TRIGONOMETRY – UNIT 5 (COMPLETE NOTES)

5.0 INTRODUCTION TO TRIGONOMETRY

Trigonometry is the branch of mathematics that deals with the relationship between angles and sides of a triangle, especially a right-angled triangle.

In a right-angled triangle:

One angle is 90°

The side opposite 90° is called the Hypotenuse (h)

The side opposite the reference angle is called Perpendicular (p)

The remaining side is called Base (b)

5.1 MEASUREMENT OF ANGLES

Definition of Angle

An angle is the amount of rotation of a line about a fixed point.

Anti–clockwise rotation → Positive angle (+θ)

Clockwise rotation → Negative angle (−θ)

SYSTEMS OF MEASURING ANGLES

There are three systems of measuring angles:

Sexagesimal System (Degree measure)

Centesimal System (Grade measure)

Circular System (Radian measure)

(i) SEXAGESIMAL SYSTEM (DEGREE MEASURE)

Right angle = 90°

1° = 60 minutes (')

1' = 60 seconds (")

Important Relations

1 right angle = 90° = 5400' = 324000"

Conversion Formulas

Degrees to minutes: 1° = 60'

Degrees to seconds: 1° = 3600"

(ii) CENTESIMAL SYSTEM (GRADE MEASURE)

Right angle = 100g

1g = 100 minutes (')

1' = 100 seconds (")

Important Relations

1 right angle = 100g = 10000' = 1000000"

(iii) CIRCULAR SYSTEM (RADIAN MEASURE)

Definition of Radian

An angle subtended at the center of a circle by an arc whose length is equal to the radius is called 1 radian.

Important Facts

180° = π radian

90° = π/2 radian

1 radian = 180°/π

Formula

θ (in radian) = Arc length (l) / Radius (r)

RELATION BETWEEN DEGREE, GRADE AND RADIAN

90° = 100g = π/2 radian

1° = (10/9) g

1° = π/180 radian

1g = (9/10)°

1g = π/200 radian

IMPORTANT FIGURES (DIAGRAMS)

FIGURE 1: RIGHT ANGLED TRIANGLE (BASIC TRIGONOMETRY)

Angle at B = θ

Hypotenuse (h): Side opposite 90°

Perpendicular (p): Side opposite angle θ

Base (b): Side adjacent to angle θ

This figure is used to define all six trigonometric ratios.

FIGURE 2: UNIT CIRCLE FOR RADIAN MEASURE

Center = O

Radius = r

Arc length = l

Angle in radian: θ = l / r

Used to understand radian measure and conversion.

FIGURE 3: STANDARD ANGLES (30°, 45°, 60°)

(a) 30°–60°–90° Triangle

Ratios:

sin 30° = 1/2

cos 30° = √3/2

tan 30° = 1/√3

(b) 45°–45°–90° Triangle

Ratios:

sin 45° = 1/√2

cos 45° = 1/√2

tan 45° = 1

FIGURE 4: QUADRANTS AND SIGNS OF TRIGONOMETRIC RATIOS

Mnemonic: ASTC Rule

Quadrant I: All ratios +ve

Quadrant II: Sin +ve

Quadrant III: Tan +ve

Quadrant IV: Cos +ve

5.2 TRIGONOMETRIC RATIOS

Trigonometric Ratios in Right Angled Triangle

Let θ be the reference angle:

sin θ = p / h

cos θ = b / h

tan θ = p / b

cot θ = b / p

sec θ = h / b

cosec θ = h / p

RECIPROCAL RELATIONS

sin θ × cosec θ = 1

cos θ × sec θ = 1

tan θ × cot θ = 1

QUOTIENT RELATIONS

tan θ = sin θ / cos θ

cot θ = cos θ / sin θ

PYTHAGOREAN IDENTITIES

sin²θ + cos²θ = 1

sec²θ − tan²θ = 1

cosec²θ − cot²θ = 1

5.3 TRIGONOMETRIC RATIOS OF STANDARD ANGLES

Values of Trigonometric Ratios

5.4 IMPORTANT THEOREMS IN TRIGONOMETRY

THEOREM 1: PYTHAGOREAN IDENTITY

Statement:

In a right-angled triangle,

sin²θ + cos²θ = 1

Proof:

Let the perpendicular = p, base = b and hypotenuse = h

sin θ = p/h and cos θ = b/h

sin²θ + cos²θ

= p²/h² + b²/h²

= (p² + b²)/h²

Using Pythagoras theorem:

p² + b² = h²

So,

= h²/h²

= 1

Hence proved.

THEOREM 2: SECOND PYTHAGOREAN IDENTITY

Statement:

1 + tan²θ = sec²θ

Proof:

We know:

tan θ = p/b and sec θ = h/b

LHS = 1 + tan²θ

= 1 + p²/b²

= (b² + p²)/b²

Using Pythagoras theorem:

= h²/b²

= (h/b)²

= sec²θ

Hence proved.

