IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read

Simple Trigonometric Identities

Two identities link sin, cos, and tan — the quotient identity tan θ = sin θ / cos θ and the Pythagorean identity sin²θ + cos²θ = 1. They let you swap one trig function for another to simplify expressions, prove results, or turn messy equations into solvable quadratics.

📘 What you need to know

Quotient identity: tan θ = sin θ / cos θ.
Pythagorean identity: sin²θ + cos²θ = 1.
Both are in the formula booklet — you don’t memorise them, you use them.
Notation: sin²θ means (sin θ)² — it’s the square of the value, not the angle.
Useful rearrangements: sin²θ = 1 − cos²θ and cos²θ = 1 − sin²θ.
When to use: an expression has both sin and cos² (or vice versa) — swap one to match.
If tan θ appears alongside sin/cos, replace it with sin θ / cos θ.

The two identities

Quotient identity

tan θ = sin θcos θ

use to remove tan from an expression

Pythagorean identity

sin²θ + cos²θ = 1

use to swap sin² for cos² or vice versa

The Pythagorean identity comes straight from a right-angled triangle with hypotenuse 1: the opposite side is sin θ, the adjacent side is cos θ, so by Pythagoras (sin θ)² + (cos θ)² = 1². The quotient identity comes from SOHCAHTOA: tan θ = opposite/adjacent = sin θ / cos θ.

Useful rearrangements

From the Pythagorean identity sin²θ = 1 − cos²θ cos²θ = 1 − sin²θ

These two are the workhorses. If an equation contains both sin x and cos²x, swap the cos²x for 1 − sin²x — and now everything is in sin x, ready to factor as a quadratic.

If you’re asked to “show that” one form equals another, look at what’s missing in the target. If tan has gone, you must have substituted sin/cos for it. If only cos² appears, you must have swapped the sin² out using the Pythagorean identity.

🧭 Recipe — using identities to simplify or solve

Spot what mixes: are both sin and cos appearing? Is tan in the mix? Are squares involved?
Choose your swap: tan → sin/cos; sin² → 1 − cos² (or the reverse); cos² → 1 − sin².
Substitute into the expression or equation.
Simplify — expand, combine like terms, factor, or rearrange to standard form.
Solve or state the result; verify by substituting a known value.

Worked examples

WE 1

Simplify using the Pythagorean identity

Simplify the expression 8 sin²θ + 8 cos²θ − 5.

Step 1: Factor out the common 8 8(sin²θ + cos²θ) − 5 Step 2: Apply sin²θ + cos²θ = 1 8(1) − 5 = 3 3 the angle θ has disappeared — the expression is constant for every value of θ

WE 2

Simplify using both identities

Simplify the expression sin θ tan θ + cos θ. Give your answer as a single fraction.

Step 1: Replace tan θ with sin θ / cos θ sin θ · (sin θ / cos θ) + cos θ = sin²θ / cos θ + cos θ Step 2: Common denominator cos θ = sin²θ / cos θ + cos²θ / cos θ = (sin²θ + cos²θ) / cos θ Step 3: Apply sin²θ + cos²θ = 1 = 1 / cos θ 1cos θ

WE 3

Rewrite as a quadratic in cos x

Show that the equation 3 sin²x + 4 cos x = 4 can be written in the form a cos²x + b cos x + c = 0, where a, b, c are integers to be found.

Equation has both sin² and cos — swap sin² to match Step 1: Use sin²x = 1 − cos²x 3(1 − cos²x) + 4 cos x = 4 Step 2: Expand 3 − 3 cos²x + 4 cos x = 4 Step 3: Move everything to one side −3 cos²x + 4 cos x − 1 = 0 3 cos²x − 4 cos x + 1 = 0 a = 3, b = −4, c = 1

WE 4

Find sin θ and tan θ given cos θ

Given that cos θ = 5/13 and θ is acute, find the exact values of sin θ and tan θ.

Step 1: Apply sin²θ = 1 − cos²θ sin²θ = 1 − (5/13)² = 1 − 25/169 = 144/169 Step 2: Take the square root — θ acute, so positive sin θ = √(144/169) = 12/13 Step 3: Use tan θ = sin θ / cos θ tan θ = (12/13) ÷ (5/13) = 12/5 sin θ = 1213, tan θ = 125 classic 5-12-13 right triangle ✓

WE 5

Prove an identity

Prove that (sin θ + cos θ)² + (sin θ − cos θ)² = 2 for all values of θ.

Step 1: Expand both squares (LHS) (sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ (sin θ − cos θ)² = sin²θ − 2 sin θ cos θ + cos²θ Step 2: Add — the cross terms cancel LHS = 2 sin²θ + 2 cos²θ Step 3: Factor and apply Pythagorean identity = 2(sin²θ + cos²θ) = 2(1) = 2 = RHS ✓ proved

WE 6

Solve a trig equation using an identity

Solve 2 cos²x + sin x = 1 for 0° ≤ x ≤ 360°.

Mixed sin and cos² — swap cos² for 1 − sin²x Step 1: Substitute 2(1 − sin²x) + sin x = 1 2 − 2 sin²x + sin x = 1 Step 2: Rearrange to a quadratic in sin x 2 sin²x − sin x − 1 = 0 Step 3: Factor (2 sin x + 1)(sin x − 1) = 0 sin x = −1/2 or sin x = 1 Step 4: Solve each in [0°, 360°] sin x = −1/2 → x = 210°, 330° sin x = 1 → x = 90° x = 90°, 210°, 330° check x = 90°: 2(0) + 1 = 1 ✓

💡 Top tips

Aim for one trig function. If you have sin and cos², swap the cos² to 1 − sin² so the whole equation is in sin only.
Match the linear term. If the equation has cos x linearly and sin² as the square, change sin² → 1 − cos² (not the other way).
Both identities are in the formula booklet — don’t waste exam time memorising; just look them up if needed.
For “find sin/cos given the other”, draw a right triangle with the given ratio — quickest way to see signs and the third side.
Always check the sign when taking a square root — does the angle’s region make sin/cos positive or negative?

⚠ Common mistakes

Reading sin²θ as sin(θ²). They’re different — sin²θ means (sin θ)², the square of the value.
Swapping the wrong square. If the linear term is cos x, you swap sin² → 1 − cos² (so everything ends up in cos).
Losing solutions. If sin x = 1, that’s one solution per period (90° only in [0°, 360°]) — not two.
Sign errors when finding sin from cos. The square root gives ±; the quadrant tells you which sign.
Forgetting to factor. Once you have a quadratic in sin x or cos x, factor or use the formula — don’t try to solve it as a single equation.

Next note: Compound Angle Formulae. The formulas for sin(A ± B), cos(A ± B), and tan(A ± B) — they let you split angles like sin 75° as sin(45° + 30°) and find exact values you couldn’t before.

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