Section 1 · Foundation

Basic Trigonometric Ratios

Grade 9 / Foundation

Trigonometry — from the Greek trigonon (triangle) and metron (measure) — studies the relationships between the angles and side lengths of triangles. In the CBC curriculum, it builds from basic right-triangle ratios in Grade 9 all the way to signal analysis in university mathematics. Every architecture, GPS system, bridge, and aircraft navigation system relies on these relationships.

The Right-Angled Triangle

Consider a right-angled triangle with an acute angle θ. The three sides are named relative to θ:

Side Definitions (relative to angle θ)

Hypotenuse (H): longest side — always opposite the right angle (90°)
Opposite (O): side directly across from the angle θ
Adjacent (A): side next to angle θ (not the hypotenuse)

Memory aid: SOH CAH TOA — the most reliable mnemonic in mathematics. Say it aloud: "Some Old Hags Can Always Hide Their Old Age" — the initials give SOH-CAH-TOA.

The Three Primary Ratios

Primary Trigonometric Ratios

sin θ (Sine) = Opposite / Hypotenuse = O/H

cos θ (Cosine) = Adjacent / Hypotenuse = A/H

tan θ (Tangent) = Opposite / Adjacent = O/A = sin θ / cos θ

The Three Reciprocal Ratios

Reciprocal Ratios

cosec θ (Cosecant) = 1 / sin θ = H / O

sec θ (Secant) = 1 / cos θ = H / A

cot θ (Cotangent) = 1 / tan θ = A / O = cos θ / sin θ

Worked Examples — Basic Ratios

Ex 1A right triangle has opposite = 5 cm, hypotenuse = 13 cm. Find all six trig ratios of θ.

Step 1Find the adjacent: by Pythagoras, A = √(H² − O²) = √(169 − 25) = √144 = 12 cm.

Step 2Primary ratios: sin θ = 5/13, cos θ = 12/13, tan θ = 5/12.

Step 3Reciprocal ratios: cosec θ = 13/5, sec θ = 13/12, cot θ = 12/5.

sin θ = 5/13 · cos θ = 12/13 · tan θ = 5/12 · cosec θ = 13/5 · sec θ = 13/12 · cot θ = 12/5

Ex 2In triangle ABC, angle B = 90°, BC = 8 m, angle A = 35°. Find AC (hypotenuse) and AB.

Step 1Identify: angle A = 35°, opposite = BC = 8 m, hypotenuse = AC (unknown), adjacent = AB (unknown).

Step 2Use sin A: sin 35° = BC/AC → AC = BC/sin 35° = 8/0.5736 = 13.95 m.

Step 3Use cos A: cos 35° = AB/AC → AB = 13.95 × cos 35° = 13.95 × 0.8192 = 11.43 m.

AC ≈ 13.95 m · AB ≈ 11.43 m

Ex 3 — ApplicationA ladder 8 m long leans against a vertical wall making a 65° angle with the ground. How high up the wall does it reach?

Step 1The ladder is the hypotenuse (8 m). The height h is opposite to the 65° angle.

Step 2sin 65° = h/8 → h = 8 × sin 65° = 8 × 0.9063 = 7.25 m.

CheckThe foot of the ladder is 8 × cos 65° = 8 × 0.4226 = 3.38 m from the wall. By Pythagoras: √(7.25² + 3.38²) = √(52.56 + 11.42) = √64 = 8 ✔

Height reached = 7.25 m

Section 2 · Foundation

Special Angles (0°, 30°, 45°, 60°, 90°)

Grade 9 / Foundation

Special angles produce exact, simple surd values for all six trig ratios. Memorising these avoids calculator dependence in non-calculator KCSE sections and is essential for solving algebraic trig problems at A-Level and university level.

These values are derived from two special right triangles: the 30-60-90 triangle (half an equilateral triangle with side 2) and the 45-45-90 isosceles right triangle (legs = 1).

Angle θ

sin θ

cos θ

tan θ

0°

0

1

0

30° (π/6)

1/2

√3/2

1/√3 = √3/3

45° (π/4)

√2/2 = 1/√2

1

60° (π/3)

√3/2

1/2

√3

90° (π/2)

1

0

undefined

tan 90° is undefined because cos 90° = 0, and division by zero is undefined. In graph form, tan θ has a vertical asymptote at θ = 90° (and every 90° thereafter, in opposite directions).

