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Class 10 Mathematics Chapter 8 of 15

Chapter 8 — Introduction To Trigonometry

Overview

Chapter 'Introduction to Trigonometry' (NCERT Class 10 Mathematics) introduces trigonometric ratios for acute angles using right-angled triangles and triangle similarity. It defines sine, cosine and tangent (and their reciprocals cosecant, secant and cotangent), shows that these ratios are independent of the size of the triangle, and derives basic identities such as sin^2θ + cos^2θ = 1 and tanθ = sinθ/cosθ. The chapter gives exact values for standard angles (30°, 45°, 60°), explores complementary-angle relations, and applies trigonometry to solve practical problems on heights and distances using angles of elevation and depression. Importance: it builds a fundamental toolset for solving geometric measurement problems and is widely used in physics, engineering, navigation and surveying. Overall, students learn to define and compute trigonometric ratios, verify simple identities, and apply these concepts to find unknown sides, angles and real-life measures.

Learning Objectives

  • Define trigonometric ratios (sine, cosine, tangent) for an acute angle in a right triangle
  • State reciprocal trigonometric ratios (cosecant, secant, cotangent) and their relation to primary ratios
  • Explain the relation between trigonometric ratios of complementary angles (co-function identities)
  • Derive and apply the Pythagorean identity sin^2θ + cos^2θ = 1
  • Prove quotient identities such as tanθ = sinθ / cosθ and cotθ = cosθ / sinθ
  • Evaluate exact values of trigonometric ratios for standard angles 0°, 30°, 45°, 60° and 90°
  • Apply trigonometric ratios to find unknown sides and angles in right-angled triangles
  • Solve numerical problems on heights and distances using angles of elevation and depression

Topics in this chapter

12 topics · tap a topic title to jump straight to it.

🔢1

Introduction and Motivation

What is trigonometry? Trigonometry is the branch of mathematics that studies the relationships between the angles and sides of triangles, especially right-angled triangles. It introduces trigonometric ratios (sine, cosine, tangent and their reciprocals) that connect an acute angle of a right triangle to the lengths of its sides.

Why learn it? (Motivation)

  • Many practical problems ask for heights, distances or slopes which are difficult to measure directly. Trigonometry provides simple ratio-based methods to find these using an angle and one known length.
  • It is fundamental in fields such as engineering, architecture, surveying, navigation, physics and computer graphics — anywhere angles and distances interact.
  • Extending trigonometric ratios to all angles and using their graphs helps model periodic phenomena (waves, oscillations) and solve equations involving angles.

Basic idea (right-angled triangle)

Given a right triangle with acute angle θ, label the side opposite θ as the opposite side, the side adjacent to θ (but not the hypotenuse) as the adjacent side, and the longest side as the hypotenuse. Trigonometric ratios are defined as:

  • sin θ = (length of opposite) / (length of hypotenuse)
  • cos θ = (length of adjacent) / (length of hypotenuse)
  • tan θ = (length of opposite) / (length of adjacent)

How this helps in real life: If you can measure an angle of elevation (or depression) and one length (for example, the distance from the base), you can compute unknown heights using these ratios. This avoids climbing, digging or direct measuring.

Extension & graphs: The same ratios can be extended to any angle using the unit circle (point on circle of radius 1 has coordinates (cos θ, sin θ)). Graphs of sin, cos and tan versus angle show periodic behaviour and help visualize how these ratios change with angle.

Remember: The initial introduction focuses on acute angles in right triangles and practical applications (heights and distances). Later topics extend definitions to all angles, add identities and study graphs.

📌 Examples
  • Height of a tree: Measure distance d from the tree and angle of elevation θ of the top. Height h = d * tan θ (if ground is level and d is horizontal distance).
  • Building height from shadow: Measure shadow length s and angle of elevation θ of sun. Height = s * tan θ.
  • Ladder problem: If a ladder of length L leans against a wall making angle θ with ground, height reached on wall = L * sin θ, base distance = L * cos θ.
  • Across a river: From two points on the same bank separated by known distance, measure angles to a point on the opposite bank and use trigonometric ratios (or triangulation) to find the river width.
  • Slope/roof pitch: If roof slope makes angle θ with horizontal and horizontal run is r, rise = r * tan θ; used in architecture and construction.
🧮 Formulas
  1. sin θ = opposite / hypotenuse
  2. cos θ = adjacent / hypotenuse
  3. tan θ = opposite / adjacent = sin θ / cos θ
  4. cosec θ = 1 / sin θ, sec θ = 1 / cos θ, cot θ = 1 / tan θ
  5. Pythagorean identity: sin^2 θ + cos^2 θ = 1
  6. 1 + tan^2 θ = sec^2 θ
📊 Visual ideas
Right-triangle diagram: Draw a right triangle with angle θ, label opposite, adjacent and hypotenuse; write sin θ, cos θ, tan θ as ratios of these sides.
Unit circle: Circle radius 1 with an angle θ from positive x-axis. Mark point (cos θ, sin θ) and show that coordinates give cos and sin. Show right triangle inside the circle to link to triangle definitions.
y = sin x and y = cos x: Plot from 0° to 360° (or 0 to 2π). Show amplitude 1, period 360° (2π), key points at 0°, 30°, 45°, 60°, 90° etc., and labels for maxima/minima.
y = tan x: Plot from -90° to 90° showing vertical asymptote at ±90° and zero at 0°. Indicate repeating pattern with period 180°.
📐2

Right Triangle and Angle Notation

Right triangle: A triangle with one angle equal to 90°. If triangle ABC is right-angled at B (∠B = 90°), the side opposite the right angle (AC) is called the hypotenuse. The other two sides (AB and BC) are the legs.

Angle notation and labeling: In trigonometry we commonly denote an acute angle by a Greek letter such as θ (theta). For triangle ABC with ∠B = 90° and ∠A = θ, the sides are named relative to θ as:

  • Opposite side: the side opposite θ (here BC).
  • Adjacent side: the side next to θ but not the hypotenuse (here AB).
  • Hypotenuse: the side opposite the right angle (here AC).

Basic idea: Trigonometric ratios relate the angles of a right triangle to ratios of its sides. They depend only on the angle, not on the size of the triangle (similar triangles principle).

Complementary angles: In a right triangle the two acute angles are complementary: ∠A + ∠C = 90°. This gives relations such as sin(90° − θ) = cos θ and tan(90° − θ) = cot θ.

Degrees and radians: In Class 10 we usually use degrees (°). If needed, convert degrees to radians by multiplying degrees by π/180.

Use in problems: To find an unknown side, choose the appropriate ratio (sin, cos or tan) involving the known angle and known side, set up the equation, and solve. To find an angle from two known sides, take the inverse trigonometric function (e.g., θ = sin⁻¹(opposite/hypotenuse)).

