Why trigonometric identities matter
Ever tried to solve a triangle and felt stuck because the angles looked weird? Trigonometric identities are the shortcuts that turn those weird angles into familiar numbers. They save you time in class tests and make the whole subject feel less like a puzzle.
💡 In Simple Words: Trigonometric identities are like secret cheat‑codes for angles. They let you replace one trig expression with another that’s easier to work with, so you can finish problems faster.
Getting the basics right
Before we jump into the identities, let’s recall the three main ratios that define a right‑angled triangle:
- sine (sin) – the length of the side opposite the angle divided by the hypotenuse (the longest side).
- cosine (cos) – the length of the adjacent side divided by the hypotenuse.
- tangent (tan) – the opposite side divided by the adjacent side. It can also be written as sin θ / cos θ.
Remember these pictures: imagine water flowing through a pipe. The pipe’s width is the hypotenuse, and the amount of water that goes through a side is like the opposite or adjacent side. The ratios tell you how much of the total flow goes one way or the other.
Fundamental identities – the first set to master
The word “fundamental” isn’t fancy here; it just means these are the building blocks. They hold true for any angle θ.
- sin²θ + cos²θ = 1
- 1 + tan²θ = sec²θ
- 1 + cot²θ = csc²θ
In the first line, sin²θ means (sin θ)² – the sine of the angle, then squared. The same idea applies to cos²θ. This identity is like the Pythagorean theorem for the unit circle.
Pythagorean identities – a quick reminder
From the fundamental one, we can rearrange to get two more useful forms:
- tan²θ = sec²θ – 1
- cot²θ = csc²θ – 1
These are handy when a problem gives you sec θ (the reciprocal of cosine) and you need tan θ.
Co‑function identities – angles that add up to 90°
When two angles are complementary (their sum is 90°), their trig functions swap places:
- sin(90° – θ) = cos θ
- cos(90° – θ) = sin θ
- tan(90° – θ) = cot θ
- cot(90° – θ) = tan θ
- sec(90° – θ) = csc θ
- csc(90° – θ) = sec θ
Think of a seesaw: when one side goes up, the other goes down. The same swap happens with these functions.
Angle sum and difference identities – combining angles
These let you break a big angle into two smaller ones you already know.
- sin(α + β) = sinα·cosβ + cosα·sinβ
- sin(α – β) = sinα·cosβ – cosα·sinβ
- cos(α + β) = cosα·cosβ – sinα·sinβ
- cos(α – β) = cosα·cosβ + sinα·sinβ
- tan(α + β) = (tanα + tanβ) / (1 – tanα·tanβ)
- tan(α – β) = (tanα – tanβ) / (1 + tanα·tanβ)
Imagine you have two Lego blocks of different lengths (the angles). Adding them together creates a new shape whose dimensions you can calculate using the formulas above.
Double‑angle and half‑angle identities – when the angle repeats
Double‑angle means the angle is twice a smaller one. Half‑angle is the opposite.
- sin2θ = 2·sinθ·cosθ
- cos2θ = cos²θ – sin²θ (also = 2·cos²θ – 1 or = 1 – 2·sin²θ)
- tan2θ = 2·tanθ / (1 – tan²θ)
These are useful when a problem gives you sin 2θ and asks for sin θ, for example.
Quick reference table
| Identity Type | Formula |
|---|---|
| Fundamental | sin²θ + cos²θ = 1 |
| Pythagorean | tan²θ = sec²θ – 1 |
| Co‑function | sin(90°–θ) = cosθ |
| Sum (sine) | sin(α+β) = sinα·cosβ + cosα·sinβ |
| Sum (cosine) | cos(α+β) = cosα·cosβ – sinα·sinβ |
| Double‑angle (sine) | sin2θ = 2·sinθ·cosθ |
| Double‑angle (cosine) | cos2θ = cos²θ – sin²θ |
Worked example – simplifying an expression
Problem: Simplify sin θ · sec θ + cos θ · csc θ.
Step 1: Write sec θ and csc θ as reciprocals.
sec θ = 1/cos θ, csc θ = 1/sin θ.
Step 2: Substitute. \[ sinθ·\frac{1}{cosθ} + cosθ·\frac{1}{sinθ} = \frac{sinθ}{cosθ} + \frac{cosθ}{sinθ} \]
Step 3: Recognise the fractions as tan θ and cot θ. \[ tanθ + cotθ \]
Step 4: Use the Pythagorean identity tan²θ + 1 = sec²θ, but we just need a simpler form. Notice tanθ + cotθ = (sin²θ + cos²θ) / (sinθ·cosθ) = 1 / (sinθ·cosθ).
Step 5: The final simplified result is 1 / (sinθ·cosθ). If the question asks for a single trig function, you can write it as secθ·cscθ.
That’s it – one line of work after the substitution!
📝 Likely Exam Questions
- Prove that sin²θ + cos²θ = 1.
Model answer: Start with a right‑angled triangle on the unit circle where the radius = 1. By the Pythagorean theorem, (adjacent)² + (opposite)² = (hypotenuse)² → cos²θ + sin²θ = 1. - Simplify: tanθ + secθ·sinθ.
Model answer: secθ·sinθ = (1/cosθ)·sinθ = tanθ. So expression becomes tanθ + tanθ = 2tanθ. - Find the value of cos 2θ if sin θ = 3/5 and θ is acute.
Model answer: 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 sin (α + β) in terms of sin α, cos α, sin β, cos β.
Model answer: Use the sum formula: sin(α+β) = sinα·cosβ + cosα·sinβ. - If tan θ = 2, find sec²θ.
Model answer: From 1 + tan²θ = sec²θ, we get sec²θ = 1 + 2² = 5.