Mathematics (9709)
Topic 3 of 9Cambridge A Levels

Pure 1: Trigonometry

Trigonometric ratios and identities, solving trig equations, radians, arc length, sector area — core P1 and P3 content for 9709.

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Pure 1: Trigonometry — CIE A Level Mathematics 9709


1. Radians and Arc/Sector Formulae


Radian measure: 2π rad = 360°. Key conversions:

  • π/6 = 30°, π/4 = 45°, π/3 = 60°, π/2 = 90°, π = 180°

Arc length: s = rθ (θ in radians)

Sector area: A = ½r²θ


Example: A sector has radius 6 cm and angle 0.8 rad. Find arc length and area.

s = 6 × 0.8 = 4.8 cm; A = ½ × 36 × 0.8 = 14.4 cm²


Segment area = Sector area − Triangle area = ½r²θ − ½r²sinθ = ½r²(θ − sinθ)


2. Exact Values


| θ | sin θ | cos θ | tan θ |

|---|---|---|---|

| 0° (0) | 0 | 1 | 0 |

| 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 |

| 90° (π/2) | 1 | 0 | undefined |


3. Key Identities


Pythagorean identity: sin²θ + cos²θ = 1


Dividing by cos²θ: tan²θ + 1 = sec²θ

Dividing by sin²θ: 1 + cot²θ = cosec²θ


Double angle formulae:

  • sin 2A = 2 sin A cos A
  • cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A
  • tan 2A = 2tan A/(1 − tan²A)

Addition formulae:

  • sin(A ± B) = sin A cos B ± cos A sin B
  • cos(A ± B) = cos A cos B ∓ sin A sin B
  • tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)

4. Solving Trigonometric Equations

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Method:

  1. Rearrange to sin/cos/tan = value
  2. Find the principal value (inverse trig)
  3. Use CAST diagram or symmetry to find all solutions in the given interval

CAST diagram: Reading anticlockwise from positive x-axis:

  • All positive (0°–90°)
  • Sine positive (90°–180°)
  • Tangent positive (180°–270°)
  • Cosine positive (270°–360°)

Example: Solve 2sin²θ − sinθ − 1 = 0 for 0° ≤ θ ≤ 360°.

Let u = sinθ: 2u² − u − 1 = 0 → (2u + 1)(u − 1) = 0 → u = −1/2 or u = 1.

  • sinθ = 1: θ = 90°
  • sinθ = −1/2: θ = 210°, 330°

Solutions: θ = 90°, 210°, 330°


Example: Solve cos 2θ = sin θ for 0 ≤ θ ≤ π.

cos 2θ = 1 − 2sin²θ → 1 − 2sin²θ = sinθ → 2sin²θ + sinθ − 1 = 0.

(2sinθ − 1)(sinθ + 1) = 0 → sinθ = 1/2 or sinθ = −1.

In [0, π]: θ = π/6, 5π/6 (sinθ = 1/2); θ = 3π/2 outside range, sinθ = −1 → θ = 3π/2, not in [0,π].

Solutions: θ = π/6, 5π/6


5. The R Form: a sinθ + b cosθ = R sin(θ + α)


Express a sinθ + b cosθ in the form R sin(θ ± α) or R cos(θ ± α).


R = √(a² + b²), and α found from expansion and matching coefficients.


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

R sin(θ + α) = R sinθ cosα + R cosθ sinα → R cosα = 3, R sinα = 4.

R = √(9+16) = 5; tanα = 4/3 → α = 53.1°.

3sinθ + 4cosθ = 5sin(θ + 53.1°)


Uses: finding maximum/minimum values, solving equations.

Maximum value = R (when sin = 1), minimum = −R.


6. Small Angle Approximations


For small θ (in radians):

  • sin θ ≈ θ
  • cos θ ≈ 1 − θ²/2
  • tan θ ≈ θ

Example: Estimate sin(0.1). sin(0.1) ≈ 0.1 (actual: 0.0998 — very close).

Key Points to Remember

  • 1Radian conversions: π rad = 180°; arc length s=rθ, sector area A=½r²θ, segment = ½r²(θ−sinθ)
  • 2Key identities: sin²θ+cos²θ=1; tan²θ+1=sec²θ; double angle: cos2A = 2cos²A−1 = 1−2sin²A
  • 3Addition formulae: sin(A±B), cos(A±B), tan(A±B) — memorise or derive
  • 4CAST diagram for all solutions; solve by factoring as quadratic in sinθ or cosθ
  • 5R-form: acosθ+bsinθ = Rcos(θ−α); R=√(a²+b²); gives max/min instantly
  • 6Small angle: sinθ≈θ, cosθ≈1−θ²/2, tanθ≈θ for θ in radians

Pakistan Example

Trigonometry in Pakistani Architecture and Navigation

Pakistan's rich Islamic geometric architecture uses trigonometric principles. The octagonal minarets of the Badshahi Mosque in Lahore rely on exact values (sin 45°, cos 45°) in their construction. Sailors navigating the Arabian Sea from Karachi Port use the sine rule to determine their position triangulated from known landmarks — if two lighthouses are 50 km apart and the angles of sight from a ship are 40° and 70° respectively, the sine rule gives the distances. Civil engineers calculating the gradient and bank angle of the Karakoram Highway switchbacks use the R-form to model the combined effect of gradient and cross-slope, finding the maximum resultant gradient to check vehicle safety specifications.

Quick Revision Infographic

Mathematics — Quick Revision

Pure 1: Trigonometry

Key Concepts

1Radian conversions: π rad = 180°; arc length s=rθ, sector area A=½r²θ, segment = ½r²(θ−sinθ)
2Key identities: sin²θ+cos²θ=1; tan²θ+1=sec²θ; double angle: cos2A = 2cos²A−1 = 1−2sin²A
3Addition formulae: sin(A±B), cos(A±B), tan(A±B) — memorise or derive
4CAST diagram for all solutions; solve by factoring as quadratic in sinθ or cosθ
5R-form: acosθ+bsinθ = Rcos(θ−α); R=√(a²+b²); gives max/min instantly
6Small angle: sinθ≈θ, cosθ≈1−θ²/2, tanθ≈θ for θ in radians

Formulas to Know

A=½r²θ, segment = ½r²(θ−sinθ)
A = 2cos²A−1 = 1−2sin²A
Rcos(θ−α); R=√(a²+b²); gives max/min instantly
Pakistan Example

Trigonometry in Pakistani Architecture and Navigation

Pakistan's rich Islamic geometric architecture uses trigonometric principles. The octagonal minarets of the Badshahi Mosque in Lahore rely on exact values (sin 45°, cos 45°) in their construction. Sailors navigating the Arabian Sea from Karachi Port use the sine rule to determine their position triangulated from known landmarks — if two lighthouses are 50 km apart and the angles of sight from a ship are 40° and 70° respectively, the sine rule gives the distances. Civil engineers calculating the gradient and bank angle of the Karakoram Highway switchbacks use the R-form to model the combined effect of gradient and cross-slope, finding the maximum resultant gradient to check vehicle safety specifications.

SeekhoAsaan.com — Free RevisionPure 1: Trigonometry Infographic

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