CBSE Class 10 Mathematics Introduction to Trigonometry Notes
About This Chapter
Chapter 8 - Introduction to Trigonometry opens one of the most powerful and widely used branches of mathematics. Trigonometry means the measurement of triangles, and this chapter introduces students to the fundamental ratios that relate the angles and sides of a right-angled triangle. These ratios form the foundation of all trigonometric work studied in higher classes.
Trigonometry has immense practical importance. It is used by engineers to calculate forces and structures, by architects to design roofs and ramps, by astronomers to measure distances to stars, by navigators to determine direction at sea, and by physicists to analyse wave motion and oscillations. Understanding it at the Class 10 level gives you a strong foundation for all science and mathematics in higher education.
This chapter carries 8 to 10 marks in the CBSE board examination. It includes direct value-based questions, identity-proof questions, and multi-step simplification problems. Mastering the six trigonometric ratios, their standard values at key angles, and the two fundamental identities is essential for full marks.
What You Will Learn:
• The six trigonometric ratios (sin, cos, tan, cosec, sec, cot) and how to define them for a right-angled triangle
• Standard trigonometric values for angles 0, 30, 45, 60, and 90 degrees and how to remember them
• Reciprocal relationships between the six ratios and quotient relations between them
• The two fundamental Pythagorean identities and how to use them to prove expressions
• Solving problems involving trigonometric ratios, angles of elevation, and complementary angles
These notes include complete concept explanations, the full standard values table, all key identities, step-by-step solved examples, a formula summary table, and practice questions categorised by marks as per the CBSE pattern.
A detailed and printable PDF version of these notes is attached below for your convenience. Download, save, and use it for your revision!
1. Introduction and Basic Definitions
The word 'trigonometry' comes from the Greek words trigonon (triangle) and metron (measure). Trigonometry is the branch of mathematics that studies the relationship between the angles and sides of triangles, particularly right-angled triangles.
1.1 Right-Angled Triangle Terminology
In a right-angled triangle, the sides are named with respect to a reference acute angle (let us call it angle A or angle theta):
• Hypotenuse: The side opposite the right angle. It is always the longest side of a right-angled triangle.
• Opposite side (Perpendicular): The side directly opposite to the reference angle (angle A).
• Adjacent side (Base): The side that forms the reference angle along with the hypotenuse (not the opposite side).
>> Important Note The labelling of Opposite and Adjacent sides CHANGES depending on which angle you choose as the reference angle. The Hypotenuse always stays the same (opposite the right angle), but Opposite and Adjacent swap when you switch the reference angle. |
1.2 Definition of Trigonometric Ratios
For an acute angle A in a right-angled triangle, the six trigonometric ratios are defined as follows. These definitions must be memorised precisely.
Six Trigonometric Ratios (Definitions) sin A = Opposite / Hypotenuse = Perpendicular / Hypotenuse cos A = Adjacent / Hypotenuse = Base / Hypotenuse tan A = Opposite / Adjacent = Perpendicular / Base
cosec A = Hypotenuse / Opposite = 1 / sin A sec A = Hypotenuse / Adjacent = 1 / cos A cot A = Adjacent / Opposite = 1 / tan A |
>> Memory Trick - SOH CAH TOA SOH: Sin = Opposite / Hypotenuse CAH: Cos = Adjacent / Hypotenuse TOA: Tan = Opposite / Adjacent cosec, sec, and cot are simply the reciprocals of sin, cos, and tan respectively. |
2. Key Concepts and Relationships
2.1 Reciprocal Relations
The six trigonometric ratios occur in three reciprocal pairs. These are fundamental relationships used constantly in simplification and proof questions.
Reciprocal Relations sin A x cosec A = 1 => cosec A = 1 / sin A cos A x sec A = 1 => sec A = 1 / cos A tan A x cot A = 1 => cot A = 1 / tan A |
2.2 Quotient Relations
Two important quotient relations connect sin, cos, tan, and cot. These are derived directly from the definitions.
