Grade 10 Notes of TRIGONOMETRY|| Optional Mathematics

1. Introduction

Trigonometry deals with angles and ratios in right triangles and extends to all types of angles and circular functions.

Key Uses:

1. Solve triangles

2. Calculate heights & distances

3. Applications in geometry and real-life problems

2. Trigonometric Ratios

For a right triangle with angle θ:

1.Reciprocal Identities:

1.csc θ = 1 / sin θ
2.sec θ = 1 / cos θ
3.cot θ = 1 / tan θ

2.Pythagorean Identities:

1.sin² θ + cos² θ = 1
2.1 + tan² θ = sec² θ
3.1 + cot² θ = csc² θ

3. Compound Angles (A ± B)

1.sin(A ± B) = sin A cos B ± cos A sin B
2.cos(A ± B) = cos A cos B ∓ sin A sin B
3.tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)
4.cot(A ± B) = (cot A cot B ∓ 1)/(cot B ± cot A)

4. Multiple Angles

1.Double Angle:2A

1.sin 2A = 2 sin A cos A
2.cos 2A = cos² A − sin² A = 2 cos² A − 1 = 1 − 2 sin² A
3.tan 2A = 2 tan A / (1 − tan² A)

2.Triple Angle:3A

1.sin 3A = 3 sin A − 4 sin³ A
2.cos 3A = 4 cos³ A − 3 cos A
3.tan 3A = (3 tan A − tan³ A)/(1 − 3 tan² A)

3.Special Formulas:

1.sin² A = (1 − cos 2A)/2
2.cos² A = (1 + cos 2A)/2
3.tan² A = (1 − cos 2A)/(1 + cos 2A)

5. Submultiple Angles (A/2, A/3)

1.Half-Angle Identities:

1.sin θ = 2 sin(θ/2) cos(θ/2)
2.cos θ = 2 cos²(θ/2) − 1 = 1 − 2 sin²(θ/2)
3.tan θ = 2 tan(θ/2)/(1 − tan²(θ/2))
4.sin θ = 2 tan(θ/2)/(1 + tan²(θ/2))
5.cos θ = (1 − tan²(θ/2))/(1 + tan²(θ/2))

2.Third-Angle Formulas:

1.sin A = 3 sin(A/3) − 4 sin³(A/3)
2.cos A = 4 cos³(A/3) − 3 cos(A/3)
3.tan A = (3 tan(A/3) − tan³(A/3))/(1 − 3 tan²(A/3))

6. Transformation Formulas

1.Product to Sum:

1.2 sin A cos B = sin(A + B) + sin(A − B)
2.2 cos A sin B = sin(A + B) − sin(A − B)
3.2 cos A cos B = cos(A + B) + cos(A − B)
4.2 sin A sin B = cos(A − B) − cos(A + B)

2.Sum to Product:

1.sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2)
2.sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2)
3.cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2)
4.cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2)

7. Conditional Identities (A + B + C = 180°)

1.tan A + tan B + tan C = tan A tan B tan C
2.sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
3.cos 2A + cos 2B + cos 2C = −1 − 2 cos 2A cos 2B cos 2C
4.sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2)

8. Trigonometric Equations

Steps to Solve:

1.Determine sign → Quadrant (CAST Rule)
2.Find reference angle
3.Adjust based on quadrant:

2nd: 180° − θ

3rd: 180° + θ

4th: 360° − θ
4.General solution: +360°k
5.Check range for sin/cos/tan

9. Height and Distance Applications

1.Angle of Elevation: Angle above horizontal to a higher point.
2.Angle of Depression: Angle below horizontal to a lower point.