📌 Examples
  • Example 1 (using a 5-12-13 triangle): In right triangle ABC, ∠B = 90° and sides are AB = 5, BC = 12, AC = 13. If ∠A = θ, then sin θ = opposite/hypotenuse = BC/AC = 12/13, cos θ = adjacent/hypotenuse = AB/AC = 5/13, tan θ = opposite/adjacent = 12/5.
  • Example 2 (ladder problem): A ladder 5 m long leans against a wall making an angle of 60° with the ground. The height reached on the wall is h = 5 × sin 60° = 5 × (√3/2) = 5√3/2 ≈ 4.33 m.
  • Example 3 (find angle): In a right triangle, the lengths of the legs are 7 and 24. Taking θ as the angle opposite the side of length 7, tan θ = 7/24, so θ = arctan(7/24) ≈ 16.26°.
  • Example 4 (angle from hypotenuse and opposite): If in a right triangle the opposite side to θ is 3 and hypotenuse is 5, then sin θ = 3/5 and θ = sin⁻¹(3/5) ≈ 36.87°.
🧮 Formulas
  1. Definitions (for angle θ in a right triangle): sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent
  2. Reciprocal ratios: cosec θ = 1/sin θ = hypotenuse/opposite, sec θ = 1/cos θ = hypotenuse/adjacent, cot θ = 1/tan θ = adjacent/opposite
  3. Pythagorean identity: sin^2 θ + cos^2 θ = 1
  4. Derived identities: 1 + tan^2 θ = sec^2 θ, 1 + cot^2 θ = cosec^2 θ
  5. Complementary-angle relations: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ
  6. Degree-radian conversion: radians = degrees × π/180 (useful if required)
📊 Visual ideas
Right-triangle schematic: draw triangle ABC with ∠B = 90°, label ∠A = θ, mark opposite, adjacent and hypotenuse — use this diagram to show how each ratio is formed.
Unit-circle relation: draw a unit circle and show a radius making angle θ with the positive x-axis. Project the point on the circle to x and y axes: x = cos θ, y = sin θ. Show the right triangle formed by the radius, the x-axis and the vertical drop.
Graph of sin θ and cos θ on 0° to 90°: plot sin θ rising from 0 to 1 and cos θ falling from 1 to 0; label a few standard angles (0°, 30°, 45°, 60°, 90°) with their values.
Graph of tan θ on 0° to 90°: plot tan θ increasing from 0 to ∞ with an asymptote at 90°; annotate that tan θ = sin θ / cos θ and why it blows up near 90°.
📐3

Trigonometric Ratios — Definitions

Definition (right‑triangle approach): In a right‑angled triangle, for an acute angle θ we call the side opposite to θ the opposite, the side next to θ (but not the hypotenuse) the adjacent, and the longest side the hypotenuse. The six trigonometric ratios of θ are defined as ratios of these sides:

  • sin θ = opposite / hypotenuse
  • cos θ = adjacent / hypotenuse
  • tan θ = opposite / adjacent
  • cosec θ = hypotenuse / opposite
  • sec θ = hypotenuse / adjacent
  • cot θ = adjacent / opposite

Memorise with SOH‑CAH‑TOA for sin, cos, tan. The last three are reciprocals of the first three: cosec = 1/sin, sec = 1/cos, cot = 1/tan (where defined).

Important identities: From Pythagoras (if hypotenuse = h, opposite = o, adjacent = a) we get o² + a² = h². Dividing by h² yields the fundamental identity: sin²θ + cos²θ = 1. Using sin and cos, tan and sec satisfy 1 + tan²θ = sec²θ. Also tan θ = sin θ / cos θ.

Complementary angles: In a right triangle the two acute angles sum to 90°. Hence sin θ = cos(90° − θ) and cos θ = sin(90° − θ); similarly tan θ = cot(90° − θ).

Range and domain (in right‑triangle context): For acute angles (0° < θ < 90°) the ratios are positive. To work with angles outside this range we extend the definitions using the unit circle, which also introduces periodicity and sign changes for sine, cosine and tangent.

How to use: Given any two sides of a right triangle, choose θ, identify opposite/adjacent/hypotenuse, and apply the appropriate ratio to find the third side or the angle (use inverse trig functions for angles).

📌 Examples
  • Given right triangle with hypotenuse 13 and opposite side 5 to angle θ, find sin θ, cos θ, tan θ. (sin θ = 5/13, cos θ = 12/13, tan θ = 5/12.)
  • A surveyor measures the angle of elevation to the top of a tower as 30°. If the surveyor is 40 m from the tower base, estimate the tower height (assume flat ground). Use tan 30° = 1/√3 so height ≈ 40 × tan 30° ≈ 40/√3 ≈ 23.09 m.
  • A ladder 10 m long leans against a wall and makes an angle of 60° with the ground. Find the height reached on the wall: height = 10 × sin 60° = 10 × (√3/2) = 5√3 ≈ 8.66 m.
  • If in a right triangle tan θ = 3/4, find sin θ and cos θ. Let opposite = 3k, adjacent = 4k => hypotenuse = 5k. So sin θ = 3/5, cos θ = 4/5.
🧮 Formulas
  1. sin θ = opposite / hypotenuse
  2. cos θ = adjacent / hypotenuse
  3. tan θ = opposite / adjacent
  4. cosec θ = 1 / sin θ = hypotenuse / opposite
  5. sec θ = 1 / cos θ = hypotenuse / adjacent
  6. cot θ = 1 / tan θ = adjacent / opposite
📊 Visual ideas
Right‑triangle diagram showing an angle θ with labelled opposite, adjacent, hypotenuse (use colour‑coded sides and arrow to indicate θ).
Unit circle diagram: show an angle θ from positive x‑axis, point (cos θ, sin θ) on circle, and projection lines to x and y axes to illustrate cos and sin as coordinates.
Plot of y = sin x over one or two periods (e.g., −2π to 2π): mark amplitude ±1, zeros at nπ, maxima at π/2 + 2nπ and minima at −π/2 + 2nπ.
Plot of y = cos x over −2π to 2π: mark amplitude ±1, zeros at π/2 + nπ, maximum at x = 0.
📐4

Ratios Depend Only on the Angle (Similarity)

Idea in one line: In right triangles, the ratios of corresponding sides (sine, cosine, tangent, etc.) depend only on the acute angle, not on the size of the triangle. This follows from similarity of right triangles.

Reason / Proof (using similarity):

  • Consider two right triangles △ABC and △PQR with right angles at B and Q respectively, and suppose ∠A = ∠P = θ (same acute angle).
  • Because both triangles have the same angles (θ, 90° and the remaining angle), they are similar by AA (angle–angle) similarity.
  • From similarity, corresponding sides are proportional. If in △ABC the side opposite θ is a, adjacent is b and hypotenuse is c, and in △PQR the corresponding sides are ka, kb, kc (scaled by the same factor k), then

a/c = (ka)/(kc),   a/b = (ka)/(kb),   b/c = (kb)/(kc)

  • Thus the ratios a/c, a/b, b/c are unchanged by scaling. Therefore, for a given acute angle θ the ratios opposite/hypotenuse, adjacent/hypotenuse and opposite/adjacent are fixed numbers that depend only on θ.

Consequence — Definitions of primary trigonometric ratios (0° < θ < 90°):

  • sin θ = (side opposite θ) / (hypotenuse)
  • cos θ = (side adjacent to θ) / (hypotenuse)
  • tan θ = (side opposite θ) / (side adjacent to θ) = sin θ / cos θ

Remarks: Because these ratios depend only on θ, any two right triangles sharing the same acute angle θ give the same values of sin θ, cos θ and tan θ, even if their absolute sizes differ. This is the foundational idea behind trigonometric tables and calculators.