Quotient Relations tan A = sin A / cos A cot A = cos A / sin A |
2.3 Pythagorean Identities
These three identities follow from the Pythagoras theorem and are the most used identities in trigonometry. They allow you to convert between sin, cos, tan, sec, cosec, and cot in any expression.
Three Pythagorean Identities Identity 1: sin^2 A + cos^2 A = 1 sin^2 A = 1 - cos^2 A cos^2 A = 1 - sin^2 A
Identity 2: 1 + tan^2 A = sec^2 A tan^2 A = sec^2 A - 1 sec^2 A - tan^2 A = 1
Identity 3: 1 + cot^2 A = cosec^2 A cot^2 A = cosec^2 A - 1 cosec^2 A - cot^2 A = 1 |
>> Derivation of Identity 1 In right triangle with right angle at C: AC^2 + BC^2 = AB^2. Dividing both sides by AB^2: (AC/AB)^2 + (BC/AB)^2 = 1, which gives cos^2 A + sin^2 A = 1. Dividing by cos^2 A gives Identity 2; dividing by sin^2 A gives Identity 3. |
2.4 Complementary Angle Relations
Two angles are complementary if they add up to 90 degrees. In a right-angled triangle, the two acute angles are always complementary. The following relations hold for complementary angles:
Complementary Angle Relations sin(90 - A) = cos A cos(90 - A) = sin A tan(90 - A) = cot A cot(90 - A) = tan A sec(90 - A) = cosec A cosec(90 - A) = sec A
Memory Pattern: Each ratio of (90 - A) equals its CO-ratio of A. sin <-> cos, tan <-> cot, sec <-> cosec |
>> Key Insight on Complementary Angles The prefix 'co' in cosine, cotangent, and cosecant literally stands for 'complementary'. So cosine of an angle = sine of its complement. This is why these pairs exist. |
3. Standard Trigonometric Values
3.1 Values at Standard Angles
The trigonometric values at 0, 30, 45, 60, and 90 degrees must be memorised for CBSE board exams. These are used directly in calculation and proof questions.
Ratio | 0 deg | 30 deg | 45 deg | 60 deg | 90 deg | Key Pattern |
sin | 0 | 1/2 | 1/sqrt(2) | sqrt(3)/2 | 1 | 0, 1/2, 1/sqrt2, sqrt3/2, 1 |
cos | 1 | sqrt(3)/2 | 1/sqrt(2) | 1/2 | 0 | 1, sqrt3/2, 1/sqrt2, 1/2, 0 |
tan | 0 | 1/sqrt(3) | 1 | sqrt(3) | Not Defined | 0, 1/sqrt3, 1, sqrt3, ND |
cosec | Not Defined | 2 | sqrt(2) | 2/sqrt(3) | 1 | Reciprocal of sin |
sec | 1 | 2/sqrt(3) | sqrt(2) | 2 | Not Defined | Reciprocal of cos |
cot | Not Defined | sqrt(3) | 1 | 1/sqrt(3) | 0 | Reciprocal of tan |
3.2 Memory Trick for sin Values
>> The 0-1-2-3-4 Trick for sin (and reverse for cos) Write the numbers 0, 1, 2, 3, 4 under 0 deg, 30 deg, 45 deg, 60 deg, 90 deg. Divide each by 4, then take the square root: sin 0 = sqrt(0/4) = 0 sin 30 = sqrt(1/4) = 1/2 sin 45 = sqrt(2/4) = 1/sqrt(2) sin 60 = sqrt(3/4) = sqrt(3)/2 sin 90 = sqrt(4/4) = 1 For cos, use the reverse order: 4, 3, 2, 1, 0. |
3.3 Derived Values for tan, cosec, sec, cot
Once sin and cos values are known, all other values follow directly from the reciprocal and quotient relations. tan = sin/cos, cosec = 1/sin, sec = 1/cos, cot = 1/tan.
