cambridge 9709/23 (adapted)

Prove that (1 − cos2θ)/(sin2θ) ≡ tanθ. [3]

LHS = (1−(1−2sin²θ))/(2sinθcosθ) [M1 using cos2θ = 1−2sin²θ and sin2θ = 2sinθcosθ]
= 2sin²θ/(2sinθcosθ) [A1]
= sinθ/cosθ = tanθ = RHS ✓ [A1]

Past Paper 3 — Cambridge 9709/21 (adapted)

Solve the equation 4cos(2x) + 2cosx = 1 for 0° ≤ x ≤ 360°. [5]

4(2cos²x−1) + 2cosx = 1 [M1]
8cos²x + 2cosx − 5 = 0 [A1]
cosx = (−2 ± √(4+160))/16 = (−2 ± √164)/16 [M1]
cosx = (−2+12.806)/16 = 0.6754 or cosx = (−2−12.806)/16 (rejected, < −1) [A1]
x = 47.5° or x = 312.5° [A1]

Past Paper 4 — Cambridge 9709/22 (adapted)

Given that sin(A+B)/cos(A−B) = 3/5, find the value of (tanA + tanB)/(1 + tanA tanB). [3]

Recognise that (tanA+tanB)/(1+tanAtanB) = sin(A+B)/cos(A+B)... wait, let us expand:
sin(A+B) = sinAcosB + cosAsinB; cos(A−B) = cosAcosB + sinAsinB [M1]
Dividing numerator and denominator of sin(A+B)/cos(A+B) by cosAcosB gives (tanA+tanB)/(1−tanAtanB) = tan(A+B)
But the given expression (tanA+tanB)/(1+tanAtanB): noting this equals sin(A+B)/cos(A−B) after dividing by cosAcosB [M1]
Therefore the value = 3/5 [A1]

Past Paper 5 — Cambridge 9709/23 (adapted)

f(θ) = 2sinθ + 3cosθ. Express f(θ) in the form R sin(θ+α). Find the minimum value of [f(θ)]² and the corresponding value of θ in [0, 2π]. [6]

R = √(4+9) = √13, α = arctan(3/2) ≈ 0.9828 rad [M1 A1]
f(θ) = √13 sin(θ+0.9828) [A1]
[f(θ)]² = 13 sin²(θ+0.9828)
Minimum of sin² = 0, so minimum of [f(θ)]² = 0 [M1 A1]
Occurs when sin(θ+0.9828) = 0 → θ+0.9828 = π → θ = π−0.9828 ≈ 2.159 rad [A1]