Where this is valid: For right triangles and for acute angles (0° < θ < 90°). By extending definitions (unit circle) these ratios are defined for other angles too.

📌 Examples
  • Example 1 — Tree height (angle of elevation): If the angle of elevation to the top of a tree is 30° when you stand 20 m from the tree, height h of the tree satisfies tan 30° = h / 20. Since tan 30° = 1/√3, h = 20 × (1/√3) ≈ 11.55 m.
  • Example 2 — Ladder against a wall: A ladder makes an angle of 60° with the ground. The foot of the ladder is 2 m from the wall. The ladder length L is hypotenuse: cos 60° = adjacent/hypotenuse = 2 / L. Since cos 60° = 1/2, L = 2 / (1/2) = 4 m.
  • Example 3 — Scaled triangles show same ratios: Triangle A has sides 3–4–5 (right triangle). Triangle B is a scaled copy with sides 6–8–10. For the acute angle opposite side 3 (or 6) we get sin θ = opposite/hypotenuse = 3/5 = 0.6 and 6/10 = 0.6 — same value because triangles are similar.
  • Example 4 — Shadow and sun angle: A pole casts a 5 m shadow. If the sun's elevation angle is 40°, pole height = shadow × tan 40° = 5 × tan 40° ≈ 5 × 0.8391 ≈ 4.20 m.
🧮 Formulas
  1. sin θ = (opposite) / (hypotenuse)
  2. cos θ = (adjacent) / (hypotenuse)
  3. tan θ = (opposite) / (adjacent) = sin θ / cos θ
  4. cosec θ = 1 / sin θ, sec θ = 1 / cos θ, cot θ = 1 / tan θ
  5. Pythagorean identity: sin^2 θ + cos^2 θ = 1
  6. tan^2 θ + 1 = sec^2 θ, 1 + cot^2 θ = cosec^2 θ
📊 Visual ideas
Diagram showing two right triangles of different sizes but with the same acute angle θ. Label corresponding sides (opposite, adjacent, hypotenuse) to illustrate proportionality and constant ratios.
Unit circle diagram: show an angle θ in standard position, mark the point (cos θ, sin θ) on the circle to link trig ratios to coordinates (so sin = y, cos = x).
Graphs of y = sin θ and y = cos θ for θ from 0° to 90° (0 to π/2 radians) to show how values vary with angle; annotate values at common angles (0°, 30°, 45°, 60°, 90°).
Graph of y = tan θ from 0° up to just before 90° to show it increases and tends to infinity near 90° (vertical asymptote), with labeled values at common angles (30°, 45°, 60°).
🔢5

Reciprocal and Quotient Relations

Overview
In right-angled triangles and on the unit circle, trigonometric functions are related by simple ratio and reciprocal rules. Quotient relations express one trig function as a ratio of two others; reciprocal relations express a function as the multiplicative inverse of another.

Definitions (from sides of a right triangle)
For an angle θ with opposite side = Opp, adjacent side = Adj and hypotenuse = Hyp:

  • sin θ = Opp / Hyp
  • cos θ = Adj / Hyp
  • tan θ = Opp / Adj
  • cosec θ = Hyp / Opp
  • sec θ = Hyp / Adj
  • cot θ = Adj / Opp

Quotient relations (derived directly)
Using the basic ratios,

  • tan θ = (Opp/Hyp) / (Adj/Hyp) = sin θ / cos θ
  • cot θ = (Adj/Hyp) / (Opp/Hyp) = cos θ / sin θ

Reciprocal relations
From definitions:

  • cosec θ = 1 / sin θ
  • sec θ = 1 / cos θ
  • cot θ = 1 / tan θ

Important derived identities

  • sin2θ + cos2θ = 1
  • 1 + tan2θ = sec2θ (divide sin2θ + cos2θ = 1 by cos2θ)
  • 1 + cot2θ = cosec2θ (divide by sin2θ)

Domain notes / where undefined
Because of division by zero: sin θ = 0 ⇒ cosec θ undefined; cos θ = 0 ⇒ sec θ undefined; cos θ = 0 ⇒ tan θ undefined; sin θ = 0 ⇒ cot θ undefined. On the unit-circle form these correspond to specific angles (e.g. cos 90° = 0 ⇒ sec 90° undefined).

Common exact values (useful for Class 10)

θsin θcos θtan θsec θcosec θcot θ
0°0101undefinedundefined
30°1/2√3/21/√32/√32√3
45°√2/2√2/21√2√21
60°√3/21/2√322/√31/√3
90°10undefinedundefined10

How to use these relations
- To simplify expressions: replace tan θ by sin θ / cos θ or replace sec θ by 1 / cos θ.
- To solve equations: convert everything to sin and cos, use sin2θ + cos2θ = 1 and algebraic methods.
- To understand behavior: reciprocal functions (sec, cosec, cot) have vertical asymptotes where their base function (cos, sin, tan) is zero.

📌 Examples
  • Ramp gradient: If a ramp makes angle θ with the ground, its slope (rise/run) = tan θ. Knowing sin and cos from geometry, tan = sin/cos gives the slope directly.
  • Measuring a building height: From a distance d, if the angle of elevation to the top is θ, height = d · tan θ (quotient relation used).
  • Ladder problem: A ladder of length L leaning against a wall making angle θ with ground has height up the wall = L · sin θ and distance from wall = L · cos θ. Then cot θ = (distance from wall)/(height).
  • Understanding large values: Near 90° (cos θ → 0) sec θ = 1/cos θ becomes very large — useful to explain sudden increases in computed values when denominator is small.
🧮 Formulas
  1. sin θ = Opp / Hyp
  2. cos θ = Adj / Hyp
  3. tan θ = Opp / Adj
  4. tan θ = sin θ / cos θ
  5. cot θ = cos θ / sin θ
  6. sec θ = 1 / cos θ
📊 Visual ideas
Plot sin θ and cos θ on the same axes for x in [-2π, 2π]. Show amplitude ±1, period 2π, and mark zeros at integer multiples of π/2 as appropriate.
Plot tan θ on x in [-π, π] (or [-2π, 2π]) with vertical dashed asymptotes at x = π/2 + kπ. Note that tan has period π, no amplitude limit, and crosses origin.
Plot sec θ and cosec θ as reciprocals of cos and sin respectively. Show vertical asymptotes where cos or sin = 0 (sec: x = π/2 + kπ; cosec: x = kπ). Use the same x-range [-2π, 2π] and y-range approx [-5, 5] to visualize main features.
Plot cot θ with vertical asymptotes at x = kπ, zeros at x = π/2 + kπ, and period π. Emphasize cot = cos/sin and cot = 1/tan by comparing graphs of cot and tan (reflected/inverted shapes).
🔢6

Pythagorean Identities

What they are: Pythagorean identities are fundamental trigonometric identities that relate squares of sine, cosine, tangent and their reciprocals. They follow from the Pythagorean theorem applied to a right triangle and the unit circle.