Worked Examples — Special Angles

Ex 4Find the exact value of: sin 60° × cos 30° − sin 30° × cos 60°

Step 1Substitute exact values: (√3/2)(√3/2) − (1/2)(1/2).

Step 2= 3/4 − 1/4 = 2/4 = 1/2.

NoteRecognise this as sin(60° − 30°) = sin 30° = 1/2. This previews the compound angle formula.

sin 60° cos 30° − sin 30° cos 60° = 1/2

Ex 5Without a calculator, simplify: (tan 45° + sin 30°) / cos 60°

Step 1Substitute: (1 + 1/2) / (1/2).

Step 2= (3/2) ÷ (1/2) = (3/2) × 2 = 3.

(tan 45° + sin 30°) / cos 60° = 3

Section 3 · Secondary

The Unit Circle & CAST Diagram

Secondary / KCSE

The unit circle extends trigonometry beyond acute angles to any angle — positive or negative, greater than 360°. It is the conceptual bridge between right-triangle trigonometry and the full periodic behaviour of trig functions used in calculus, physics, and engineering.

Definition of the Unit Circle

The unit circle is a circle of radius 1 centred at the origin (0, 0) in the Cartesian plane. For any angle θ measured anticlockwise from the positive x-axis, the corresponding point P on the circle has coordinates (cos θ, sin θ). This is how sine and cosine are formally defined for all angles.

Unit Circle Equation

x² + y² = 1 where x = cos θ and y = sin θ

CAST Diagram — Signs in Each Quadrant

The CAST rule tells you which trig functions are positive in each quadrant. Negative angles and reflex angles are handled automatically once you know CAST.

Quadrant

Angle Range

Positive Functions

Sign Summary

I (All)

0° – 90°

sin, cos, tan all +ve

All positive

II (Sin)

90° – 180°

sin only +ve

x < 0, y > 0

III (Tan)

180° – 270°

tan only +ve

x < 0, y < 0

IV (Cos)

270° – 360°

cos only +ve

x > 0, y < 0

Memory: "All Students Take Calculus" — the first letter of each word gives: All (Q1), Sin (Q2), Tan (Q3), Cos (Q4).

Reference Angle Method

To find the trig value of any angle: (1) find the reference angle (acute angle to the nearest x-axis), (2) use CAST to determine the sign, (3) evaluate the ratio for the reference angle.

Worked Examples — Unit Circle & CAST

Ex 6Find sin 150°, cos 240°, and tan 315° exactly.

sin 150°Q2 → sin is +ve. Reference angle = 180° − 150° = 30°. sin 150° = +sin 30° = +1/2.

cos 240°Q3 → cos is −ve. Reference angle = 240° − 180° = 60°. cos 240° = −cos 60° = −1/2.

tan 315°Q4 → tan is −ve. Reference angle = 360° − 315° = 45°. tan 315° = −tan 45° = −1.

sin 150° = 1/2 · cos 240° = −1/2 · tan 315° = −1

Ex 7Find all angles θ in [0°, 360°] such that cos θ = −√3/2.

Step 1cos is negative in Q2 and Q3. Reference angle: cos⁻¹(√3/2) = 30°.

Step 2Q2: θ = 180° − 30° = 150°. Q3: θ = 180° + 30° = 210°.

Checkcos 150° = −cos 30° = −√3/2 ✔ cos 210° = −cos 30° = −√3/2 ✔

θ = 150° or θ = 210°

Section 4 · Secondary

Trigonometric Identities

Secondary / KCSE

A trigonometric identity is an equation that is true for all values of the variable for which both sides are defined. Proving identities is a core KCSE skill and the foundation for all integration and differentiation of trig functions at A-Level and university.

Pythagorean Identities

These three identities all come from the equation of the unit circle x² + y² = 1. Substituting x = cos θ, y = sin θ gives the first identity; dividing through by cos²θ or sin²θ yields the other two.