Angle | sin | cos | tan |
0 deg | 0 | 1 | 0 |
30 deg | 1/2 | sqrt(3)/2 | 1/sqrt(3) |
45 deg | 1/sqrt(2) | 1/sqrt(2) | 1 |
60 deg | sqrt(3)/2 | 1/2 | sqrt(3) |
90 deg | 1 | 0 | Not Defined |
4. Solved Examples
4.1 Finding Trigonometric Ratios from a Side
Example 4.1 - In right triangle ABC (right angle at B), AB = 3, BC = 4. Find all six trig ratios for angle A. Step 1: Find the hypotenuse AC using Pythagoras theorem. AC^2 = AB^2 + BC^2 = 9 + 16 = 25 => AC = 5
With reference to angle A: Opposite = BC = 4, Adjacent = AB = 3, Hypotenuse = AC = 5
sin A = 4/5 cos A = 3/5 tan A = 4/3 cosec A = 5/4 sec A = 5/3 cot A = 3/4 |
Example 4.2 - If sin A = 5/13, find cos A and tan A. sin A = Opposite/Hypotenuse = 5/13 Using Pythagoras: Adjacent^2 = 13^2 - 5^2 = 169 - 25 = 144 => Adjacent = 12
cos A = 12/13 tan A = 5/12 |
4.2 Using Standard Values
Example 4.3 - Evaluate: 2 tan^2(45) + cos^2(30) - sin^2(60) Substituting standard values: tan 45 = 1 => tan^2(45) = 1 cos 30 = sqrt(3)/2 => cos^2(30) = 3/4 sin 60 = sqrt(3)/2 => sin^2(60) = 3/4
= 2(1) + 3/4 - 3/4 = 2 + 0 = 2
Answer: 2 |
Example 4.4 - Evaluate: (sin 30 + tan 45 - cosec 60) / (sec 30 + cos 60 + cot 45) Numerator: sin 30 + tan 45 - cosec 60 = 1/2 + 1 - 2/sqrt(3) = 3/2 - 2/sqrt(3)
Denominator: sec 30 + cos 60 + cot 45 = 2/sqrt(3) + 1/2 + 1 = 2/sqrt(3) + 3/2
Rationalising and simplifying: Answer = (3*sqrt(3) - 4) / (4 + 3*sqrt(3)) Multiply numerator and denominator by (4 - 3*sqrt(3)) to rationalise if needed. |
4.3 Using Trigonometric Identities
Example 4.5 - Prove: (1 - sin^2 A) / cos A = cos A LHS = (1 - sin^2 A) / cos A
Using Identity 1: 1 - sin^2 A = cos^2 A
LHS = cos^2 A / cos A = cos A = RHS
Hence proved. |
Example 4.6 - Prove: (sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 A LHS = sin^2 A + 2*sin A*cosec A + cosec^2 A + cos^2 A + 2*cos A*sec A + sec^2 A = sin^2 A + cos^2 A + 2(1) + 2(1) + cosec^2 A + sec^2 A = 1 + 4 + (1 + cot^2 A) + (1 + tan^2 A) = 1 + 4 + 1 + cot^2 A + 1 + tan^2 A = 7 + tan^2 A + cot^2 A = RHS
Hence proved. |
4.4 Complementary Angle Problems
Example 4.7 - Evaluate: cos 38 / sin 52 + sin 15 * sec 75 cos 38 = cos(90 - 52) = sin 52 => cos 38 / sin 52 = sin 52 / sin 52 = 1
sec 75 = sec(90 - 15) = cosec 15 = 1/sin 15 => sin 15 sec 75 = sin 15 (1/sin 15) = 1
Answer: 1 + 1 = 2 |
Example 4.8 - If tan 2A = cot(A - 18), find the value of A. tan 2A = cot(A - 18) We know: cot(90 - x) = tan x => tan x = cot(90 - x)
So: cot(90 - 2A) = cot(A - 18) 90 - 2A = A - 18 108 = 3A A = 36 degrees
Answer: A = 36 degrees |
5. Applications and Special Cases
5.1 Real-Life Applications
• Engineering: Calculating forces, angles of inclination of ramps and bridges, structural load analysis.