Derivation from a right triangle / unit circle: Consider a right triangle with hypotenuse 1 (unit circle). If the angle at the origin is θ, the horizontal coordinate is cosθ and the vertical coordinate is sinθ. By the Pythagorean theorem: cos2θ + sin2θ = 1.

Deriving the other identities: Divide cos2θ + sin2θ = 1 by cos2θ (where cosθ ≠ 0) to get 1 + tan2θ = sec2θ. Divide by sin2θ (where sinθ ≠ 0) to get cot2θ + 1 = csc2θ. These are the three standard Pythagorean identities:

  • sin2θ + cos2θ = 1
  • 1 + tan2θ = sec2θ
  • 1 + cot2θ = csc2θ

Uses and remarks: These identities let you replace squares of one function by expressions in another (e.g., sin2θ = 1 − cos2θ), simplify trigonometric expressions and solve equations. Note domain restrictions: tan and sec undefined where cosθ = 0; cot and csc undefined where sinθ = 0.

📌 Examples
  • Numeric check: If sinθ = 3/5 for an acute angle, then cosθ = 4/5. Check: (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 25/25 = 1.
  • Ladder problem: A ladder of length 5 m leans making angle θ with the ground. Height reached = 5·sinθ and base distance = 5·cosθ. Then (height/5)^2 + (base/5)^2 = sin^2θ + cos^2θ = 1, consistent with Pythagoras.
  • Projectile components: If a velocity vector v has components vx = v·cosθ and vy = v·sinθ, then (vx/v)^2 + (vy/v)^2 = cos^2θ + sin^2θ = 1, so v = sqrt(vx^2 + vy^2).
  • Algebraic simplification: To simplify 1 − sin^2θ, use sin^2θ + cos^2θ = 1 to get 1 − sin^2θ = cos^2θ.
🧮 Formulas
  1. sin^2 θ + cos^2 θ = 1
  2. 1 + tan^2 θ = sec^2 θ (valid when cos θ ≠ 0)
  3. 1 + cot^2 θ = csc^2 θ (valid when sin θ ≠ 0)
  4. sin^2 θ = 1 − cos^2 θ
  5. cos^2 θ = 1 − sin^2 θ
  6. sec θ = 1 / cos θ, csc θ = 1 / sin θ, tan θ = sin θ / cos θ, cot θ = cos θ / sin θ
📊 Visual ideas
Unit circle diagram: show circle x^2 + y^2 = 1, point (cos θ, sin θ), right triangle from origin to point and projection on x-axis. Animate θ to show coordinates changing but x^2 + y^2 stays 1.
Plot sin^2 θ (in one color), cos^2 θ (another color) and their sum (a horizontal line at 1) over θ ∈ [−2π, 2π]. This visually confirms sin^2 θ + cos^2 θ = 1 at every θ.
Plot tan^2 θ and sec^2 θ over an interval that avoids cos θ = 0 (e.g., (−π/2 + 0.1, π/2 − 0.1)). Show vertical asymptotes where cos θ = 0 and plot 1 + tan^2 θ overlapping sec^2 θ to confirm equality.
Plot cot^2 θ and csc^2 θ similarly, marking asymptotes where sin θ = 0 and showing cot^2 θ + 1 = csc^2 θ.
📐7

Trigonometric Ratios of Complementary Angles

Definition: Two angles are complementary if their sum is 90° (or π/2 radians). In a right triangle the two acute angles are complementary. The trigonometric ratios of complementary angles are related by simple identities: each sine equals the cosine of its complement, each tangent equals the cotangent of its complement, and so on.

Core identities (degree & radian forms):

  • sin(90° − θ) = cos θ    (or sin(π/2 − x) = cos x)
  • cos(90° − θ) = sin θ    (or cos(π/2 − x) = sin x)
  • tan(90° − θ) = cot θ    (or tan(π/2 − x) = cot x)
  • cot(90° − θ) = tan θ    (or cot(π/2 − x) = tan x)
  • sec(90° − θ) = cosec θ    (or sec(π/2 − x) = csc x)
  • cosec(90° − θ) = sec θ    (or csc(π/2 − x) = sec x)

Proof (using a right triangle): Consider right triangle ABC with ∠C = 90°, ∠A = θ and ∠B = 90° − θ. Let the sides be: opposite A = a, opposite B = b, hypotenuse = h. Then by definition sin A = a/h and cos A = b/h. But sin B = (opposite B)/h = b/h = cos A. Thus sin(90° − θ) = cos θ. Similar reasoning gives the other identities.

Notes: These identities hold for angles measured in degrees or radians (replace 90° with π/2). They are useful to convert problems involving one trigonometric ratio into another and simplify calculations involving complementary angles.

📌 Examples
  • Ladder against a wall: If a ladder makes an angle θ with the ground, it makes angle 90° − θ with the wall. If the ladder length (hypotenuse) is L, the height reached on the wall = L·sin θ = L·cos(90° − θ). So height can be computed using either sin θ or cos(90° − θ).
  • Sun elevation and shadow: If the sun's elevation angle is 60°, the angle between the sun's ray and the vertical is 30° (complement). The tangent of the elevation (tan 60°) equals the cotangent of 30°, so shadow length calculations using tan or cot give matching results.
  • Surveying a tower: If angle of elevation to the top is 30°, the angle between the line of sight and the vertical is 60°. Therefore sin 30° (height/hypotenuse) equals cos 60° — the same numerical value can be used depending on which angle is referenced.
  • Numeric check: θ = 30°. sin 30° = 1/2. The complement is 60° and cos 60° = 1/2. Also tan 30° = 1/√3 and cot 60° = 1/√3.
🧮 Formulas
  1. sin(90° − θ) = cos θ (sin(π/2 − x) = cos x)
  2. cos(90° − θ) = sin θ (cos(π/2 − x) = sin x)
  3. tan(90° − θ) = cot θ (tan(π/2 − x) = cot x)
  4. cot(90° − θ) = tan θ (cot(π/2 − x) = tan x)
  5. sec(90° − θ) = cosec θ (sec(π/2 − x) = csc x)
  6. cosec(90° − θ) = sec θ (csc(π/2 − x) = sec x)
📊 Visual ideas
Plot sin(x) and cos(x) on the same axes for x from 0° to 90° (or 0 to π/2 radians). Observe that the curves are mirror images about x = 45° (π/4); sin(x) = cos(90° − x). Mark points x = 30° and x = 60° showing sin30 = cos60 = 1/2 and sin60 = cos30 = √3/2.
Plot tan(x) and cot(x) on (0°, 90°). Note tan(x) increases from 0 to ∞, cot(x) decreases from ∞ to 0, and tan(x) = cot(90° − x). Draw the vertical asymptote at 90° (π/2).
Unit circle sketch: Draw the unit circle and two radii making angles θ and 90° − θ with the x-axis. Show that the x-coordinate of one equals the y-coordinate of the other (cos θ = sin(90° − θ)), illustrating coordinate swap.
Suggested plotting tools/settings: Use Desmos or GeoGebra. In Desmos, enter f(x)=sin(x) and g(x)=cos(x) with x in radians, or convert degrees by using sin(x·π/180). Set x-range [0, π/2] and add points (π/6, 1/2) and (π/3, √3/2) to illustrate complementary pairs.
📐8

Standard Angle Values

What are standard angles? In Class 10 trigonometry, standard angles usually mean the commonly used angles 0°, 30°, 45°, 60° and 90°. For these angles we have exact, simple trigonometric values that are learned and used frequently. These values arise from two special right triangles (30°–60°–90° and 45°–45°–90°) and the unit circle.