Pythagorean Identities (must be memorised)

Identity 1sin²θ + cos²θ = 1

Identity 2 (÷ cos²θ)tan²θ + 1 = sec²θ

Identity 3 (÷ sin²θ)1 + cot²θ = cosec²θ

Quotient Identities

Tangenttan θ = sin θ / cos θ

Cotangentcot θ = cos θ / sin θ

Co-function Identities

Co-function Identities (complementary angles)

sin(90° − θ)= cos θ

cos(90° − θ)= sin θ

tan(90° − θ)= cot θ

Even / Odd (Negative Angle) Identities

Negative Angle Identities

sin(−θ)= −sin θ (odd function)

cos(−θ)= +cos θ (even function)

tan(−θ)= −tan θ (odd function)

Strategy for Proving Identities

Work on one side only — usually the more complex side. Do not cross-multiply or move terms across the equals sign.
Convert everything to sin and cos if the expression mixes different trig functions.
Look for Pythagorean patterns — if you see 1 − sin²θ, replace with cos²θ immediately.
Factorise or expand as needed to create matching factors.
Conclude with "= RHS ✓" (or "= LHS ✓") once both sides match.

Worked Examples — Proving Identities

Ex 8Prove: (1 − sin²θ) / cos²θ = 1

Step 1Work on LHS. Recognise 1 − sin²θ = cos²θ (Pythagorean identity).

Step 2LHS = cos²θ / cos²θ = 1 = RHS ✓

LHS = RHS = 1 (proved)

Ex 9Prove: (sin θ + cos θ)² = 1 + 2 sin θ cos θ

Step 1Expand LHS: sin²θ + 2 sin θ cos θ + cos²θ.

Step 2= (sin²θ + cos²θ) + 2 sin θ cos θ = 1 + 2 sin θ cos θ = RHS ✓

Proved: (sin θ + cos θ)² = 1 + 2 sin θ cos θ

Ex 10If sin θ = 5/13 and θ is in the first quadrant, find the exact values of all six trig ratios.

Step 1cos θ = √(1 − sin²θ) = √(1 − 25/169) = √(144/169) = 12/13. (positive in Q1)

Step 2tan θ = sin θ / cos θ = (5/13) / (12/13) = 5/12.

Step 3Reciprocals: cosec θ = 13/5, sec θ = 13/12, cot θ = 12/5.

sin = 5/13 · cos = 12/13 · tan = 5/12 · cosec = 13/5 · sec = 13/12 · cot = 12/5

Section 5 · Secondary / KCSE

Sine Rule, Cosine Rule & Triangle Area

Secondary / KCSE

The sine and cosine rules extend trigonometry to any triangle — not just right-angled ones. They appear heavily in KCSE Paper 1 and are applied extensively in surveying, navigation, and engineering in Kenya and globally.

The Sine Rule

In triangle ABC with sides a, b, c opposite to angles A, B, C respectively, the sine rule states:

Sine Rule

a / sin A = b / sin B = c / sin C = 2R

where R = circumradius of the triangle

Use Sine Rule when you know:

Two angles + one side (AAS or ASA), or Two sides + a non-included angle (SSA — beware ambiguous case)

Ambiguous Case (SSA):

When given two sides and a non-included angle, there may be 0, 1, or 2 valid triangles. Always check whether sin of the calculated angle exceeds 1 (no solution) or produces two valid angles.

The Cosine Rule

Cosine Rule (two forms)

Finding a side (SAS)a² = b² + c² − 2bc cos A

Finding an angle (SSS)cos A = (b² + c² − a²) / (2bc)

The cosine rule is a generalisation of Pythagoras: when A = 90°, cos A = 0, so the term −2bc cos A disappears and we recover a² = b² + c².

Area of a Triangle

Area Formulas

Two sides + included angleArea = ½ ab sin C = ½ bc sin A = ½ ca sin B

Heron's formula (SSS)Area = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2

Worked Examples — Sine & Cosine Rules

Ex 11 — Sine RuleIn triangle PQR, p = 9 cm, P = 42°, Q = 75°. Find q and the area.

Step 1Find R: R = 180° − 42° − 75° = 63°.

Step 2Sine rule: q/sin Q = p/sin P → q = 9 × sin 75° / sin 42° = 9 × 0.9659 / 0.6691 = 12.99 cm.

