Why Trig Identities Matter in Your ICSE Exams
Ever wondered why a triangle can hide so many shortcuts? Those shortcuts are trig identities, and they can turn a messy problem into a quick win.
💡 In Simple Words: Trigonometric identities are like secret codes that show different ways to write the same sine, cosine or tangent expression. Knowing them lets you swap one form for another, making calculations easier.
What Are Trigonometric Identities?
In plain language, an identity is an equation that’s true for every angle you plug in. It’s not a one‑time answer; it works forever, just like 2 + 2 = 4.
Basic Ratios First
Remember the three basic ratios of a right‑angled triangle?
- sin θ = opposite ÷ hypotenuse
- cos θ = adjacent ÷ hypotenuse
- tan θ = opposite ÷ adjacent
These are the building blocks for every identity.
Fundamental Identities
These four are the heart of everything else.
- Reciprocal identities: 1/sin θ = csc θ, 1/cos θ = sec θ, 1/tan θ = cot θ
- Pythagorean identity: sin²θ + cos²θ = 1
- Co‑function identities: sin(90°‑θ) = cos θ, cos(90°‑θ) = sin θ, tan(90°‑θ) = cot θ
- Quotient identities: tan θ = sin θ / cos θ, cot θ = cos θ / sin θ
Pythagorean Identities – More Than One
If you divide the basic Pythagorean identity by cos²θ or sin²θ you get two handy versions.
- 1 + tan²θ = sec²θ
- 1 + cot²θ = csc²θ
Sum and Difference Formulas
These let you find the sine or cosine of a sum or difference of two angles.
- sin(α ± β) = sinα·cosβ ± cosα·sinβ
- cos(α ± β) = cosα·cosβ ∓ sinα·sinβ
- tan(α ± β) = (tanα ± tanβ) / (1 ∓ tanα·tanβ)
Double‑Angle Formulas
Just plug the same angle into the sum formulas.
- sin2θ = 2·sinθ·cosθ
- cos2θ = cos²θ ‑ sin²θ (or 2·cos²θ ‑ 1, or 1 ‑ 2·sin²θ)
- tan2θ = 2·tanθ / (1 ‑ tan²θ)
How to Use Identities in an Exam
Step‑by‑step, here’s a quick recipe:
- Read the question. Spot if a sine, cosine or tangent appears twice.
- Choose an identity that links those two appearances.
- Replace one side using the identity.
- Simplify the algebra – often the terms cancel.
- Check if the answer matches the required form (e.g., no radicals in denominator).
graph TD
A[Start with basic ratios] --> B[Apply Pythagorean identity] --> C[Use co‑function rules] --> D[Get sum/difference formulas] --> E[Simplify problem] --> F[Write final answer]
Quick Reference Table
| Identity Type | Formula |
|---|---|
| Reciprocal | csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ |
| Pythagorean | sin²θ + cos²θ = 1 |
| Derived Pythagorean | 1 + tan²θ = sec²θ, 1 + cot²θ = csc²θ |
| Co‑function | sin(90°‑θ)=cos θ, cos(90°‑θ)=sin θ |
| Sum/Difference | sin(α±β)=sinα·cosβ±cosα·sinβ |
| Double‑Angle | sin2θ=2·sinθ·cosθ |
📝 Likely Exam Questions
- Prove that 1 + tan²θ = sec²θ.
Use sin²θ + cos²θ = 1, divide by cos²θ, and simplify. - Simplify (sin θ · sec θ) + (cos θ · csc θ).
Replace sec θ with 1/cos θ and csc θ with 1/sin θ, then the expression becomes 1 + 1 = 2. - Find the value of cos 2θ if sin θ = 3/5 and θ is acute.
Use cos²θ = 1 ‑ sin²θ = 1 ‑ 9/25 = 16/25, so cos θ = 4/5. Then cos 2θ = cos²θ ‑ sin²θ = 16/25 ‑ 9/25 = 7/25. - Express tan (45° + θ) in terms of tan θ.
Apply tan(α+β) formula: (tan45° + tanθ)/(1 ‑ tan45°·tanθ) = (1 + tanθ)/(1 ‑ tanθ). - Solve for θ (0°
Set sinθ = cosθ → tanθ = 1 → θ = 45°.
#ICSE#Class 10#Mathematics#Trigonometry#Identities
More from us