Derivation (special triangles):

  • 45°–45°–90° triangle: take legs = 1, hypotenuse = √2. So sin45° = cos45° = 1/√2 = √2/2.
  • 30°–60°–90° triangle: take hypotenuse = 2, shorter side = 1 (opposite 30°), longer side = √3 (opposite 60°). So sin30° = 1/2, cos30° = √3/2, sin60° = √3/2, cos60° = 1/2, tan30° = 1/√3 = √3/3, tan60° = √3.

Unit circle view: On the unit circle (radius 1) the coordinates of the point at angle θ are (cos θ, sin θ). The standard-angle values are just these coordinates for θ = 0°, 30°, 45°, 60°, 90°. Using symmetry and signs of coordinates you can extend values to 90°–360°.

Signs by quadrant (useful when extending values): Quadrant I (0°–90°): sin, cos, tan all +. Quadrant II (90°–180°): sin +, cos −, tan −. Quadrant III (180°–270°): sin −, cos −, tan +. Quadrant IV (270°–360°): sin −, cos +, tan −. Use the reference angle (acute angle to the x-axis) to get magnitudes and apply sign per quadrant.

Table of standard values:

θsin θcos θtan θ
0°010
30°1/2√3/21/√3 = √3/3
45°√2/2√2/21
60°√3/21/2√3
90°10undefined

Practical tips: Memorise the table and derivations (special triangles + unit circle). To find values for other standard angles (like 120°, 135°, 150°) find the corresponding acute reference angle (e.g., 180°−θ) and apply quadrant sign rules.

📌 Examples
  • Ramp: A ramp of length 10 m is inclined at 30° to the horizontal. Vertical rise = 10 × sin30° = 10 × 1/2 = 5 m.
  • Roof slope: If a roof forms a 45° angle with the horizontal, a horizontal run of 4 m gives a rise = 4 × tan45° = 4 × 1 = 4 m.
  • Surveying height: From a point 20 m from the base of a tower the angle of elevation is 60°. Tower height ≈ 20 × tan60° = 20 × √3 ≈ 34.64 m.
  • Shadow and sun angle: If the sun’s elevation is 30° and a pole is 2 m tall, shadow length = 2 / tan30° = 2 × √3 ≈ 3.46 m.
🧮 Formulas
  1. sin^2θ + cos^2θ = 1 (Pythagorean identity)
  2. tan θ = sin θ / cos θ (where cos θ ≠ 0)
  3. cosec θ = 1 / sin θ, sec θ = 1 / cos θ, cot θ = 1 / tan θ
  4. Co-function identities: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ
  5. Reference angle method: value(θ) = ± value(reference angle), sign from quadrant
📊 Visual ideas
Unit circle diagram: circle of radius 1 with angles 0°, 30°, 45°, 60°, 90° marked; label coordinates (cos θ, sin θ) at each point.
Right-triangle diagrams: one 45°–45°–90° triangle (1,1,√2) and one 30°–60°–90° triangle (1,√3,2) showing opposite/adjacent/hypotenuse and the resulting sin/cos/tan values.
sin θ and cos θ graphs from 0° to 360°: plot both curves on same axes and mark values at standard angles (0°,30°,45°,60°,90°, etc.).
tan θ graph between −90° and 90° with asymptotes at 90° and −90°; mark tan values at 0°, 30°, 45°, 60° and show undefined behavior at 90°.
🔢9

Evaluation and Simplification of Expressions

What it is: Evaluation and simplification of trigonometric expressions means (a) finding numerical values of trigonometric expressions for given angles and (b) reducing complex trig expressions into simpler or standard forms using identities and algebraic techniques.

Key ideas:

  • Use standard values of trig ratios for special angles 0°, 30°, 45°, 60°, 90° and their multiples as building blocks.
  • Convert all functions to sines and cosines when simplifying (for example tan = sin/cos, cot = cos/sin) so common factors and identities can be applied.
  • Apply fundamental identities (reciprocal, Pythagorean, quotient, co-function, even–odd) to rewrite and reduce expressions.
  • Use algebraic manipulation: common denominator, factorization, multiplying by conjugate (rationalization), grouping, and cancellation, being careful about domain restrictions (no division by zero).

Typical steps for simplification or verification:

  1. Decide whether to evaluate numerically or to simplify symbolically.
  2. If numeric evaluation, reduce angle to a known reference angle and use sign rules from quadrants; substitute exact values where possible.
  3. If symbolic simplification, rewrite in terms of sin and cos, apply identities (eg sin2θ + cos2θ = 1), perform algebraic simplification, and simplify the result to a standard form.
  4. Always check domain restrictions (angles where denominators vanish) and, when verifying identities, show both sides are equal for all allowable angles.

Common pitfalls: forgetting sign changes in different quadrants, dividing by zero (cos θ = 0 makes tan undefined), and assuming identities hold without checking domain.

Degrees vs radians: In Class 10 you usually work in degrees (0°–360°). In later work radians are used; the algebraic simplification techniques are the same.

Connection to geometry and unit circle: Many simplifications follow from the unit circle definition: a point on unit circle at angle θ has coordinates (cos θ, sin θ). The Pythagorean identity sin2θ + cos2θ = 1 is the circle equation.

📌 Examples
  • Numeric evaluation: Evaluate sin 30&deg; + cos 60&deg;. Using standard values sin 30&deg; = 1/2 and cos 60&deg; = 1/2, so the sum = 1/2 + 1/2 = 1.
  • Simplification using identities: Simplify (1 - sin<sup>2</sup>&theta;)/cos &theta;. Use Pythagorean identity 1 - sin<sup>2</sup>&theta; = cos<sup>2</sup>&theta;. Then (cos<sup>2</sup>&theta;)/cos &theta; = cos &theta; (provided cos &theta; ≠ 0).
  • Verify identity: Show tan &theta; + cot &theta; = 1/(sin &theta; cos &theta;). Convert to sin and cos: tan &theta; + cot &theta; = sin/cos + cos/sin = (sin<sup>2</sup> + cos<sup>2</sup>)/(sin cos) = 1/(sin &theta; cos &theta;), using sin<sup>2</sup> + cos<sup>2</sup> = 1.
🧮 Formulas
  1. Reciprocal identities: sin &theta; = 1/cosec &theta;, cos &theta; = 1/sec &theta;, tan &theta; = 1/cot &theta;, cosec &theta; = 1/sin &theta;, sec &theta; = 1/cos &theta;, cot &theta; = 1/tan &theta;.
  2. Quotient identities: tan &theta; = sin &theta;/cos &theta;, cot &theta; = cos &theta;/sin &theta; (where denominators ≠ 0).
  3. Pythagorean identities: sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1; 1 + tan<sup>2</sup>&theta; = sec<sup>2</sup>&theta;; 1 + cot<sup>2</sup>&theta; = cosec<sup>2</sup>&theta;.
  4. Co-function (complement) identities: sin(90&deg; - &theta;) = cos &theta;, cos(90&deg; - &theta;) = sin &theta;, tan(90&deg; - &theta;) = cot &theta;.
  5. Even-odd identities: sin(-&theta;) = -sin &theta;, cos(-&theta;) = cos &theta;, tan(-&theta;) = -tan &theta;.
  6. Standard values: sin 0&deg; = 0, sin 30&deg; = 1/2, sin 45&deg; = √2/2, sin 60&deg; = √3/2, sin 90&deg; = 1; cos values mirror accordingly; tan 45&deg; = 1, tan 30&deg; = 1/√3, tan 60&deg; = √3.
📊 Visual ideas
Plot of y = sin x and y = cos x on the same axes from -360&deg; to 360&deg; (or -2π to 2π). Use different colors and mark key points at 0&deg;, 30&deg;, 45&deg;, 60&deg;, 90&deg; to connect numeric values and wave shape.
Plot of y = tan x from -90&deg; to 90&deg; (with vertical asymptotes at ±90&deg;), showing how tan grows and crosses zero at 0. Use this to illustrate where expressions with tan become undefined.
Unit circle diagram: draw the unit circle and a radius making angle &theta; with x-axis. Label coordinates (cos &theta;, sin &theta;) and show how sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta; = 1 follows from the circle equation.
Graphical verification of identities: overlay y = sin<sup>2</sup> x + cos<sup>2</sup> x and y = 1 to show they coincide; plot y = 1 + tan<sup>2</sup> x and y = sec<sup>2</sup> x to show equality (away from asymptotes).
📐10

