Trigonometric Functions

Angles: Definition and Measurement

**Angle** is the measure of rotation of a ray (half-line) about its initial point (vertex). The original ray is called the **initial side** and its final position after rotation is the **terminal side**.

**Positive angle**: Rotation in the anticlockwise direction

**Negative angle**: Rotation in the clockwise direction

The measure of an angle indicates the amount of rotation performed. The simplest unit considers one complete revolution as the basic unit.

Degree Measure

If a rotation from initial side to terminal side is 1/360th of a complete revolution, the angle is said to have a measure of **one degree (1°)**.

**Subdivisions of degree measure:**

1° = 60 minutes (60′)

1′ = 60 seconds (60″)

Therefore: 1° = 3600″

**Example conversions:** 40° 20′ = 40 + 20/60 degrees = 40 1/3 degrees = 121/3 degrees

**Worked Example 1:** Convert 40° 20′ into radian measure

40° 20′ = 121/3 degree

Using: Radian measure = (π/180) × Degree measure

Radian measure = (π/180) × (121/3) = 121π/540 radians

Radian Measure

**Radian** is a unit of angular measurement based on arc length. The angle subtended at the centre by an arc of length 1 unit in a **unit circle** (circle of radius 1 unit) is said to have a measure of **1 radian**.

**Fundamental relationship:** For a circle of radius r, if an arc of length l subtends an angle θ (in radians) at the centre, then:

**θ = l/r** or equivalently **l = rθ**

This is one of the most important formulas in trigonometry and applies to any circle, not just unit circles.

**Complete revolution:** The circumference of a unit circle is 2π. Therefore, one complete revolution subtends an angle of **2π radians** at the centre.

**Key principle:** Equal arcs of a circle subtend equal angles at the centre. Since an arc of length r subtends 1 radian, an arc of length l subtends l/r radians.

**Worked Example 2:** Find the radius of circle where central angle 60° intercepts arc of length 37.4 cm

Convert: θ = 60° = (60π/180) = π/3 radians

Using l = rθ: 37.4 = r × (π/3)

r = (37.4 × 3)/π = (37.4 × 3 × 7)/22 = 35.7 cm

**Worked Example 3:** Minute hand of watch is 1.5 cm long. Distance tip moves in 40 minutes?

In 60 minutes, hand completes 360° = 2π radians

In 40 minutes: θ = (40/60) × 2π = 4π/3 radians

Distance l = rθ = 1.5 × (4π/3) = 2π = 6.28 cm

Relation Between Radian and Real Numbers

Consider a unit circle with centre at origin O and point A on the circle. The tangent line PAQ at point A can be identified with the real number line where:

Point A represents zero

Positive real numbers are represented along AP (upward direction)

Negative real numbers are represented along AQ (downward direction)

When we wrap the positive real line along the circle in anticlockwise direction and the negative real line in clockwise direction, **every real number corresponds to a unique radian measure and vice versa**. This establishes a one-to-one correspondence between real numbers and radian measures.

**Notational Convention:**

When angle is written as θ°, its degree measure is θ

When angle is written as β (without degree symbol), its radian measure is β

The word "radian" is frequently omitted in notation

Relation Between Degree and Radian

Since a complete revolution has radian measure 2π and degree measure 360°:

**2π radians = 360°**

**π radians = 180°**

**Conversion formulas:**

**Radian measure = (π/180) × Degree measure**

**Degree measure = (180/π) × Radian measure**

**Common approximate values:**

1 radian = 180°/π ≈ 57° 16′ (approximately)

1° = π/180 radians ≈ 0.01746 radians

**Table of Standard Angles (Essential for memorization):**

| Degree | 30° | 45° | 60° | 90° | 180° | 270° | 360° |

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

| Radian | π/6 | π/4 | π/3 | π/2 | π | 3π/2 | 2π |

**Worked Example 4:** Convert 6 radians into degree measure

Using: Degree measure = (180/π) × Radian measure

Degree measure = (180/π) × 6 = 1080/π degrees

Using π ≈ 22/7: = (1080 × 7)/22 = 343 7/11 degrees

= 343° + (7/11) × 60′ = 343° 38′ + (2/11) × 60″

≈ 343° 38′ 11″

**Worked Example 5:** Two circles have arcs of equal length subtending angles 65° and 110° at their centres. Find ratio of radii.

