Strand 2 — Geometry & Trigonometry

Trigonometry for Leaving Certificate

5th Year · 6th Year (Leaving Cert)

✓By the end of this lesson students will be able to convert between radian and degree measure and apply radian measure to arc length and area of a sector.
✓By the end of this lesson students will be able to understand and use the unit circle to determine trigonometric ratios for all angles and recall exact values.
✓By the end of this lesson students will be able to apply trigonometric identities, including Pythagorean, compound angle, and double angle identities, to simplify expressions and prove results.
✓By the end of this lesson students will be able to solve trigonometric equations within a given domain.
✓By the end of this lesson students will be able to apply the Sine Rule, Cosine Rule, and Area formula (½ab sin C) to solve problems involving triangles, including 3D contexts.

Key concepts

Radian Measure

Radian measure is an alternative unit for angles, where one radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. The relationship between radians and degrees is π radians = 180°. Radian measure is often used in calculus and advanced trigonometry.

π radians = 180° Arc length, s = rθ (where θ is in radians) Area of sector, A = (1/2)r²θ (where θ is in radians)

Unit Circle

The unit circle is a circle with a radius of 1 unit, centred at the origin (0,0) of a Cartesian coordinate system. For any point (x,y) on the unit circle, the angle θ (measured anti-clockwise from the positive x-axis) has cos θ = x and sin θ = y. The tangent of the angle is tan θ = y/x. The unit circle helps to visualise trigonometric ratios for all angles, including those greater than 90° or negative angles, and to understand the signs of ratios in different quadrants (CAST rule).

x = cos θ, y = sin θ, tan θ = y/x

Exact Values

Certain angles have exact trigonometric ratios that should be memorised or easily derived. These include 0°, 30° (π/6), 45° (π/4), 60° (π/3), and 90° (π/2), and their multiples.

Trigonometric Identities (HL)

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables for which the functions are defined. They are crucial for simplifying expressions, proving other identities, and solving trigonometric equations.

Pythagorean Identity: sin²θ + cos²θ = 1 Compound Angle Identities: 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) Double Angle Identities: sin(2A) = 2 sin A cos A cos(2A) = cos²A - sin²A = 2cos²A - 1 = 1 - 2sin²A tan(2A) = (2 tan A) / (1 - tan²A)

Sine Rule

The Sine Rule relates the sides of a triangle to the sines of its opposite angles. It is used to find unknown sides or angles when you have a side and its opposite angle, along with one other piece of information (ASA, AAS, or SSA - the ambiguous case).

a / sin A = b / sin B = c / sin C

Cosine Rule

The Cosine Rule relates the sides of a triangle to the cosine of one of its angles. It is used to find an unknown side when two sides and the included angle are known (SAS), or to find an unknown angle when all three sides are known (SSS).

a² = b² + c² - 2bc cos A

Area of a Triangle

The area of a triangle can be calculated using two sides and the included angle.

Area = (1/2)ab sin C

3D Problems (HL)

Trigonometric rules and concepts can be applied to solve problems in three dimensions. This often involves identifying right-angled triangles or non-right-angled triangles within a 3D structure and applying Pythagoras' theorem, the Sine Rule, or the Cosine Rule.

Solving Trigonometric Equations

Solving trigonometric equations involves finding the values of the angle that satisfy the given equation. This often requires using identities to simplify the equation, finding a reference angle, and then determining all solutions within a specified domain by considering the signs of the trigonometric functions in different quadrants.

General solutions: If sin x = k, then x = nπ + (-1)ⁿα, where α = sin⁻¹(k) If cos x = k, then x = 2nπ ± α, where α = cos⁻¹(k) If tan x = k, then x = nπ + α, where α = tan⁻¹(k) (n ∈ ℤ and α is the reference angle)

Key facts to remember

1π radians = 180°. To convert degrees to radians, multiply by π/180. To convert radians to degrees, multiply by 180/π.
2Arc length s = rθ and Area of sector A = (1/2)r²θ, where θ must be in radians.
3The exact values for sin, cos, and tan for 0°, 30°, 45°, 60°, and 90° (and their radian equivalents) should be known.
4The Pythagorean identity is sin²θ + cos²θ = 1.
5The CAST rule helps determine the sign of trigonometric ratios in each quadrant.
6Sine Rule: a/sin A = b/sin B = c/sin C. Used for ASA, AAS, and SSA (ambiguous case).
7Cosine Rule: a² = b² + c² - 2bc cos A. Used for SAS and SSS.
8Area of a triangle = (1/2)ab sin C.

Worked examples

Example 1

A sector of a circle has a radius of 10 cm and the angle subtended at the centre is 2π/3 radians. Calculate the arc length and the area of the sector.

IIdentify the given values: radius r = 10 cm, angle θ = 2π/3 radians.

IIUse the formula for arc length: s = rθ.

IIISubstitute the values: s = (10)(2π/3) = 20π/3 cm.

IVUse the formula for the area of a sector: A = (1/2)r²θ.

VSubstitute the values: A = (1/2)(10)²(2π/3) = (1/2)(100)(2π/3) = 100π/3 cm².