Step 3Area = ½ × p × q × sin R = ½ × 9 × 12.99 × sin 63° = ½ × 9 × 12.99 × 0.891 = 52.14 cm².

q ≈ 12.99 cm · Area ≈ 52.14 cm²

Ex 12 — Cosine Rule SASIn triangle ABC, b = 7 cm, c = 10 cm, A = 55°. Find a.

Step 1a² = b² + c² − 2bc cos A = 49 + 100 − 2(7)(10) cos 55°.

Step 2= 149 − 140 × 0.5736 = 149 − 80.30 = 68.70.

Step 3a = √68.70 = 8.29 cm.

a ≈ 8.29 cm

Ex 13 — Cosine Rule SSSFind angle B in a triangle with a = 5, b = 8, c = 6.

Step 1cos B = (a² + c² − b²) / (2ac) = (25 + 36 − 64) / (2 × 5 × 6) = −3/60 = −0.05.

Step 2B = cos⁻¹(−0.05) = 92.87° (obtuse — side b is the longest, so B is the largest angle).

B ≈ 92.87°

Section 6 · A-Level / Form 4

Compound & Multiple Angle Formulas

A-Level / Form 4

Compound angle formulas express the trig ratio of a sum or difference of two angles in terms of the ratios of the individual angles. They are among the most powerful tools in all of mathematics — used to derive differentiation and integration formulas for trig functions, and to prove every other advanced trig result.

Addition Formulas

Addition & Subtraction Formulas (must be memorised)

sin(A + B)= sin A cos B + cos A sin B

sin(A − B)= sin A cos B − cos A sin B

cos(A + B)= cos A cos B − sin 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)

tan(A − B)= (tan A − tan B) / (1 + tan A tan B)

Double Angle Formulas

Setting A = B in the addition formulas gives the double angle formulas. The cosine double angle has three equivalent forms — each useful in different contexts.

Double Angle Formulas (A = B)

sin 2A= 2 sin A cos A

cos 2A (form 1)= cos²A − sin²A

cos 2A (form 2)= 2 cos²A − 1

cos 2A (form 3)= 1 − 2 sin²A

tan 2A= 2 tan A / (1 − tan²A)

Half-Angle Formulas

Rearranging cos 2A forms 2 and 3 gives the half-angle formulas, essential for integration of trig powers.

Half-Angle Formulas (derived from cos 2A)

sin²A= (1 − cos 2A) / 2

cos²A= (1 + cos 2A) / 2

sin(A/2)= ±√[(1 − cos A)/2]

cos(A/2)= ±√[(1 + cos A)/2]

Product-to-Sum Formulas

Product-to-Sum (useful in integration)

sin A + sin B= 2 sin[(A+B)/2] cos[(A−B)/2]

sin A − sin B= 2 cos[(A+B)/2] sin[(A−B)/2]

cos A + cos B= 2 cos[(A+B)/2] cos[(A−B)/2]

cos A − cos B= −2 sin[(A+B)/2] sin[(A−B)/2]

Worked Examples — Compound Angles

Ex 14Find sin 75° exactly without a calculator.

Step 1Write 75° = 45° + 30°. Use sin(A + B) = sin A cos B + cos A sin B.

Step 2= sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2).

Step 3= √6/4 + √2/4 = (√6 + √2)/4 ≈ 0.9659.

sin 75° = (√6 + √2)/4

Ex 15If sin θ = 3/5 (θ acute), find sin 2θ, cos 2θ, and tan 2θ.

Step 1Find cos θ: cos θ = √(1 − 9/25) = 4/5. tan θ = 3/4.

Step 2sin 2θ = 2 sin θ cos θ = 2(3/5)(4/5) = 24/25.

Step 3cos 2θ = cos²θ − sin²θ = 16/25 − 9/25 = 7/25.

Step 4tan 2θ = sin 2θ / cos 2θ = (24/25)/(7/25) = 24/7. (Or: 2 × 3/4 / (1 − 9/16) = (3/2)/(7/16) = 24/7 ✔)

sin 2θ = 24/25 · cos 2θ = 7/25 · tan 2θ = 24/7

Section 7 · Secondary

Trigonometric Graphs & Transformations

Secondary / KCSE

Trig graphs are periodic waves. Understanding their transformations — amplitude, period, phase shift, and vertical shift — is essential for modelling real-world phenomena such as tides, sound waves, alternating current, and seasonal temperature patterns.