θ₁ = 65° = (65π/180) = 13π/36 radians

θ₂ = 110° = (110π/180) = 22π/36 radians

Since l = r₁θ₁ = r₂θ₂ (equal arc lengths): r₁(13π/36) = r₂(22π/36)

r₁/r₂ = 22/13

Therefore **r₁ : r₂ = 22 : 13**

**Important conceptual points:**

Arc length, radius, and central angle are fundamentally connected through l = rθ

This relationship is independent of units as long as angle is in radians

The radian measure is the ratio of arc length to radius, making it dimensionless

Trigonometric Functions: Definition on Unit Circle

**Definition:** Consider a unit circle with centre at origin O. Let P(a, b) be any point on this circle such that the angle AOP = x radians (where A is the point (1, 0) and arc AP has length x).

**Primary definitions:**

**cos x = a** (x-coordinate of point P)

**sin x = b** (y-coordinate of point P)

Since P lies on the unit circle: a² + b² = 1

**Fundamental Identity (Most Important):**

**sin²x + cos²x = 1** for all real x

This identity is the foundation for all trigonometric identities.

**Quadrantal Angles:** Angles that are integer multiples of π/2 are called quadrantal angles. For these angles:

| Angle | 0 | π/2 | π | 3π/2 | 2π |

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

| sin | 0 | 1 | 0 | -1 | 0 |

| cos | 1 | 0 | -1 | 0 | 1 |

**Periodicity of Sine and Cosine:**

**sin(2nπ + x) = sin x** for all integers n

**cos(2nπ + x) = cos x** for all integers n

Sine and cosine have period 2π, meaning their values repeat every 2π radians (or 360°).

**Zeros of trigonometric functions:**

**sin x = 0** when x = nπ where n ∈ ℤ (any integer multiple of π)

**cos x = 0** when x = (2n + 1)π/2 where n ∈ ℤ (any odd multiple of π/2)

Other Trigonometric Functions

Based on sine and cosine, four additional trigonometric functions are defined:

**cosec x = 1/sin x** (defined when sin x ≠ 0, i.e., x ≠ nπ, n ∈ ℤ)

**sec x = 1/cos x** (defined when cos x ≠ 0, i.e., x ≠ (2n+1)π/2, n ∈ ℤ)

**tan x = sin x/cos x** (defined when cos x ≠ 0, i.e., x ≠ (2n+1)π/2, n ∈ ℤ)

**cot x = cos x/sin x** (defined when sin x ≠ 0, i.e., x ≠ nπ, n ∈ ℤ)

**Derived Identities:**

From sin²x + cos²x = 1, dividing both sides by cos²x:

**1 + tan²x = sec²x** (valid when cos x ≠ 0)

Similarly, dividing by sin²x:

**1 + cot²x = cosec²x** (valid when sin x ≠ 0)

**Table of Values for Standard Angles:**

| Angle | 0 | π/6 | π/4 | π/3 | π/2 |

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

| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |

| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |

| tan | 0 | 1/√3 | 1 | √3 | ∞ |

**Worked Example 6:** If cos x = -3/5 and x lies in third quadrant, find all trigonometric functions.

sec x = 1/cos x = -5/3

From sin²x + cos²x = 1: sin²x = 1 - 9/25 = 16/25

sin x = ±4/5. Since x is in third quadrant (both sin and cos negative): **sin x = -4/5**