Solving Right Triangle Problems

What it means: Solving right triangle problems means finding unknown sides or angles of a triangle that has one 90° angle using trigonometric ratios and the Pythagorean theorem.

Key ideas:

  • Label the triangle: hypotenuse (the side opposite 90°), opposite (side opposite the angle of interest), adjacent (other non‑hypotenuse side).
  • Use trigonometric ratios for an acute angle θ: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.
  • Use the Pythagorean theorem when two sides are known: hypotenuse^2 = opposite^2 + adjacent^2.
  • If you need the angle, use inverse trigonometric functions: θ = sin^(-1)(...), cos^(-1)(...), or tan^(-1)(...).
  • Use special triangles to shortcut calculations: 45°-45°-90° (legs equal, hypotenuse = leg·√2) and 30°-60°-90° (sides 1 : √3 : 2).

Step-by-step method:

  1. Draw the figure and mark the right angle and the known values.
  2. Identify the angle of interest θ and label opposite, adjacent, hypotenuse.
  3. Choose the appropriate relation (sin, cos, tan or Pythagoras).
  4. Set up the equation and solve for the unknown (algebra or inverse trig as required).
  5. Check units and reasonableness (e.g., sin and cos values must lie between 0 and 1).

Common pitfalls: mixing up opposite/adjacent, using hypotenuse instead of adjacent in tan/cos, forgetting to use inverse trig when solving for angles, and not checking calculator mode (degrees vs radians).

📌 Examples
  • Example 1 — Ladder problem: A 5 m ladder leans against a wall making an angle of 60° with the ground. How high does the ladder reach on the wall? Solution: height = hypotenuse × sin60 = 5 × (√3/2) = (5√3)/2 ≈ 4.33 m.
  • Example 2 — Angle of elevation: From a point on the ground a man sees the top of a tower at an angle of elevation 30°. If he is 50 m from the tower base, find the tower height. Solution: height = distance × tan30 = 50 × (1/√3) ≈ 50 × 0.5774 = 28.87 m.
  • Example 3 — Find an angle from sides: In a right triangle the opposite side to angle θ is 7 cm and the adjacent side is 24 cm. Find θ. Solution: tan θ = 7/24 ⇒ θ = arctan(7/24) ≈ 16.26° (use calculator in degree mode).
  • Example 4 — Using Pythagoras: A right triangle has legs 6 cm and 8 cm. Find the hypotenuse and the angles. Solution: hypotenuse = √(6^2 + 8^2) = √100 = 10 cm. For angle opposite 6 cm: sin θ = 6/10 = 3/5 ⇒ θ = sin^(-1)(0.6) ≈ 36.87°; the other angle ≈ 53.13°.
  • Example 5 — Special triangle: Hypotenuse of a 45°-45°-90° triangle is 10 cm. Find each leg. Solution: leg = hypotenuse/√2 = 10/√2 = 5√2 ≈ 7.07 cm.
🧮 Formulas
  1. sin θ = opposite / hypotenuse
  2. cos θ = adjacent / hypotenuse
  3. tan θ = opposite / adjacent
  4. cot θ = adjacent / opposite = 1 / tan θ
  5. sec θ = hypotenuse / adjacent = 1 / cos θ
  6. cosec θ = hypotenuse / opposite = 1 / sin θ
📊 Visual ideas
Draw a right triangle and label sides (hypotenuse, opposite, adjacent) and angle θ; annotate the chosen ratio (sin/cos/tan) to solve for the unknown — this is the basic diagram used in all problems.
Plot sin θ and cos θ for θ from 0° to 90° (sin increases from 0 to 1; cos decreases from 1 to 0). Use vertical axis for ratio value and horizontal axis for angle. Mark typical values at 30°, 45°, 60°.
Plot tan θ from 0° to 80° (it increases and tends to infinity near 90°). Show why tan becomes large as θ approaches 90° (adjacent → 0).
Interactive sketch suggestion: a movable angle θ in a right triangle where sliders change hypotenuse length; display live values of opposite, adjacent, sin θ, cos θ, tan θ and numeric results to help visual reasoning.
🔢11

Proofs and Reasoning

What this topic means
"Proofs and Reasoning" in Introduction to Trigonometry explains why trigonometric ratios (sin, cos, tan, etc.) are well defined for an acute angle and how basic trig identities follow from geometry (right triangles and the Pythagorean theorem). It teaches how to set up logical steps to derive identities and to solve height-and-distance problems.

1. Why trig ratios are well defined (proof using similar triangles)
Consider two right triangles that have one acute angle equal to θ. By AA (angle–angle) similarity the triangles are similar, so corresponding sides are proportional. If in one triangle the sides adjacent, opposite and hypotenuse are a, b, c and in the other are ka, kb, kc (k>0), then opposite/hypotenuse = a/c = (ka)/(kc). Hence sinθ = opposite/hypotenuse is independent of triangle size and depends only on θ. The same argument holds for cosθ and tanθ. This justifies the definitions:

  • sinθ = opposite/hypotenuse
  • cosθ = adjacent/hypotenuse
  • tanθ = opposite/adjacent

2. Complementary angle relations (geometric reasoning)
In a right triangle, the two acute angles add to 90°. If one acute angle is θ, the other is 90° − θ. The side opposite θ is the side adjacent to 90° − θ, and vice versa. Therefore sinθ = cos(90° − θ) and cosθ = sin(90° − θ). Similarly tanθ = cot(90° − θ).

3. Pythagorean identity (proof from Pythagoras)
In a right triangle with sides opposite = p, adjacent = q, hypotenuse = r, Pythagoras gives p^2 + q^2 = r^2. Divide both sides by r^2 to get (p/r)^2 + (q/r)^2 = 1, i.e. sin^2θ + cos^2θ = 1. This is the fundamental identity from which other identities follow (see below).

