Introduction to Trigonometry

Class 10 Maths - Chapter 8 | NCERT Formula Sheet

1. Trigonometric Ratios

In right triangle ABC, right-angled at B:
• sin ? = Opposite/Hypotenuse = AB/AC
• cos ? = Adjacent/Hypotenuse = BC/AC
• tan ? = Opposite/Adjacent = AB/BC
• cosec ? = 1/sin ? = AC/AB
• sec ? = 1/cos ? = AC/BC
• cot ? = 1/tan ? = BC/AB
Mnemonic: Some People Have Curly Black Hair Through Proper Brushing
Sin = P/H, Cos = B/H, Tan = P/B

2. Trigonometric Ratios of Specific Angles

Angle (?) sin ? cos ? tan ? cosec ? sec ? cot ?
0 1 0 Not defined 1 Not defined
30° 1/2 v3/2 1/v3 2 2/v3 v3
45° 1/v2 1/v2 1 v2 v2 1
60° v3/2 1/2 v3 2/v3 2 1/v3
90° 1 0 Not defined 1 Not defined 0

3. Trigonometric Identities

Pythagorean Identities:
1. sin˛ ? + cos˛ ? = 1
2. 1 + tan˛ ? = sec˛ ?
3. 1 + cot˛ ? = cosec˛ ?
Reciprocal Identities:
• cosec ? = 1/sin ?
• sec ? = 1/cos ?
• cot ? = 1/tan ?
Quotient Identities:
• tan ? = sin ?/cos ?
• cot ? = cos ?/sin ?

4. Complementary Angles

sin(90° - ?) = cos ?
cos(90° - ?) = sin ?
tan(90° - ?) = cot ?
cot(90° - ?) = tan ?
sec(90° - ?) = cosec ?
cosec(90° - ?) = sec ?
Note: 90° - ? and ? are complementary angles

5. Trigonometric Ratios of 0° and 90°

At 0°

sin 0° = 0
cos 0° = 1
tan 0° = 0

At 90°

sin 90° = 1
cos 90° = 0
tan 90° = 8 (undefined)

6. Important Formulas

• sin˛ ? = 1 - cos˛ ?
• cos˛ ? = 1 - sin˛ ?
• tan˛ ? = sec˛ ? - 1
• cot˛ ? = cosec˛ ? - 1
• sin ? = v(1 - cos˛ ?)
• cos ? = v(1 - sin˛ ?)

7. Trigonometric Ratios Table (Simplified)

Remember:
sin 0° = 0/4 = 0
sin 30° = 1/4 = 1/2
sin 45° = 2/4 = 1/v2
sin 60° = 3/4 = v3/2
sin 90° = 4/4 = 1

8. Solving Trigonometric Equations

  1. Use identities to simplify equation
  2. Express everything in terms of sin ? and cos ?
  3. Solve the algebraic equation
  4. Find angle using trigonometric table
  5. Check if solution is within valid range (0° to 90°)

9. Quick Reference

Basic Ratios

sin = P/H
cos = B/H
tan = P/B

Key Identity

sin˛? + cos˛? = 1

Complementary

sin(90°-?) = cos ?

10. Example (NCERT Style)

Problem: If sin A = 3/5, find cos A and tan A

Solution:
Using sin˛A + cos˛A = 1
(3/5)˛ + cos˛A = 1
9/25 + cos˛A = 1
cos˛A = 1 - 9/25 = 16/25
cos A = 4/5 (positive in first quadrant)
tan A = sin A/cos A = (3/5)/(4/5) = 3/4
Problem: Evaluate: 2 tan˛ 45° + cos˛ 30° - sin˛ 60°

Solution:
tan 45° = 1, cos 30° = v3/2, sin 60° = v3/2
= 2(1)˛ + (v3/2)˛ - (v3/2)˛
= 2 + 3/4 - 3/4 = 2

11. Special Values Memory Aid

For sin: 0°, 30°, 45°, 60°, 90°
Values: v0/2, v1/2, v2/2, v3/2, v4/2
Which gives: 0, 1/2, 1/v2, v3/2, 1
For cos: Reverse of sin
cos 0° = sin 90° = 1
cos 30° = sin 60° = v3/2
cos 45° = sin 45° = 1/v2
cos 60° = sin 30° = 1/2
cos 90° = sin 0° = 0