**cosec x = -5/4**

tan x = sin x/cos x = (-4/5)/(-3/5) = **4/3**

**cot x = 3/4**

**Worked Example 7:** If cot x = -5/12 and x lies in second quadrant, find all trigonometric functions.

tan x = -12/5

sec²x = 1 + tan²x = 1 + 144/25 = 169/25

sec x = ±13/5. In second quadrant, sec x < 0: **sec x = -13/5**

**cos x = -5/13**

sin x = tan x × cos x = (-12/5) × (-5/13) = **12/13**

**cosec x = 13/12**

Sign of Trigonometric Functions in Different Quadrants

**Quadrant Analysis:**

Consider point P(a, b) on unit circle. If P is in:

**First quadrant (0 < x < π/2):** a > 0, b > 0 → sin x > 0, cos x > 0

**Second quadrant (π/2 < x < π):** a < 0, b > 0 → sin x > 0, cos x < 0

**Third quadrant (π < x < 3π/2):** a < 0, b < 0 → sin x < 0, cos x < 0

**Fourth quadrant (3π/2 < x < 2π):** a > 0, b < 0 → sin x < 0, cos x > 0

**Sign Table for All Trigonometric Functions:**

| Quadrant | I | II | III | IV |

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

| sin | + | + | - | - |

| cos | + | - | - | + |

| tan | + | - | + | - |

| cosec | + | + | - | - |

| sec | + | - | - | + |

| cot | + | - | + | - |

**Memory Aid: "All Students Take Calculus"** — In Q1 all functions are positive, in Q2 only sin and cosec are positive, in Q3 only tan and cot are positive, in Q4 only cos and sec are positive.

**Relationship: cos(-x) = cos x and sin(-x) = -sin x**

These show cosine is an **even function** while sine is an **odd function**.

Domain and Range of Trigonometric Functions

**Sine and Cosine:**

Domain: All real numbers ℝ

Range: [-1, 1], i.e., -1 ≤ sin x ≤ 1 and -1 ≤ cos x ≤ 1

**Cosecant and Secant:**

Domain of cosec x: {x ∈ ℝ : x ≠ nπ, n ∈ ℤ}

Range of cosec x: {y ∈ ℝ : y ≥ 1 or y ≤ -1}, written as ℝ - (-1, 1)

Domain of sec x: {x ∈ ℝ : x ≠ (2n+1)π/2, n ∈ ℤ}

Range of sec x: {y ∈ ℝ : y ≥ 1 or y ≤ -1}, written as ℝ - (-1, 1)

**Tangent and Cotangent:**

Domain of tan x: {x ∈ ℝ : x ≠ (2n+1)π/2, n ∈ ℤ}

Range of tan x: All real numbers ℝ

Domain of cot x: {x ∈ ℝ : x ≠ nπ, n ∈ ℤ}

Range of cot x: All real numbers ℝ

**Behaviour of Functions in Each Quadrant:**

| Quadrant | sin | cos | tan | cot | sec | cosec |

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

| I (0 to π/2) | 0→1 | 1→0 | 0→∞ | ∞→0 | 1→∞ | ∞→1 |

| II (π/2 to π) | 1→0 | 0→-1 | -∞→0 | 0→-∞ | -∞→-1 | 1→∞ |

| III (π to 3π/2) | 0→-1 | -1→0 | 0→∞ | ∞→0 | -∞→-1 | -1→-∞ |

| IV (3π/2 to 2π) | -1→0 | 0→1 | -∞→0 | 0→-∞ | 1→∞ | -∞→-1 |

**Periodicity:**

sin x and cos x have period **2π** (or 360°)

tan x and cot x have period **π** (or 180°)

sec x and cosec x have period **2π**

This means:

sin(x + 2π) = sin x and cos(x + 2π) = cos x

tan(x + π) = tan x and cot(x + π) = cot x

**Worked Example 8:** Find sin(31π/3)

Use periodicity: sin(31π/3) = sin(31π/3 - 2πn) where 31π/3 = 10π + π/3

So 31π/3 = 10π + π/3, and 10π = 5(2π)

Therefore: sin(31π/3) = sin(π/3) = **√3/2**

**Worked Example 9:** Find cos(-1710°)

Use periodicity with 360°: cos(-1710°) = cos(-1710° + 5×360°) = cos(-1710° + 1800°) = cos(90°) = **0**