General Form: y = a sin(bx + c) + d

Parameter

Name

Effect

Formula

a

Amplitude

Vertical stretch; max = a, min = −a

|a|

b

Angular frequency

Horizontal compression

Period = 2π/b (or 360°/b)

c

Phase shift

Horizontal translation

Shift = −c/b (left if c > 0)

d

Vertical shift

Moves midline up or down

Midline: y = d

Key Features of y = sin x and y = cos x

Standard Sine and Cosine (a=1, b=1, c=0, d=0)

DomainAll real numbers (−∞, +∞)

Range[−1, 1]

Period360° (or 2π radians)

y = sin x zerosx = 0°, 180°, 360°, ... (multiples of 180°)

y = cos x zerosx = 90°, 270°, ... (odd multiples of 90°)

The Tangent Graph — Special Properties

y = tan x

Period180° (or π radians)

Range(−∞, +∞) — unbounded

Asymptotesx = 90°, 270°, 450°, ... (odd multiples of 90°)

Worked Examples — Graphs

Ex 16For y = 3 sin(2x − 60°) + 1, state the amplitude, period, phase shift, vertical shift, maximum, and minimum values.

a = 3Amplitude = 3. The graph is stretched by factor 3 vertically.

b = 2Period = 360°/2 = 180°.

c = 60°Phase shift = +60°/2 = 30° to the right (since the form is bx − c, shift is c/b positive).

d = 1Vertical shift: midline y = 1. Maximum = 1 + 3 = 4. Minimum = 1 − 3 = −2.

Amplitude=3 · Period=180° · Phase shift=30° right · Max=4 · Min=−2

Section 8 · Secondary / KCSE

Solving Trigonometric Equations

Secondary / KCSE

Solving trig equations requires combining CAST knowledge, inverse trig functions, and — where a restricted domain is given — counting all valid solutions within that interval. This is one of the most commonly examined skills in KCSE Paper 2.

General Solution Formulas

General Solutions (n ∈ ℤ)

sin x = kx = arcsin(k) + 360°n or x = 180° − arcsin(k) + 360°n

cos x = kx = ±arccos(k) + 360°n

tan x = kx = arctan(k) + 180°n

Rearrange to isolate the trig function: get sin θ = k, cos θ = k, or tan θ = k.
Find the principal value (reference angle) using the inverse trig function.
Apply CAST to find all solutions in the given interval (usually 0° to 360°, or 0 to 2π).
Handle double angles by first expanding the interval — if solving cos 2x in [0°, 360°], work with 2x in [0°, 720°].
Check all solutions by substituting back into the original equation.

Worked Examples — Solving Equations

Ex 17Solve 2 sin x − 1 = 0 for 0° ≤ x ≤ 360°.

Step 1Rearrange: sin x = 1/2.

Step 2Principal value: arcsin(1/2) = 30°. sin is positive in Q1 and Q2.

Step 3Q1: x = 30°. Q2: x = 180° − 30° = 150°. Verify: sin 30° = 1/2 ✔ sin 150° = 1/2 ✔

x = 30° or x = 150°

Ex 18Solve cos 2x = √3/2 for 0° ≤ x ≤ 360°.

Step 1Let u = 2x. Since x ∈ [0°, 360°], we have u ∈ [0°, 720°].

Step 2cos u = √3/2. Principal value: arccos(√3/2) = 30°. cos positive in Q1 and Q4.

Step 3u = 30°, 330°, 390° (= 30°+360°), 690° (= 330°+360°).

Step 4x = u/2: x = 15°, 165°, 195°, 345°.

x = 15°, 165°, 195°, 345°

Ex 19Solve 2 cos²x − cos x − 1 = 0 for 0° ≤ x ≤ 360°.

Step 1This is a quadratic in cos x. Let c = cos x: 2c² − c − 1 = 0.

Step 2Factorise: (2c + 1)(c − 1) = 0 → cos x = −1/2 or cos x = 1.