THEOREM 3: THIRD PYTHAGOREAN IDENTITY

Statement:

1 + cot²θ = cosec²θ

Proof:

We know:

cot θ = b/p and cosec θ = h/p

LHS = 1 + cot²θ

= 1 + b²/p²

= (p² + b²)/p²

Using Pythagoras theorem:

= h²/p²

= (h/p)²

= cosec²θ

Hence proved.

5.5 TRIGONOMETRIC IDENTITIES

A trigonometric identity is an equation that is true for all values of the angle.

Common Identities

1 + tan²θ = sec²θ

1 + cot²θ = cosec²θ

1 − sin²θ = cos²θ

1 − cos²θ = sin²θ

5.5 TRIGONOMETRIC RATIOS OF ANY ANGLE

Signs of Trigonometric Ratios in Quadrants

IMPORTANT QUESTIONS WITH SOLUTIONS

Q1. Convert 30° into radian

Solution:

30° = 30 × π/180

= π/6 radian

Q2. If sin θ = 3/5, find cos θ and tan θ

Step 1: sin θ = p/h = 3/5

Let p = 3, h = 5

Step 2: Using Pythagoras theorem

b² = h² − p²

b² = 25 − 9 = 16

b = 4

Step 3: Find required ratios

cos θ = b/h = 4/5

tan θ = p/b = 3/4

Q3. Find the value of (sin 60° + cos 30°) tan 30°

Step 1: Write values

sin 60° = √3/2

cos 30° = √3/2

tan 30° = 1/√3

Step 2: Substitute

= (√3/2 + √3/2) × 1/√3

= √3 × 1/√3

= 1

Q4. Prove: sin²θ + cos²θ = 1

Proof:

In right triangle:

sin θ = p/h, cos θ = b/h

sin²θ + cos²θ = p²/h² + b²/h²

= (p² + b²)/h²

= h²/h² = 1

Hence proved.

Q5. Find the third angle of a right-angled triangle if one angle is 30° (in grade)

Step 1: Convert 30° into grade

30° = 30 × 10/9 = 33.33g

Step 2: Sum of angles in triangle = 200g

Third angle = 200g − (100g + 33.33g)

= 66.67g

MORE IMPORTANT QUESTIONS WITH SOLUTIONS

Q6. If cos θ = 5/13, find all other trigonometric ratios

Step 1:
cos θ = b/h = 5/13

Let b = 5, h = 13

Step 2: Find perpendicular using Pythagoras theorem

p² = h² − b²

p² = 169 − 25 = 144

p = 12

Step 3: Find remaining ratios

sin θ = p/h = 12/13

tan θ = p/b = 12/5

cot θ = b/p = 5/12

sec θ = h/b = 13/5

cosec θ = h/p = 13/12

Q7. Evaluate: (sin 30° cos 60° + cos 30° sin 60°)

Step 1: Use identity

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

Step 2: Substitute values

= sin (30° + 60°)

= sin 90°

= 1

Q8. Prove: (1 + tan²θ) / (1 + sec²θ) = sin²θ

Proof:

LHS = (1 + tan²θ) / (1 + sec²θ)

Using identities:

1 + tan²θ = sec²θ

1 + sec²θ = 1 + sec²θ

LHS = sec²θ / (1 + sec²θ)

Divide numerator and denominator by sec²θ

= 1 / (1/sec²θ + 1)

= 1 / (cos²θ + 1)

= sin²θ

Hence proved.

Q9. Find the value of tan 45° + 2 sin 30°

Step 1: Write values

tan 45° = 1

sin 30° = 1/2

Step 2: Substitute

= 1 + 2 × 1/2

= 1 + 1

= 2

Q10. Convert π/3 radian into degree

Step 1: Use relation

180° = π radian

Step 2: Convert

π/3 = (π/3) × 180°/π

= 60°

Q11. If tan θ = 1, find the value of sin θ + cos θ

Step 1:
tan θ = 1 = p/b

So p = b

Step 2: Assume p = b = 1

h = √(1² + 1²) = √2

Step 3: Find sin θ + cos θ

= p/h + b/h

= 1/√2 + 1/√2

= √2

Q12. Prove: cosec θ (1 − sin²θ) = sin θ sec θ

Proof:

LHS = cosec θ (1 − sin²θ)

= (1/sin θ) × cos²θ

= cos²θ / sin θ

= cos θ × (cos θ / sin θ)

= cos θ × cot θ

= sin θ sec θ

Hence proved.

Q13.One angle of a right-angled triangle is 27°. Find its third angle in grade measure.(Fig 5)

Solution In a right-angled triangle ABC, ∠B = 100ᵍ,

∠C = 27° = 27 × (10/9)ᵍ = 30ᵍ, ∠A = ?

Now, we know that ∠A + ∠B + ∠C = 200ᵍ [ ∵ Sum of interior angles of triangle in grade ]

or, ∠A + 100ᵍ + 30ᵍ = 200ᵍ

or, ∠A = 200ᵍ – 130ᵍ = 70ᵍ

∴ The required third angle of the triangle is 70ᵍ.

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