Trigonometric Functions of Sum and Difference of Angles

**Theorem 1: Cosine of Sum**

**cos(x + y) = cos x cos y - sin x sin y**

**Proof:** Consider unit circle with centre O. Let:

∠P₄OP₁ = x (P₁ is at angle x)

∠P₁OP₂ = y (P₂ is at angle x+y)

∠P₄OP₃ = -y (P₃ is at angle -y)

P₄ is at (1, 0)

Coordinates: P₁(cos x, sin x), P₂(cos(x+y), sin(x+y)), P₃(cos y, -sin y)

Triangles P₁OP₃ and P₂OP₄ are congruent (same angles, same radii). Therefore P₁P₃ = P₂P₄.

Using distance formula:

P₁P₃² = (cos x - cos y)² + (sin x + sin y)²

= cos²x + cos²y - 2cos x cos y + sin²x + sin²y + 2sin x sin y

= 2 - 2(cos x cos y - sin x sin y)

P₂P₄² = (1 - cos(x+y))² + (0 - sin(x+y))²

= 1 - 2cos(x+y) + cos²(x+y) + sin²(x+y)

= 2 - 2cos(x+y)

Setting P₁P₃² = P₂P₄²:

2 - 2(cos x cos y - sin x sin y) = 2 - 2cos(x+y)

**Therefore: cos(x + y) = cos x cos y - sin x sin y** ✓

**Theorem 2: Cosine of Difference**

**cos(x - y) = cos x cos y + sin x sin y**

**Proof:** Replace y with -y in Theorem 1:

cos(x - y) = cos(x + (-y)) = cos x cos(-y) - sin x sin(-y) = cos x cos y - sin x(-sin y)

**Therefore: cos(x - y) = cos x cos y + sin x sin y** ✓

**Theorem 3: Complementary Angle Identities**

**cos(π/2 - x) = sin x**

**Proof:** Using cos(x - y) with x = π/2, y = x:

cos(π/2 - x) = cos(π/2)cos x + sin(π/2)sin x = 0·cos x + 1·sin x = sin x ✓

**sin(π/2 - x) = cos x**

**Proof:** Using identity cos(π/2 - x) = sin x, replace x with (π/2 - x):

cos(π/2 - (π/2 - x)) = sin(π/2 - x)

cos x = sin(π/2 - x)

**Therefore: sin(π/2 - x) = cos x** ✓

**Theorem 4: Sine of Sum**

**sin(x + y) = sin x cos y + cos x sin y**

**Proof:** Using complementary identity:

sin(x + y) = cos(π/2 - (x + y)) = cos((π/2 - x) - y)

= cos(π/2 - x)cos y + sin(π/2 - x)sin y

= sin x cos y + cos x sin y ✓

**Theorem 5: Sine of Difference**

**sin(x - y) = sin x cos y - cos x sin y**

**Proof:** Replace y with -y in Theorem 4:

sin(x - y) = sin(x + (-y)) = sin x cos(-y) + cos x sin(-y) = sin x cos y + cos x(-sin y)

**Therefore: sin(x - y) = sin x cos y - cos x sin y** ✓

**Summary of Sum and Difference Identities:**

1. cos(x + y) = cos x cos y - sin x sin y

2. cos(x - y) = cos x cos y + sin x sin y

3. sin(x + y) = sin x cos y + cos x sin y

4. sin(x - y) = sin x cos y - cos x sin y

**Working with these identities:**

Note the sign pattern: cos formula has minus, sin formula has plus for (x+y)

For difference (x-y): cos formula has plus, sin formula has minus

Verify: cos(x + y) has minus between the two products; sin(x + y) has plus

Each product involves sin of one angle and cos of the other

**Worked Example 10:** Find sin 75°

75° = 45° + 30°

sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

= (1/√2)(√3/2) + (1/√2)(1/2)

= (√3)/(2√2) + 1/(2√2)

= (√3 + 1)/(2√2)