Step 3cos x = 1: x = 0° (or 360°). cos x = −1/2: reference = 60°, Q2 → 120°, Q3 → 240°.

Check2(1)² − 1 − 1 = 0 ✔ 2(1/4) − (−1/2) − 1 = 1/2 + 1/2 − 1 = 0 ✔

x = 0°, 120°, 240°, 360°

Section 9 · University

Inverse Trigonometric Functions

University

Inverse trig functions answer the question "given a ratio, what is the angle?" Because trig functions are periodic, the inverse is only a true function on a restricted domain. Their derivatives produce important integration formulas that appear in engineering, physics, and probability integrals.

Domains and Ranges (Principal Value Branches)

Function

Domain

Range (principal values)

Notes

arcsin x (sin⁻¹ x)

[−1, 1]

[−π/2, π/2]

Odd function

arccos x (cos⁻¹ x)

[−1, 1]

[0, π]

arcsin + arccos = π/2

arctan x (tan⁻¹ x)

(−∞, ∞)

(−π/2, π/2)

Odd; horizontal asymptotes at ±π/2

Derivatives of Inverse Trig Functions

Derivatives (essential for calculus)

d/dx (arcsin x)= 1 / √(1 − x²) |x| < 1

d/dx (arccos x)= −1 / √(1 − x²) |x| < 1

d/dx (arctan x)= 1 / (1 + x²)

d/dx (arccot x)= −1 / (1 + x²)

d/dx (arcsec x)= 1 / (|x| √(x² − 1)) |x| > 1

d/dx (arccsc x)= −1 / (|x| √(x² − 1)) |x| > 1

Standard Integrals Involving Inverse Trig

Integration Results

∫ 1/√(1−x²) dx= arcsin x + C

∫ 1/(1+x²) dx= arctan x + C

∫ 1/√(a²−x²) dx= arcsin(x/a) + C

∫ 1/(a²+x²) dx= (1/a) arctan(x/a) + C

Worked Examples — Inverse Trig

Ex 20Differentiate y = arctan(3x²).

Chain ruleLet u = 3x². Then y = arctan(u), so dy/dx = (1/(1+u²)) × du/dx.

Resultdu/dx = 6x. dy/dx = 6x / (1 + 9x⁴).

dy/dx = 6x / (1 + 9x⁴)

Ex 21Evaluate ∫₀¹ dx/(4 + x²).

Step 1Write as ∫₀¹ 1/(4 + x²) dx = ∫₀¹ 1/(2² + x²) dx.

Step 2Use ∫ 1/(a²+x²) dx = (1/a) arctan(x/a) + C with a = 2.

Step 3= [(1/2) arctan(x/2)]₀¹ = (1/2)[arctan(1/2) − arctan(0)] = (1/2)[arctan(0.5) − 0] ≈ (1/2)(0.4636) ≈ 0.2318.

∫₀¹ dx/(4 + x²) = (1/2) arctan(1/2) ≈ 0.2318

Section 10 · University

Complex Exponentials & Hyperbolic Functions

University

Euler's formula is one of the most celebrated results in mathematics, connecting trigonometry, exponential functions, and complex numbers in one elegant identity. Hyperbolic functions are the "complex twins" of circular trig functions and arise naturally in mechanics, electrical engineering, and special relativity.

Euler's Formula

Euler's Formula (Leonhard Euler, 1748)

e^(iθ) = cos θ + i sin θ

e^(−iθ) = cos θ − i sin θ

Adding and subtracting these two expressions yields the exponential forms of sine and cosine:

Trig Functions as Complex Exponentials

cos θ= (e^(iθ) + e^(−iθ)) / 2

sin θ= (e^(iθ) − e^(−iθ)) / (2i)

Setting θ = π in Euler's formula gives e^(iπ) + 1 = 0 — Euler's identity, connecting the five most important constants in mathematics: e, i, π, 1, and 0.

De Moivre's Theorem

(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)

De Moivre's theorem is used to compute powers and roots of complex numbers, and to derive multiple-angle formulas in terms of powers of sin and cos.

Hyperbolic Functions

Replacing the imaginary unit in trig gives real-valued analogues — the hyperbolic functions. They describe the shape of a hanging cable (catenary), and appear throughout advanced mechanics and relativity.