4. Deriving quotient and reciprocal identities
From definitions: tanθ = (opposite/hypotenuse)/(adjacent/hypotenuse) = sinθ/cosθ. Reciprocals: cosecθ = 1/sinθ, secθ = 1/cosθ, cotθ = 1/tanθ.

5. Derived identities by algebraic reasoning
Divide sin^2θ + cos^2θ = 1 by cos^2θ to obtain tan^2θ + 1 = sec^2θ. Divide by sin^2θ to obtain 1 + cot^2θ = cosec^2θ. These are obtained by algebraic manipulation based on the basic Pythagorean identity.

6. Reasoning in problem solving (strategy)
When solving heights and distances: (a) draw a clear right triangle, (b) label the angle and known sides, (c) choose the trigonometric ratio matching the angle and known side, (d) solve algebraically, and (e) check units and reasonableness.

Why the proofs matter
They guarantee that trig ratios are consistent (do not depend on triangle size) and provide tools to derive many identities used in algebraic simplification and applied problems.

📌 Examples
  • Example 1 — Prove trig ratios depend only on the angle: Take two right triangles with equal acute angle θ. By similarity corresponding sides are proportional. If sides in first triangle are (opposite=a, adjacent=b, hypotenuse=c) and in second are (ka,kb,kc), then a/c = (ka)/(kc). So sinθ = a/c is same for both triangles. Similar reasoning for cosθ and tanθ.
  • Example 2 — Prove sin^2θ + cos^2θ = 1: In right triangle with opposite = p, adjacent = q, hypotenuse = r, Pythagoras gives p^2 + q^2 = r^2. Divide by r^2: (p/r)^2 + (q/r)^2 = 1, i.e. sin^2θ + cos^2θ = 1.
  • Example 3 — Height of a tree: A person 30 m from the base of a tree measures angle of elevation to top = 35°. Height h of tree satisfies tan35° = h/30. So h = 30·tan35°. (Compute tan35° from table or calculator to get numerical height.)
🧮 Formulas
  1. sinθ = opposite / hypotenuse
  2. cosθ = adjacent / hypotenuse
  3. tanθ = opposite / adjacent = sinθ / cosθ
  4. cosecθ = 1 / sinθ, secθ = 1 / cosθ, cotθ = 1 / tanθ
  5. sin^2θ + cos^2θ = 1
  6. 1 + tan^2θ = sec^2θ
📊 Visual ideas
Right-triangle diagram: label one acute angle θ, opposite, adjacent, hypotenuse. Show two similar triangles of different sizes to illustrate proportionality.
Unit circle: show angle θ in first quadrant, mark point (cosθ, sinθ) on circle to visualize sin and cos as coordinates and to see sin^2θ + cos^2θ = 1.
Complementary-angle sketch: same right triangle with both acute angles labeled θ and 90°−θ, with arrows showing that opposite side of θ is adjacent for 90°−θ (supports sinθ = cos(90°−θ)).
Graphs of y = sin x and y = cos x from 0° to 90°: show that sin increases from 0 to 1 while cos decreases from 1 to 0; the two curves are symmetric about x = 45° in [0°,90°].
🔢12

Exercises and Examples

In Class 10 Trigonometry the Exercises and Examples focus on using right triangles to define trigonometric ratios (sine, cosine, tangent and their reciprocals) for acute angles and applying identities to solve numerical and word problems. Typical tasks include calculating trig ratios from given side lengths, finding missing sides or angles, using Pythagorean identity to obtain other ratios, and solving height & distance problems using angle of elevation/depression.

Core ideas:

  • Given a right triangle with an acute angle θ, define: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.
  • Use reciprocal relations: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.
  • Use Pythagorean identity: sin^2θ + cos^2θ = 1 to find one ratio from another.
  • Apply complementary-angle relations: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ, etc.
  • Translate real situations (height of a pole, ladder, slope) into right triangles and apply trig ratios to compute unknown lengths or angles.

Strategy for exercises:

  • Sketch the figure and mark the right angle, the angle θ, hypotenuse, opposite and adjacent sides.
  • Choose the appropriate trig ratio that relates known and unknown quantities.
  • If only one trig ratio is given, use reciprocal/quadratic identities to find others (careful about signs; for Class 10 angles are acute so positive).
  • Check units and round final numerical answers to suitable accuracy.
📌 Examples
  • Example 1 — Basic ratios from a 3–4–5 triangle: In a right triangle with sides 3 (opp), 4 (adj), 5 (hyp) for angle θ opposite the side 3: sin θ = 3/5, cos θ = 4/5, tan θ = 3/4; cosec θ = 5/3, sec θ = 5/4, cot θ = 4/3.
  • Example 2 — Find other ratios: Given sin θ = 3/5 and θ acute. Use cos θ = √(1 − sin^2θ) = √(1 − 9/25) = √(16/25) = 4/5. Then tan θ = sin θ / cos θ = (3/5)/(4/5) = 3/4. Reciprocals: cosec = 5/3, sec = 5/4, cot = 4/3.
  • Example 3 — Height from angle of elevation: From a point 50 m from the base of a tower the angle of elevation is 30°. Height h = 50 · tan 30° = 50 · (1/√3) ≈ 28.87 m.
  • Example 4 — Ladder problem: A 10 m ladder reaches a height of 8 m on a wall. For angle θ between ladder and ground: sin θ = 8/10 = 4/5, so θ = arcsin(0.8) ≈ 53.13°. The foot of the ladder is 6 m from the wall because adjacent = 10·cos θ = 10·(3/5) = 6.
  • Example 5 — Complementary angles: If θ = 30°, then sin(60°) = cos 30°. Numerically sin 60° = √3/2 and cos 30° = √3/2, showing sin(90° − θ) = cos θ.
🧮 Formulas
  1. Definitions (right triangle): sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent
  2. Reciprocal identities: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ
  3. Quotient identities: tan θ = sin θ / cos θ, cot θ = cos θ / sin θ
  4. Pythagorean identity: sin^2θ + cos^2θ = 1. Consequences: 1 + tan^2θ = sec^2θ and 1 + cot^2θ = cosec^2θ
  5. Complementary-angle relations: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ
  6. Special angle values: sin 0° = 0, cos 0° = 1; sin 30° = 1/2, cos 30° = √3/2; sin 45° = cos 45° = √2/2; sin 60° = √3/2, cos 60° = 1/2; sin 90° = 1, cos 90° = 0
📊 Visual ideas
Graph of y = sin x (0° to 360°): plot one period with amplitude 1. Mark key points: (0°,0), (90°,1), (180°,0), (270°,-1), (360°,0). Label x-axis in degrees and y-axis from -1 to 1.
Graph of y = cos x (0° to 360°): plot one period with amplitude 1. Mark (0°,1), (90°,0), (180°,-1), (270°,0), (360°,1). Compare phase shift with sine curve.
Graph of y = tan x (−90° to 90° and 90° to 270°): show basic shape with vertical asymptotes at 90° and 270°, zeros at 0° and 180°, period 180°. Scale y-axis (tan grows large near asymptotes).
Suggested right-triangle diagrams: For each problem draw a right triangle with θ at one acute vertex, clearly label hypotenuse, adjacent and opposite sides and include numerical values when given — this is often enough to set up the appropriate ratio.