= **(√6 + √2)/4**

**Worked Example 11:** Find cos 15°

15° = 45° - 30°

cos 15° = cos(45° - 30°) = cos 45° cos 30° + sin 45° sin 30°

= (1/√2)(√3/2) + (1/√2)(1/2)

= (√3 + 1)/(2√2)

= **(√6 + √2)/4**

Tangent and Cotangent of Sum and Difference

**Theorem 6: Tangent of Sum**

**tan(x + y) = (tan x + tan y)/(1 - tan x tan y)**

**Proof:**

tan(x + y) = sin(x + y)/cos(x + y)

= (sin x cos y + cos x sin y)/(cos x cos y - sin x sin y)

Dividing numerator and denominator by cos x cos y:

= [(sin x cos y + cos x sin y)/(cos x cos y)]/[(cos x cos y - sin x sin y)/(cos x cos y)]

= [(sin x/cos x)(cos y/cos y) + (cos x/cos x)(sin y/cos y)]/[(cos x/cos x)(cos y/cos y) - (sin x/cos x)(sin y/cos y)]

= **(tan x + tan y)/(1 - tan x tan y)** ✓

**Condition:** Formula valid when tan x, tan y, and tan(x+y) all exist and tan x tan y ≠ 1

**Theorem 7: Tangent of Difference**

**tan(x - y) = (tan x - tan y)/(1 + tan x tan y)**

**Proof:** Replace y with -y in Theorem 6:

tan(x - y) = (tan x + tan(-y))/(1 - tan x tan(-y))

= (tan x - tan y)/(1 + tan x tan y) ✓

**Cotangent Formulas (Similarly):**

**cot(x + y) = (cot x cot y - 1)/(cot y + cot x)**

**cot(x - y) = (cot x cot y + 1)/(cot y - cot x)**

**Worked Example 12:** Find tan 75°

75° = 45° + 30°

tan 75° = (tan 45° + tan 30°)/(1 - tan 45° tan 30°)

= (1 + 1/√3)/(1 - 1·(1/√3))

= ((√3 + 1)/√3)/((√3 - 1)/√3)

= (√3 + 1)/(√3 - 1)

Rationalize: = (√3 + 1)²/((√3 - 1)(√3 + 1))

= (3 + 2√3 + 1)/(3 - 1)

= (4 + 2√3)/2

= **2 + √3**

Multiple Angle Formulas

**Theorem 8: Double Angle Formulas**

By setting y = x in sum formulas:

**sin 2x = 2 sin x cos x**

**Proof:** sin(x + x) = sin x cos x + cos x sin x = 2 sin x cos x ✓

**cos 2x = cos² x - sin² x**

**Alternative forms (derived from sin² x + cos² x = 1):**

**cos 2x = 2 cos² x - 1**

**cos 2x = 1 - 2 sin² x**

**Proof for cos 2x = 2cos² x - 1:**

cos 2x = cos² x - sin² x = cos² x - (1 - cos² x) = 2cos² x - 1 ✓

**tan 2x = (2 tan x)/(1 - tan² x)**

**Proof:**

tan 2x = tan(x + x) = (tan x + tan x)/(1 - tan x tan x) = (2 tan x)/(1 - tan² x) ✓

**Important consequence formulas:**

**sin² x = (1 - cos 2x)/2**

**cos² x = (1 + cos 2x)/2**

**tan² x = (1 - cos 2x)/(1 + cos 2x)**

These are derived by rearranging cos 2x = 1 - 2sin² x and cos 2x = 2cos² x - 1.