Answer

Arc length = 20π/3 cm, Area = 100π/3 cm².

Ensure your calculator is in radian mode if you are evaluating numerical approximations for π.

Example 2

Solve the equation 2cos(2θ) - 3sin(θ) + 1 = 0 for 0 ≤ θ ≤ 2π.

IRecognise the double angle term cos(2θ). Choose the identity that involves sin(θ): cos(2θ) = 1 - 2sin²(θ).

IISubstitute the identity into the equation: 2(1 - 2sin²(θ)) - 3sin(θ) + 1 = 0.

IIIExpand and simplify: 2 - 4sin²(θ) - 3sin(θ) + 1 = 0.

IVRearrange into a quadratic equation in terms of sin(θ): 4sin²(θ) + 3sin(θ) - 3 = 0.

VLet x = sin(θ). Solve the quadratic equation 4x² + 3x - 3 = 0 using the quadratic formula x = (-b ± √(b² - 4ac)) / 2a.

VIx = (-3 ± √(3² - 4(4)(-3))) / (2(4)) = (-3 ± √(9 + 48)) / 8 = (-3 ± √57) / 8.

VIICalculate the two possible values for sin(θ): sin(θ) ≈ (-3 + √57) / 8 ≈ 0.5937 or sin(θ) ≈ (-3 - √57) / 8 ≈ -1.3437.

VIIIThe value sin(θ) ≈ -1.3437 is not possible as the range of sin(θ) is [-1, 1]. So, we only consider sin(θ) ≈ 0.5937.

9Find the reference angle α = sin⁻¹(0.5937) ≈ 0.636 radians (to 3 decimal places).

10Since sin(θ) is positive, θ lies in the first or second quadrant.

11First quadrant solution: θ₁ = α ≈ 0.636 radians.

12Second quadrant solution: θ₂ = π - α ≈ π - 0.636 ≈ 2.506 radians.

13Check if these solutions are within the domain [0, 2π]. Both are.

Answer

θ ≈ 0.636 radians, 2.506 radians (to 3 decimal places).

Always check the range of trigonometric functions (e.g., sin θ and cos θ must be between -1 and 1). Remember to find all solutions within the given domain.

Example 3

A vertical pole AB of height 12 m stands at point A on horizontal ground. Points C and D are on the ground. C is due East of A and D is due North of A. The angle of elevation of B from C is 60°. The angle of elevation of B from D is 45°. Find the distance CD, correct to two decimal places.

IDraw a diagram. Let A be the origin (0,0,0). B is (0,0,12). C is on the positive x-axis (East), so C = (AC, 0, 0). D is on the positive y-axis (North), so D = (0, AD, 0).

IIConsider the right-angled triangle ABC (right-angled at A). The angle of elevation of B from C is ∠BCA = 60°. We have tan(∠BCA) = AB/AC.

IIItan(60°) = 12/AC => √3 = 12/AC => AC = 12/√3 = 12√3/3 = 4√3 m.

IVConsider the right-angled triangle ABD (right-angled at A). The angle of elevation of B from D is ∠BDA = 45°. We have tan(∠BDA) = AB/AD.

Vtan(45°) = 12/AD => 1 = 12/AD => AD = 12 m.

VINow consider the triangle CAD on the horizontal ground. Since C is due East of A and D is due North of A, the angle ∠CAD is 90°. Triangle CAD is a right-angled triangle.

VIIUse Pythagoras' theorem to find CD: CD² = AC² + AD².

VIIICD² = (4√3)² + (12)² = (16 × 3) + 144 = 48 + 144 = 192.

9CD = √192 = √(64 × 3) = 8√3 m.

10Calculate the numerical value: CD ≈ 8 × 1.73205 ≈ 13.856 m.

11Round to two decimal places: CD ≈ 13.86 m.

Answer

The distance CD is approximately 13.86 m.

For 3D problems, break down the problem into 2D right-angled triangles or triangles where the Sine/Cosine Rule can be applied. Visualise the horizontal and vertical planes.

Common mistakes

✗Confusing degrees and radians: Always check your calculator mode and ensure angles are in the correct unit for formulas like s=rθ and A=½r²θ.
✗Forgetting the ambiguous case of the Sine Rule: When using SSA (Side-Side-Angle) to find an angle, there might be two possible solutions.
✗Incorrectly applying trigonometric identities: Ensure the correct compound or double angle formula is used, especially when dealing with signs.
✗Not finding all solutions when solving trigonometric equations: Solutions exist in multiple quadrants and often repeat every 2π (or π for tan).
✗Errors with signs of trigonometric ratios in different quadrants: Misapplying the CAST rule can lead to incorrect solutions.

Exam tips

★Always draw a clear, labelled diagram for geometry and trigonometry problems, especially for 3D scenarios.
★Memorise the exact values for common angles and the key trigonometric identities (Pythagorean, compound angle, double angle).
★When solving trigonometric equations, first find the reference angle, then use the CAST rule to identify all solutions within the given domain.
★Show all steps clearly in your working, particularly when proving identities or solving complex equations, to maximise part marks.

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