3.Formulas:

1.tan θ = height / distance
2.For multiple angles, calculate using difference or triangle properties

4.Common Applications:

1.Towers, poles, ladders, shadows
2.Shadow difference → height of tower
3.Observation points → distance or height

5.Diagrams: Always draw a right triangle for visualization

10. Important Formulas & Tables

11.Important Questions

1. If sin A = 3/5, find sin 2A, cos 2A, tan 2A
Solution:
or cos A = √(1 − sin² A) = √(1 − 9/25) = 4/5
or sin 2A = 2 × sin A × cos A = 2 × 3/5 × 4/5 = 24/25
or cos 2A = cos² A − sin² A = 16/25 − 9/25 = 7/25
or tan 2A = sin 2A / cos 2A = 24/25 ÷ 7/25 = 24/7

2. Prove sin 3A = 3 sin A − 4 sin³ A
Solution:
or sin 3A = sin(2A + A) = sin 2A cos A + cos 2A sin A
or Substitute sin 2A = 2 sin A cos A, cos 2A = 1 − 2 sin² A
or sin 3A = (2 sin A cos A) cos A + (1 − 2 sin² A) sin A = 2 sin A cos² A + sin A − 2 sin³ A
or cos² A = 1 − sin² A → 2 sin A (1 − sin² A) + sin A − 2 sin³ A = 3 sin A − 4 sin³ A

3. Solve sin x + cos x = √2 (0° ≤ x ≤ 360°)
Solution:
or Divide by √2 → (sin x + cos x)/√2 = 1
or sin x/√2 + cos x/√2 = 1 → sin(x + 45°) = 1
or x + 45° = 90° → x = 45°
or General solution in 0°–360° → x = 45°, 225°

4. If A + B + C = 180°, prove sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2)
Solution:
or sin A + sin B = 2 sin((A + B)/2) cos((A − B)/2)
or A + B = 180° − C → sin A + sin B = 2 cos(C/2) cos((A − B)/2)
or Add sin C → 2 cos(C/2) cos((A − B)/2) + sin C
or Simplify → 4 cos(A/2) cos(B/2) cos(C/2)

5. A tower casts a shadow 40 m longer at 30° sun elevation than at 60°. Find height
Solution:
or Let height = h, distance at 60° = x, at 30° = x + 40
or tan 60° = h / x → h = √3 x
or tan 30° = h / (x + 40) → h = (x + 40)/√3
or Equate → √3 x = (x + 40)/√3 → 3 x = x + 40 → x = 10
or Height h = √3 × 10 = 10√3 m

6. Height of a pole if shadow at 10 AM = 12 m (sun 60°), at 2 PM = 20 m (sun 30°)
Solution:
or Let height = h
or At 10 AM → tan 60° = h / 12 → h = 12√3 m
or At 2 PM → tan 30° = h / 20 → h = 20/√3 m
or Solve for h using consistent system or average if needed

7. A ladder 10 m leans on wall. Top slides 2 m down. Find change in angle of elevation
Solution:
or Initial triangle: height = h1, base = b1 → h1² + b1² = 10²
or New triangle: height = h2 = h1 − 2, base = b2 = same or recompute if base moves
or tan θ1 = h1 / b1, tan θ2 = h2 / b2
or Change in angle = θ2 − θ1 → compute using arctan

8. Angle of elevation to top of tower 30° from A, 45° from B, 50 m closer. Find height
Solution:
or Let height = h, distances from tower = x (A), x − 50 (B)
or tan 30° = h / x → h = x / √3
or tan 45° = h / (x − 50) → h = x − 50
or Equate → x / √3 = x − 50 → solve x → h ≈ 28.87 m

9. Height of tree if angles of elevation from two points 30 m apart are 45° and 30°
Solution:
or Let height = h, distances = x (45°), x + 30 (30°)
or tan 45° = h / x → h = x
or tan 30° = h / (x + 30) → h / (h + 30) = 1/√3
or Solve → h = 15 m

10. Solve tan² x − 3 tan x + 2 = 0 (0° ≤ x ≤ 360°)
Solution:
or Factor → (tan x − 1)(tan x − 2) = 0
or tan x = 1 → x = 45°, 225°
or tan x = 2 → x = arctan 2 ≈ 63.43°, 243.43°

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