**Worked Example 13:** If sin x = 3/5 and x is acute, find sin 2x and cos 2x

sin x = 3/5, so cos x = 4/5 (positive in first quadrant)

sin 2x = 2 sin x cos x = 2(3/5)(4/5) = **24/25**

cos 2x = cos² x - sin² x = 16/25 - 9/25 = **7/25**

Alternatively: cos 2x = 2cos² x - 1 = 2(16/25) - 1 = 32/25 - 25/25 = 7/25 ✓

Triple Angle Formulas

**sin 3x = 3 sin x - 4 sin³ x**

**Proof:**

sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x

= (2 sin x cos x) cos x + (1 - 2sin² x) sin x

= 2 sin x cos² x + sin x - 2 sin³ x

= 2 sin x(1 - sin² x) + sin x - 2 sin³ x

= 2 sin x - 2 sin³ x + sin x - 2 sin³ x

= **3 sin x - 4 sin³ x** ✓

**cos 3x = 4 cos³ x - 3 cos x**

**Proof:**

cos 3x = cos(2x + x) = cos 2x cos x - sin 2x sin x

= (2cos² x - 1) cos x - (2 sin x cos x) sin x

= 2 cos³ x - cos x - 2 sin² x cos x

= 2 cos³ x - cos x - 2(1 - cos² x) cos x

= 2 cos³ x - cos x - 2 cos x + 2 cos³ x

= **4 cos³ x - 3 cos x** ✓

**tan 3x = (3 tan x - tan³ x)/(1 - 3 tan² x)**

**Proof:** Using sin 3x/cos 3x and dividing by cos³ x ✓

Product to Sum Formulas

These convert products of trigonometric functions to sums (useful in integration and solving equations).

**sin x sin y = [cos(x - y) - cos(x + y)]/2**

**Proof:** We know:

cos(x - y) = cos x cos y + sin x sin y ... (1)

cos(x + y) = cos x cos y - sin x sin y ... (2)

Subtracting (2) from (1):

cos(x - y) - cos(x + y) = 2 sin x sin y

**Therefore: sin x sin y = [cos(x - y) - cos(x + y)]/2** ✓

**cos x cos y = [cos(x - y) + cos(x + y)]/2**

**Proof:** Adding equations (1) and (2):

cos(x - y) + cos(x + y) = 2 cos x cos y

**Therefore: cos x cos y = [cos(x - y) + cos(x + y)]/2** ✓

**sin x cos y = [sin(x + y) + sin(x - y)]/2**

**Proof:** We know:

sin(x + y) = sin x cos y + cos x sin y ... (3)

sin(x - y) = sin x cos y - cos x sin y ... (4)

Adding (3) and (4):

sin(x + y) + sin(x - y) = 2 sin x cos y

**Therefore: sin x cos y = [sin(x + y) + sin(x - y)]/2** ✓

Sum to Product Formulas

These convert sums/differences of trigonometric functions to products.

**sin A + sin B = 2 sin((A + B)/2) cos((A - B)/2)**

**Proof:** Let A = x + y and B = x - y, so x = (A + B)/2 and y = (A - B)/2

sin A + sin B = sin(x + y) + sin(x - y)

= (sin x cos y + cos x sin y) + (sin x cos y - cos x sin y)

= 2 sin x cos y

= **2 sin((A + B)/2) cos((A - B)/2)** ✓

**sin A - sin B = 2 cos((A + B)/2) sin((A - B)/2)**

**Proof:** Similarly, from sin(x + y) - sin(x - y) = 2 cos x sin y ✓

**cos A + cos B = 2 cos((A + B)/2) cos((A - B)/2)**

**Proof:** From cos(x + y) + cos(x - y) = 2 cos x cos y ✓

**cos A - cos B = -2 sin((A + B)/2) sin((A - B)/2)**

**Proof:** From cos(x - y) - cos(x + y) = 2 sin x sin y, which gives:

cos A - cos B = -2 sin((A + B)/2) sin((A - B)/2) ✓

**Worked Example 14:** Simplify sin 5x + sin 3x

Using sin A + sin B formula with A = 5x, B = 3x:

sin 5x + sin 3x = 2 sin((5x + 3x)/2) cos((5x - 3x)/2)

= 2 sin 4x cos x

= **2 sin 4x cos x**

**Worked Example 15:** Prove that cos 6x - cos 4x = -2 sin 5x sin x

Using cos A - cos B with A = 6x, B = 4x:

cos 6x - cos 4x = -2 sin