Key Concepts

Angle
Figure formed by two rays with a common endpoint; measured in degrees (°).
Degree
Unit for measuring angles; full circle = 360°.
Right triangle
Triangle with one 90° angle; trigonometric ratios are defined for its acute angles.
Hypotenuse
Side opposite the right angle in a right triangle; longest side.
Opposite side
Side opposite a given angle in a triangle.
Adjacent side
Side next to a given angle (but not the hypotenuse) in a right triangle.
Trigonometric ratio
Ratio of two sides of a right triangle taken with respect to an acute angle.
Sine (sin)
Ratio of the opposite side to the hypotenuse for an acute angle in a right triangle: sin θ = opposite/hypotenuse.
Cosine (cos)
Ratio of the adjacent side to the hypotenuse: cos θ = adjacent/hypotenuse.
Tangent (tan)
Ratio of the opposite side to the adjacent side: tan θ = opposite/adjacent.
Cosecant (csc)
Reciprocal of sine: csc θ = 1/sin θ (defined when sin θ ≠ 0).
Secant (sec)
Reciprocal of cosine: sec θ = 1/cos θ (defined when cos θ ≠ 0).
Cotangent (cot)
Reciprocal of tangent: cot θ = 1/tan θ = adjacent/opposite (defined when tan θ ≠ 0).
Reciprocal identities
Relationships expressing sec, csc, cot as reciprocals of cos, sin, tan respectively.
Quotient identities
Identities expressing tan and cot as ratios of sin and cos: tan θ = sin θ / cos θ, cot θ = cos θ / sin θ.
Pythagorean identity
Fundamental identity: sin^2θ + cos^2θ = 1 for all θ.
Co-function (complementary) identities
Relations between trig functions of complementary angles: sin(90°−θ)=cos θ, tan(90°−θ)=cot θ, etc.
Standard angles
Common angles with known exact trig values: 0°, 30°, 45°, 60°, 90°.
Trigonometric table
Table listing values of trig ratios for standard angles to use in problems.
Angle of elevation and depression
Angle of elevation: angle between horizontal and line of sight upward. Angle of depression: between horizontal and line of sight downward.

End-of-Chapter Trial Paper & Test Questions

Topic-wise questions to test your understanding of every concept in this chapter.

  1. Define sin θ, cos θ and tan θ for an acute angle θ in a right triangle. / समकोण त्रिभुज में न्यून कोण θ के लिए sin θ, cos θ और tan θ को परिभाषित कीजिए।
    Show answer

    For acute angle θ: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, and tan θ = opposite/adjacent. / न्यून कोण θ के लिए: sin θ = सम्मुख/कर्ण, cos θ = आसन्न/कर्ण, और tan θ = सम्मुख/आसन्न।

  2. Why do trigonometric ratios depend only on the angle and not on the size of the triangle? / त्रिकोणमितीय अनुपात केवल कोण पर ही क्यों निर्भर करते हैं, त्रिभुज के आकार पर नहीं?
    Show answer

    Two right triangles with the same acute angle θ are similar by AA, so their corresponding sides are proportional; when the sides scale by a factor k, ratios like (ka)/(kc) = a/c remain unchanged, so the ratios depend only on θ. / समान न्यून कोण θ वाले दो समकोण त्रिभुज AA से समरूप होते हैं, अतः उनकी संगत भुजाएँ समानुपाती होती हैं; जब भुजाएँ गुणांक k से बढ़ती हैं, तो (ka)/(kc) = a/c जैसे अनुपात अपरिवर्तित रहते हैं, इसलिए अनुपात केवल θ पर निर्भर करते हैं।

  3. Derive the identity sin²θ + cos²θ = 1 from a right triangle. / समकोण त्रिभुज से सर्वसमिका sin²θ + cos²θ = 1 व्युत्पन्न कीजिए।
    Show answer

    For a right triangle with opposite p, adjacent q and hypotenuse r, Pythagoras gives p² + q² = r²; dividing by r² yields (p/r)² + (q/r)² = 1, i.e., sin²θ + cos²θ = 1. / सम्मुख p, आसन्न q और कर्ण r वाले समकोण त्रिभुज के लिए पाइथागोरस से p² + q² = r²; r² से भाग देने पर (p/r)² + (q/r)² = 1, अर्थात् sin²θ + cos²θ = 1।

  4. If tan θ = 3/4, find sin θ and cos θ for acute θ. / यदि tan θ = 3/4 है, तो न्यून θ के लिए sin θ और cos θ ज्ञात कीजिए।
    Show answer

    Let opposite = 3k and adjacent = 4k, so hypotenuse = √(9k² + 16k²) = 5k; hence sin θ = 3/5 and cos θ = 4/5. / माना सम्मुख = 3k और आसन्न = 4k, तब कर्ण = √(9k² + 16k²) = 5k; अतः sin θ = 3/5 और cos θ = 4/5।

  5. State the complementary-angle relations between sine, cosine and tangent. / ज्या, कोज्या और स्पर्शज्या के बीच पूरक कोण संबंध लिखिए।
    Show answer

    Since the two acute angles of a right triangle sum to 90°: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, and tan(90° − θ) = cot θ. / चूँकि समकोण त्रिभुज के दोनों न्यून कोणों का योग 90° होता है: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, और tan(90° − θ) = cot θ।

  6. Evaluate sin 30° + cos 60° + tan 45° using standard values. / मानक मानों का प्रयोग करके sin 30° + cos 60° + tan 45° का मान ज्ञात कीजिए।
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    sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1, so the sum = 1/2 + 1/2 + 1 = 2. / sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1, अतः योग = 1/2 + 1/2 + 1 = 2।

  7. From a point 50 m from the base of a tower the angle of elevation of its top is 30°. Find the tower's height. / एक मीनार के आधार से 50 m दूर एक बिंदु से उसके शीर्ष का उन्नयन कोण 30° है। मीनार की ऊँचाई ज्ञात कीजिए।
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    Using tan 30° = height/distance, height = 50 × tan 30° = 50 × (1/√3) = 50/√3 ≈ 28.87 m. / tan 30° = ऊँचाई/दूरी का प्रयोग करते हुए, ऊँचाई = 50 × tan 30° = 50 × (1/√3) = 50/√3 ≈ 28.87 m।

  8. Prove that tan θ + cot θ = 1/(sin θ cos θ). / सिद्ध कीजिए कि tan θ + cot θ = 1/(sin θ cos θ)।
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    Converting to sine and cosine: tan θ + cot θ = sin θ/cos θ + cos θ/sin θ = (sin²θ + cos²θ)/(sin θ cos θ) = 1/(sin θ cos θ), using sin²θ + cos²θ = 1. / ज्या और कोज्या में बदलने पर: tan θ + cot θ = sin θ/cos θ + cos θ/sin θ = (sin²θ + cos²θ)/(sin θ cos θ) = 1/(sin θ cos θ), जहाँ sin²θ + cos²θ = 1 का प्रयोग किया गया है।

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