• Astronomy: Measuring distances to stars and planets using parallax angles and trigonometric ratios.
• Navigation: Determining direction and position using angles and distances.
• Architecture: Designing roof slopes, staircases, and triangular support structures.
• Physics: Analysing projectile motion, wave functions, and oscillations using trigonometric functions.
5.2 Values at Boundary Angles
The values at 0 degrees and 90 degrees require special attention because some ratios become undefined. Understanding why helps avoid errors.
Ratio | At 0 degrees | At 90 degrees | Reason for ND |
tan | 0 | Not Defined | cos 90 = 0, division by zero |
cot | Not Defined | 0 | sin 0 = 0, division by zero |
sec | 1 | Not Defined | cos 90 = 0, division by zero |
cosec | Not Defined | 1 | sin 0 = 0, division by zero |
5.3 Range of Trigonometric Ratios
For acute angles (0 to 90 degrees), the trigonometric ratios have specific ranges. These help quickly verify whether a computed value is plausible.
Ratio | Minimum Value | Maximum Value | Note |
sin A | 0 (at 0 deg) | 1 (at 90 deg) | Always between 0 and 1 |
cos A | 0 (at 90 deg) | 1 (at 0 deg) | Always between 0 and 1 |
tan A | 0 (at 0 deg) | Not Defined | Can be any positive value |
cosec A | 1 (at 90 deg) | Not Defined | Always >= 1 |
sec A | 1 (at 0 deg) | Not Defined | Always >= 1 |
cot A | 0 (at 90 deg) | Not Defined | Can be any positive value |
6. Formula Summary Table
All key formulas and identities for Introduction to Trigonometry at a glance:
Category | Formula / Relation |
Basic Definitions | sin=Opp/Hyp, cos=Adj/Hyp, tan=Opp/Adj |
Reciprocals | cosec=1/sin, sec=1/cos, cot=1/tan |
Quotient Relations | tan=sin/cos, cot=cos/sin |
Identity 1 | sin^2 A + cos^2 A = 1 |
Identity 2 | 1 + tan^2 A = sec^2 A |
Identity 3 | 1 + cot^2 A = cosec^2 A |
Comp. Angles | sin(90-A)=cos A, tan(90-A)=cot A, sec(90-A)=cosec A |
sin values | 0, 1/2, 1/sqrt(2), sqrt(3)/2, 1 for 0,30,45,60,90 deg |
cos values | 1, sqrt(3)/2, 1/sqrt(2), 1/2, 0 for 0,30,45,60,90 deg |
tan values | 0, 1/sqrt(3), 1, sqrt(3), ND for 0,30,45,60,90 deg |
7. Key Theorems and Properties
>> Theorem 1 - Pythagoras Identity Proof In right triangle ABC with right angle at C: By Pythagoras: a^2 + b^2 = c^2 (where c = hypotenuse) Dividing by c^2: (a/c)^2 + (b/c)^2 = 1 => sin^2 A + cos^2 A = 1 Dividing by b^2: (a/b)^2 + 1 = (c/b)^2 => tan^2 A + 1 = sec^2 A Dividing by a^2: 1 + (b/a)^2 = (c/a)^2 => 1 + cot^2 A = cosec^2 A |
>> Theorem 2 - Complementary Angle Proof In right triangle ABC with right angle at B, angles A and C are complementary (A + C = 90). sin C = AB/AC = cos A and cos C = BC/AC = sin A. Since C = 90 - A: sin(90 - A) = cos A and cos(90 - A) = sin A. All other complementary relations follow similarly. |
>> Theorem 3 - Values at 45 degrees In an isosceles right triangle, both acute angles are 45 degrees and the two legs are equal. If each leg = 1, hypotenuse = sqrt(2). So sin 45 = 1/sqrt(2), cos 45 = 1/sqrt(2), tan 45 = 1. This is why sin 45 = cos 45. |
>> Theorem 4 - Values at 30 and 60 degrees In an equilateral triangle with side 2, each angle = 60 deg. The altitude bisects the base, creating a right triangle with hypotenuse 2, base 1, and height sqrt(3). The angles are 30 and 60 degrees. From this: sin 30 = 1/2, cos 30 = sqrt(3)/2, sin 60 = sqrt(3)/2, cos 60 = 1/2. |
7.1 Important Identity Forms
The Pythagorean identities can be rearranged into many forms. The following are frequently used in board exam proofs:
From Identity | Rearrangements (all equivalent) |
sin^2 + cos^2 = 1 | sin^2 = 1 - cos^2, cos^2 = 1 - sin^2, (1-sin)(1+sin) = cos^2 |
sec^2 - tan^2 = 1 | sec^2 = 1 + tan^2, tan^2 = sec^2 - 1, (sec-tan)(sec+tan) = 1 |
cosec^2 - cot^2 = 1 | cosec^2 = 1 + cot^2, cot^2 = cosec^2 - 1, (cosec-cot)(cosec+cot) = 1 |
