Why Trigonometric Identities Matter in Your Exams
Ever felt stuck because a sine or cosine expression just won’t simplify? Those handy shortcuts called trigonometric identities are the secret weapons that turn a messy problem into a neat answer.
💡 In Simple Words: Trigonometric identities are equations that always hold true for any angle. They let you replace one trig function with another, making calculations easier—just like swapping a long road for a shortcut.
What Are Trigonometric Identities?
A trigonometric identity is a relationship between the six basic trig functions—sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec) and cotangent (cot)—that works for every angle. Think of it as a rulebook for angles, just like a game’s rulebook tells you what moves are allowed.
Basic Trigonometric Identities
The first set you’ll meet are the reciprocal identities. They show how each function is the inverse (or “flip”) of another.
- csc θ = 1/sin θ – Cosecant is the flip of sine.
- sec θ = 1/cos θ – Secant is the flip of cosine.
- cot θ = 1/tan θ – Cotangent is the flip of tangent.
Remember, “reciprocal” just means “one over”.
Pythagorean Identities
These come from the famous Pythagoras theorem (a² + b² = c²) applied to a right‑angled triangle drawn on the unit circle (a circle with radius 1). The most common one is:
sin²θ + cos²θ = 1
Here the superscript 2 means “squared”. If you draw a right triangle with hypotenuse 1, the legs are sin θ and cos θ, and the Pythagorean theorem tells you their squares add up to 1.
From this you can get two more:
- 1 + tan²θ = sec²θ
- 1 + cot²θ = csc²θ
Co‑function Identities
Co‑function identities link a function of an angle to a different function of its complement (90° – θ). The idea is similar to how the height of a ladder and its shadow swap roles when you rotate the ladder 90°.
- sin(90° – θ) = cos θ
- cos(90° – θ) = sin θ
- tan(90° – θ) = cot θ
- cot(90° – θ) = tan θ
Sum and Difference Identities
These let you break down the sine or cosine of a sum (θ + φ) or difference (θ – φ) into products of simpler functions. They’re like splitting a complicated Lego build into two smaller sets you already know how to handle.
sin(θ + φ) = sin θ·cos φ + cos θ·sin φ
sin(θ – φ) = sin θ·cos φ – cos θ·sin φ
cos(θ + φ) = cos θ·cos φ – sin θ·sin φ
cos(θ – φ) = cos θ·cos φ + sin θ·sin φ
Worked Example: Simplify sin θ / (1 + cos θ)
Step 1: Multiply numerator and denominator by the “conjugate” (1 – cos θ). This is similar to rationalising a denominator.
\[\frac{\sin\theta}{1+\cos\theta}\times\frac{1-\cos\theta}{1-\cos\theta}=\frac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta}\]
Step 2: Notice the denominator is a difference of squares, which becomes sin²θ because of the Pythagorean identity (sin²θ + cos²θ = 1 ⇒ 1 – cos²θ = sin²θ).
\[\frac{\sin\theta(1-\cos\theta)}{\sin^2\theta}=\frac{1-\cos\theta}{\sin\theta}\]
Step 3: The result is the co‑function form of tan (θ/2), but for class 10 it’s enough to stop at \(\frac{1-\cos\theta}{\sin\theta}\).
Quick Revision Table
| Identity Type | Formula |
|---|---|
| Reciprocal | csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ |
| Pythagorean | sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = csc²θ |
| Co‑function | sin(90° – θ) = cos θ; cos(90° – θ) = sin θ; tan(90° – θ) = cot θ |
| Sum/Difference (sine) | sin(θ ± φ) = sin θ·cos φ ± cos θ·sin φ |
| Sum/Difference (cosine) | cos(θ ± φ) = cos θ·cos φ ∓ sin θ·sin φ |
How to Use Identities in Exam Problems
1. Identify the pattern. Look for expressions like sin²θ, 1 + cos θ, or sin (θ + φ).
2. Pick the right identity. If you see a sum inside a sine, use the sum identity. If you have a square plus another square, think Pythagorean.
3. Replace and simplify. Substitute the identity, cancel common terms, and you’ll often end up with a basic ratio like tan θ or a constant.
📝 Likely Exam Questions
- Prove that sin²θ + cos²θ = 1. Model answer: Draw a right‑angled triangle on the unit circle (radius = 1). The legs are sin θ and cos θ, and the hypotenuse is 1. By Pythagoras, (sin θ)² + (cos θ)² = 1², giving the identity.
- Simplify (1 – cos θ) / sin θ. Model answer: Multiply numerator and denominator by (1 + cos θ) → (1 – cos²θ) / [sin θ(1 + cos θ)] → sin²θ / [sin θ(1 + cos θ)] = sin θ / (1 + cos θ).
- Find the value of sin(45° + 30°) using identities. Model answer: Use sin(θ + φ) = sin θ·cos φ + cos θ·sin φ. Plug θ=45°, φ=30°: sin45°·cos30° + cos45°·sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4.
- Show that tan θ = sin θ / cos θ. Model answer: By definition, tan θ = opposite/adjacent in a right triangle, while sin θ = opposite/hypotenuse and cos θ = adjacent/hypotenuse. Dividing sin θ by cos θ cancels the hypotenuse, leaving opposite/adjacent = tan θ.
- If sin θ = 3/5 and θ is acute, find sec θ. Model answer: First find cos θ using sin²θ + cos²θ = 1 → (3/5)² + cos²θ = 1 → 9/25 + cos²θ = 1 → cos²θ = 16/25 → cos θ = 4/5 (positive because acute). Then sec θ = 1/cos θ = 5/4.