Trigonometry — Study Notes for CG TET Paper II
Overview
Trigonometry forms a crucial component of the Mathematics and Science section in CG TET Paper II, which targets teachers for classes VI to VIII. This topic builds the foundation for understanding relationships between angles and sides of triangles, with direct applications in geometry, mensuration, and real-world problem-solving.
For CG TET, you must master the six trigonometric ratios, their relationships in right-angled triangles, standard angle values, and fundamental identities. Questions typically test your ability to calculate ratios from given information, simplify expressions using identities, and apply complementary angle relationships. A solid grasp of this topic also strengthens your ability to teach spatial reasoning and mathematical relationships to upper primary students.
The scope is limited to right-angled triangle trigonometry and algebraic identities—no trigonometric equations, heights and distances applications, or graphs are expected at this level.
Key Concepts
- Trigonometric ratios are defined only for acute angles in a right-angled triangle, using the relationship between the sides relative to a specific angle (not the right angle).
- The three primary ratios — sine, cosine, and tangent — along with their reciprocals (cosecant, secant, cotangent) form the complete set of six ratios.
- SOH-CAH-TOA memory aid: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent.
- Reciprocal relationships: cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ. These are not new ratios but inverses of the primary three.
- Complementary angle property: For any acute angle θ, sin θ = cos(90° - θ), tan θ = cot(90° - θ), sec θ = cosec(90° - θ). This means ratios of complementary angles are related.
- Pythagorean identities connect the ratios algebraically and are derived from the Pythagoras theorem applied to unit relationships.
- Standard angles (0°, 30°, 45°, 60°, 90°) have fixed ratio values that must be memorised — these appear in nearly every calculation-based question.
Formulas / Key Facts
Six Trigonometric Ratios (for angle θ in right triangle)
| Ratio | Formula | Reciprocal |
|---|---|---|
| sin θ | Opposite / Hypotenuse | cosec θ = Hypotenuse / Opposite |
| cos θ | Adjacent / Hypotenuse | sec θ = Hypotenuse / Adjacent |
| tan θ | Opposite / Adjacent | cot θ = Adjacent / Opposite |
Quotient Relations
- tan θ = sin θ / cos θ
- cot θ = cos θ / sin θ
Three Fundamental Identities
- sin²θ + cos²θ = 1 (Most important — appears in majority of problems)
- 1 + tan²θ = sec²θ
- 1 + cot²θ = cosec²θ
Standard Angle Values Table
| Angle | sin | cos | tan | cosec | sec | cot |
|---|---|---|---|---|---|---|
| 0° | 0 | 1 | 0 | undefined | 1 | undefined |
| 30° | 1/2 | √3/2 | 1/√3 | 2 | 2/√3 | √3 |
| 45° | 1/√2 | 1/√2 | 1 | √2 | √2 | 1 |
| 60° | √3/2 | 1/2 | √3 | 2/√3 | 2 | 1/√3 |
| 90° | 1 | 0 | undefined | 1 | undefined | 0 |
Memory trick for sine values: 0°, 30°, 45°, 60°, 90° correspond to √0/2, √1/2, √2/2, √3/2, √4/2 (simplify each).
Complementary Angle Relations
- sin(90° - θ) = cos θ
- cos(90° - θ) = sin θ
- tan(90° - θ) = cot θ
- cot(90° - θ) = tan θ
- sec(90° - θ) = cosec θ
- cosec(90° - θ) = sec θ
Worked Examples
Example 1: Finding ratios from a given triangle
Problem: In a right triangle ABC, angle B = 90°, AB = 3 cm, BC = 4 cm. Find sin A, cos A, and tan A.
Solution:
- First, find hypotenuse AC using Pythagoras: AC² = AB² + BC² = 9 + 16 = 25, so AC = 5 cm
- For angle A: Opposite side = BC = 4, Adjacent side = AB = 3, Hypotenuse = AC = 5
- sin A = Opposite/Hypotenuse = 4/5
- cos A = Adjacent/Hypotenuse = 3/5
- tan A = Opposite/Adjacent = 4/3
Example 2: Using identities to simplify
Problem: Prove that (1 - cos²θ)(1 + cot²θ) = 1
Solution:
- From identity 1: sin²θ + cos²θ = 1, so 1 - cos²θ = sin²θ
- From identity 3: 1 + cot²θ = cosec²θ
- Substituting: sin²θ × cosec²θ = sin²θ × (1/sin²θ) = 1
- Hence proved.
Example 3: Complementary angles
Problem: If sin(A + B) = 1 and cos(A - B) = 1, find A and B (where A, B are acute).
Solution:
- sin(A + B) = 1 means A + B = 90° (since sin 90° = 1)
- cos(A - B) = 1 means A - B = 0° (since cos 0° = 1)
- Solving: A + B = 90° and A - B = 0°
- Adding: 2A = 90°, so A = 45°
- Therefore B = 45°
Common Mistakes
- Confusing opposite and adjacent sides → The opposite and adjacent sides change depending on which angle you consider. Always identify the angle first, then label sides relative to that specific angle.
- Forgetting that ratios are defined for acute angles only in basic trigonometry → At this level, θ must be between 0° and 90°. Don't apply these formulas to obtuse angles.
- Mixing up reciprocal pairs — thinking cosec is reciprocal of cos → Remember: cosec goes with sin (both have 's'), sec goes with cos (both have 'c'), cot goes with tan.
- Calculation errors with √3 and √2 — treating √3/2 as greater than 1 → Always verify: √3 ≈ 1.732, so √3/2 ≈ 0.866, which is less than 1. All sine and cosine values for acute angles lie between 0 and 1.
- Applying Pythagoras theorem incorrectly — adding wrong sides → The formula is: Hypotenuse² = Base² + Perpendicular². The hypotenuse is always the longest side, opposite the right angle.
Quick Reference
- sin²θ + cos²θ = 1 — the most-used identity; rearrange as needed for sin²θ or cos²θ.
- tan θ = sin θ / cos θ — use this to convert tan into sine-cosine form for simplification.
- sin 30° = cos 60° = 1/2 and sin 60° = cos 30° = √3/2 — complementary pairs.
- sin 45° = cos 45° = 1/√2 — the only angle where sine equals cosine.
- At 0°: sin = 0, cos = 1, tan = 0. At 90°: sin = 1, cos = 0, tan = undefined.
- For any identity problem: Start by converting everything to sin and cos, then apply sin²θ + cos²θ = 1.