8. Common Mistakes and Exam Tips
8.1 Frequent Errors to Avoid
1. Writing sin A + cos A = 1. This is WRONG. The correct identity is sin^2 A + cos^2 A = 1.
2. Thinking sin 2A = 2 sin A. This is incorrect. sin 2A is a different value entirely (not in Class 10 syllabus).
3. Confusing opposite and adjacent sides when the reference angle changes. Always re-label when the reference angle switches.
4. Forgetting that tan 90 and cosec 0 are undefined - do not write a value for these.
5. In identity proofs, trying to work on both LHS and RHS simultaneously. Always start from one side and reach the other.
6. Confusing sin^2 A with sin(A^2). sin^2 A means (sin A)^2, not sin of A-squared.
8.2 Exam Strategy Tips
• In proof questions, start from the more complex side (usually LHS) and simplify towards the simpler side.
• Convert everything to sin and cos when stuck - it makes identities easier to handle.
• For complementary angle problems, write out sin(90-x) = cos(x) explicitly before substituting.
• Learn all three Pythagorean identities AND their rearrangements - you will need them in board proofs.
• When evaluating with standard values, substitute values carefully and simplify step by step.
• For ratio-given problems (e.g., sin A = 5/13), always draw the right triangle and find the third side using Pythagoras first.
9. Practice Questions
Questions are organised by marks as per CBSE board examination pattern. Attempt all categories for complete preparation.
9.1 - 1 Mark Questions (MCQ / Very Short Answer)
1. If sin A = 3/5, find the value of cos A.
2. What is the value of tan 45 + sin 90?
3. Evaluate: sin 30 cos 60 + cos 30 sin 60.
4. If sec A = 2, find the value of tan A.
5. State the value of (sin^2 63 + sin^2 27).
6. What is sin(90 - A) / cos A equal to?
9.2 - 3 Mark Questions (Short Answer)
1. In right triangle ABC with right angle at B, if tan A = 1/sqrt(3), find the values of sin A, cos A, sin B, and cos B.
2. If 3 cot A = 4, find the value of (5 sin A - 3 cos A) / (5 sin A + 3 cos A) without using trigonometric tables.
3. Evaluate: (tan 65 / cot 25) + (sin 31 / cos 59) + cot 27 * tan 63 + (sin 31 / cos 59).
4. Prove the identity: (cos A - sin A + 1) / (cos A + sin A - 1) = cosec A + cot A.
5. If sin A + cos A = p and sec A + cosec A = q, show that q(p^2 - 1) = 2p.
9.3 - 5 Mark Questions (Long Answer)
1. Prove the identity: (sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 A.
2. Prove: (1 + sec A)/sec A = sin^2 A / (1 - cos A).
3. Prove: (tan A + sin A) / (tan A - sin A) = (sec A + 1) / (sec A - 1).
4. If cos A + sin A = sqrt(2) cos A, show that cos A - sin A = sqrt(2) sin A.
5. If sin A + cos A = sqrt(3), prove that tan A + cot A = 1.
CBSE Class 10 Syllabus |
CBSE Class 10 Notes |