Hyperbolic Function Definitions

sinh x (hyperbolic sine)= (eˣ − e⁻ˣ) / 2

cosh x (hyperbolic cosine)= (eˣ + e⁻ˣ) / 2

tanh x= sinh x / cosh x = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ)

Fundamental identitycosh²x − sinh²x = 1

Hyperbolic Derivatives and Integrals

d/dx (sinh x)= cosh x

d/dx (cosh x)= sinh x

d/dx (tanh x)= sech²x

∫ sinh x dx= cosh x + C

∫ cosh x dx= sinh x + C

Worked Examples — Complex & Hyperbolic

Ex 22Use De Moivre's theorem to find all cube roots of 1 (z³ = 1).

Step 1Write 1 = e^(2πki) for k = 0, 1, 2. Then z = e^(2πki/3).

Step 2k=0: z = e⁰ = 1. k=1: z = e^(2πi/3) = cos(120°) + i sin(120°) = −1/2 + i√3/2.

Step 3k=2: z = e^(4πi/3) = cos(240°) + i sin(240°) = −1/2 − i√3/2.

1, −½ + i√3/2, −½ − i√3/2

Ex 23Show that cosh²x − sinh²x = 1.

LHScosh²x − sinh²x = [(eˣ+e⁻ˣ)/2]² − [(eˣ−e⁻ˣ)/2]².

Expand= [(e²ˣ + 2 + e⁻²ˣ) − (e²ˣ − 2 + e⁻²ˣ)] / 4 = 4/4 = 1 = RHS ✓

cosh²x − sinh²x = 1 (proved)

Section 11 · University

Fourier Series

University

Fourier series decompose any periodic function into an infinite sum of sines and cosines — transforming complicated waveforms into a spectrum of simple oscillations. This is the mathematical foundation of MP3 audio compression, JPEG image encoding, mobile phone signal processing, and the solution of the heat equation in engineering.

Fourier Series Representation

Any periodic function f(x) with period 2L satisfying the Dirichlet conditions (bounded, finite number of maxima/minima and discontinuities in one period) can be written:

Fourier Series

f(x) = a₀/2 + Σₙ₌₁^∞ [aₙ cos(nπx/L) + bₙ sin(nπx/L)]

Fourier Coefficients

Formulas for the Coefficients

a₀ (constant term)= (1/L) ∫₋ₗᴸ f(x) dx

aₙ (cosine coefficients)= (1/L) ∫₋ₗᴸ f(x) cos(nπx/L) dx

bₙ (sine coefficients)= (1/L) ∫₋ₗᴸ f(x) sin(nπx/L) dx

Odd and Even Functions — Simplifications

Even function: f(−x) = f(x)

All bₙ = 0. Only cosine terms remain. Example: f(x) = |x|, cos x, x².

Odd function: f(−x) = −f(x)

All aₙ = 0. Only sine terms remain. Example: f(x) = x, sin x, x³.

Parseval's Theorem

Parseval's Theorem (energy conservation)

(1/L) ∫₋ₗᴸ |f(x)|² dx = a₀²/2 + Σₙ₌₁^∞ (aₙ² + bₙ²)

Parseval's theorem states that the total "energy" (mean square value) of a signal equals the sum of the squared magnitudes of all its Fourier coefficients. It is used to evaluate infinite series — for example, by finding the Fourier series of f(x) = x on [−π, π] and applying Parseval's theorem, we can prove that Σ 1/n² = π²/6.

Worked Example — Fourier Series

Ex 24Find the Fourier series of the square wave: f(x) = 1 for 0 < x < π, f(x) = −1 for −π < x < 0, period 2π.

Step 1f is an odd function, so a₀ = 0 and aₙ = 0 for all n. Only bₙ terms survive.

Step 2bₙ = (1/π) ∫₋π^π f(x) sin(nx) dx = (2/π) ∫₀^π sin(nx) dx = (2/π)[−cos(nx)/n]₀^π.

Step 3= (2/πn)[1 − cos(nπ)] = (2/πn)[1 − (−1)ⁿ]. This is 4/(πn) when n is odd, and 0 when n is even.

