Trigonometry — Study Notes for Assam TET Paper II

Overview

Trigonometry is a fundamental branch of mathematics that deals with relationships between angles and sides of triangles, particularly right-angled triangles. For Assam TET Paper II, this topic carries significant weight as it forms the foundation for understanding heights, distances, and practical measurement problems that teachers must explain to upper primary students.

The scope for this exam focuses on trigonometric ratios (sine, cosine, tangent and their reciprocals) and fundamental identities. Questions typically test your ability to calculate ratios from given information, apply identities to simplify expressions, and solve problems involving complementary angles. Mastery of this topic requires memorizing the ratio definitions, standard angle values, and the three Pythagorean identities.

Understanding trigonometry is essential not just for solving direct problems but also for teaching students how mathematics connects to real-world applications like measuring building heights, calculating distances, and understanding navigation — all relevant to the Assam context with its varied terrain and river systems.

Key Concepts

• **Right-angled triangle orientation**: In any right triangle, the side opposite to the right angle is the hypotenuse (always the longest side). For any acute angle θ, identify the opposite side (facing the angle) and adjacent side (forming the angle with hypotenuse).

• **Six trigonometric ratios**: For angle θ in a right triangle — sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent. The reciprocals are cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.

• **Complementary angle relationship**: Two angles are complementary if they sum to 90°. In a right triangle, the two acute angles are always complementary. This gives rise to relations like sin(90° - θ) = cos θ.

• **Pythagorean identities**: These connect different ratios and are derived from the Pythagorean theorem. They allow conversion between ratios and simplification of complex expressions.

• **Standard angles**: The values for 0°, 30°, 45°, 60°, and 90° must be memorized as they appear in most numerical problems.

• **Domain restrictions**: tan 90° and sec 90° are undefined (division by zero). Similarly, cot 0° and cosec 0° are undefined.

Formulas / Key Facts

**Primary Trigonometric Ratios (for angle θ)**

• sin θ = Perpendicular / Hypotenuse = P/H
• cos θ = Base / Hypotenuse = B/H
• tan θ = Perpendicular / Base = P/B

**Reciprocal Ratios**

• cosec θ = H/P = 1/sin θ
• sec θ = H/B = 1/cos θ
• cot θ = B/P = 1/tan θ

**Quotient Relations**

• tan θ = sin θ / cos θ
• cot θ = cos θ / sin θ

**Pythagorean Identities**

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ

**Complementary Angle Identities**

• sin(90° - θ) = cos θ and cos(90° - θ) = sin θ
• tan(90° - θ) = cot θ and cot(90° - θ) = tan θ
• sec(90° - θ) = cosec θ and cosec(90° - θ) = sec θ

**Standard Angle Values Table**

| Angle | 0° | 30° | 45° | 60° | 90° | |-------|-----|------|------|------|------| | sin | 0 | 1/2 | 1/√2 | √3/2 | 1 | | cos | 1 | √3/2 | 1/√2 | 1/2 | 0 | | tan | 0 | 1/√3 | 1 | √3 | undefined |

**Memory trick for sin values**: Write 0, 1, 2, 3, 4 under the angles 0°, 30°, 45°, 60°, 90°. Divide each by 4 and take square root. For cos, reverse the order.

Worked Examples

**Example 1**: If tan θ = 3/4, find sin θ and cos θ.

Solution:

• tan θ = P/B = 3/4, so Perpendicular = 3k and Base = 4k (for some constant k)
• Using Pythagoras: H² = P² + B² = 9k² + 16k² = 25k²
• Therefore H = 5k
• sin θ = P/H = 3k/5k = 3/5
• cos θ = B/H = 4k/5k = 4/5

**Example 2**: Prove that (sin 30° + cos 60°) / (1 + sin 60° × cos 30°) = 4/7

Solution:

• sin 30° = 1/2, cos 60° = 1/2, sin 60° = √3/2, cos 30° = √3/2
• Numerator = 1/2 + 1/2 = 1
• Denominator = 1 + (√3/2)(√3/2) = 1 + 3/4 = 7/4
• Expression = 1 ÷ (7/4) = 4/7

**Example 3**: Simplify: sin²45° + cos²45° + tan²45°

Solution:

• sin 45° = 1/√2, so sin²45° = 1/2
• cos 45° = 1/√2, so cos²45° = 1/2
• tan 45° = 1, so tan²45° = 1
• Sum = 1/2 + 1/2 + 1 = 2

**Example 4**: If sin θ = 5/13, find sec θ + tan θ.

Solution:

• sin θ = 5/13 means P = 5, H = 13
• B² = H² - P² = 169 - 25 = 144, so B = 12
• cos θ = 12/13, hence sec θ = 13/12
• tan θ = 5/12
• sec θ + tan θ = 13/12 + 5/12 = 18/12 = 3/2

Common Mistakes

• **Confusing opposite and adjacent sides**: Students often mix up which side is opposite and which is adjacent. The fix — always mark the angle first, then identify the side directly facing it (opposite) and the side touching it that is not the hypotenuse (adjacent).

• **Writing tan 90° = infinity or some value**: tan 90° is undefined, not infinity. In MCQs, options with tan 90° having a numerical value are always wrong.

• **Forgetting to rationalize denominators**: When sin 45° = 1/√2, many students leave it unrationalized. The standard form is √2/2. Check if answer options are rationalized.

• **Applying wrong identity for simplification**: Using sin²θ - cos²θ = 1 instead of sin²θ + cos²θ = 1. Remember — the fundamental identity always has a plus sign between sin² and cos².

• **Errors with complementary angles**: Writing sin(90° - θ) = sin θ instead of cos θ. The key — "co" in cosine suggests complementary, so sin becomes cos and vice versa.

Quick Reference

• SOH-CAH-TOA: Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent

• sin²θ + cos²θ = 1 is the master identity — the other two are derived from it

• At 45°, sin and cos are equal (both 1/√2); tan = 1

• Complementary rule: sin and cos swap, tan and cot swap, sec and cosec swap

• If one ratio is given, use Pythagoras to find the third side, then compute all other ratios

• tan θ × cot θ = 1, sin θ × cosec θ = 1, cos θ × sec θ = 1