Resultf(x) = (4/π)[sin x + sin(3x)/3 + sin(5x)/5 + ...] = (4/π) Σₙ odd sin(nx)/n.

f(x) = (4/π) Σₖ₌₁^∞ sin((2k−1)x) / (2k−1)

Applications of Fourier Series

Signal processing: decomposing audio into frequency components for MP3 encoding and equalisation.
Image compression: JPEG encoding uses the Discrete Cosine Transform (DCT), a cousin of Fourier analysis.
Solving PDEs: the heat equation ∂u/∂t = α ∂²u/∂x² and wave equation are solved using Fourier series separation of variables.
Electrical engineering: AC circuits with non-sinusoidal voltages are analysed by Fourier decomposition into harmonic components.
Quantum mechanics: a particle's state is expressed as a superposition (Fourier series) of energy eigenstates.

Quick Reference

Complete Trigonometry Formula Sheet

All Levels

Basic Ratios

sin θ = O/H cos θ = A/H tan θ = O/A

cosec θ = 1/sin θ sec θ = 1/cos θ

cot θ = 1/tan θ = cos θ/sin θ

Pythagorean Identities

sin²θ + cos²θ = 1

1 + tan²θ = sec²θ

1 + cot²θ = cosec²θ

Compound Angles

sin(A±B) = sinA cosB ± cosA sinB

cos(A±B) = cosA cosB ∓ sinA sinB

tan(A±B) = (tanA ± tanB)/(1 ∓ tanA tanB)

Double Angle

sin 2A = 2 sinA cosA

cos 2A = cos²A − sin²A = 2cos²A−1 = 1−2sin²A

tan 2A = 2 tanA / (1 − tan²A)

Sine & Cosine Rules

a/sinA = b/sinB = c/sinC = 2R

a² = b² + c² − 2bc cosA

Area = ½ab sinC = √[s(s−a)(s−b)(s−c)]

Inverse Trig Derivatives

d/dx arcsin x = 1/√(1−x²)

d/dx arccos x = −1/√(1−x²)

d/dx arctan x = 1/(1+x²)

Euler & Hyperbolic

e^(iθ) = cosθ + i sinθ

sinh x = (eˣ−e⁻ˣ)/2 cosh x = (eˣ+e⁻ˣ)/2

cosh²x − sinh²x = 1

Fourier Series

f(x) = a₀/2 + Σ[aₙcos(nπx/L) + bₙsin(nπx/L)]

aₙ = (1/L)∫₋ₗᴸ f(x)cos(nπx/L) dx

bₙ = (1/L)∫₋ₗᴸ f(x)sin(nπx/L) dx

Error Analysis

Common Errors & How to Fix Them

All Levels

✗ Wrong:sin(A + B) = sin A + sin B

✓ Correct:sin(A + B) = sin A cos B + cos A sin B

Trig ratios do not distribute over addition. This is the most common single-mark error in KCSE Paper 2.

✗ Wrong:sin²θ = sin(θ²)

✓ Correct:sin²θ = (sin θ)² — the whole ratio is squared

sin²θ means square the value of sin θ. It does not mean sin(θ²).

✗ Wrong:For cos 2θ: only one form known

✓ Correct:Know all three: cos²θ−sin²θ = 2cos²θ−1 = 1−2sin²θ

Choosing the wrong form for the context leads to an identity that cannot be simplified. Match the form to the surrounding expression.

✗ Wrong:Solving 2sin²x = 1 as sin x = 1/2 only

✓ Correct:sin x = ±1/√2 = ±√2/2 (two principal values!)

Taking square roots always gives ± both positive and negative values. Missing the negative root loses half the solutions.

✗ Wrong:Forgetting to expand the interval for cos 2x = k

✓ Correct:If x ∈ [0°, 360°], then 2x ∈ [0°, 720°] — double the range

Not doubling the interval causes you to miss 2 of the 4 solutions. Always substitute u = 2x and work with u first.

✗ Wrong:Saying tan 90° = ∞ or "equals a large number"

✓ Correct:tan 90° is undefined (the limit is +∞ from the left, −∞ from the right)

Infinity is not a number. tan has a vertical asymptote at 90°, 270°, etc. — the function has no value at these points.