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

Trigonometric Proof

A trig proof = pick one side, transform it into the other using known identities. No solving — you’re showing the two sides are equal for every valid θ. The skill is choosing which identity unlocks the swap.

📘 What you need to know

Pick a side (usually the messier one) and transform it step by step into the other side.
Don’t manipulate both sides as if it’s an equation — that doesn’t prove anything.
Identities to draw on: tan = sin/cos; sin² + cos² = 1; double angles; compound angles; reciprocal identities (sec/cosec/cot); 1 + tan² = sec²; 1 + cot² = cosec².
If you see an even power (sin², cos², tan², sec², etc.), the Pythagorean identities are usually the move.
If you see different angles (θ and 2θ, or θ and 4θ), use double or compound angle formulas to match them up.
If you see a fraction, combine over a common denominator first.
Always look at the target — it tells you which identity to pick.

What to spot, what to use

Spot this	Use this
sin² and cos² in the same expression	sin²θ + cos²θ = 1
tan with sin or cos	tan θ = sin θ / cos θ
sec², cosec², or tan² alone	1 + tan²θ = sec²θ or 1 + cot²θ = cosec²θ
Different angles (θ and 2θ, etc.)	double angle: sin 2θ, cos 2θ, tan 2θ
(A + B) or (A − B) inside sin/cos/tan	compound angle formulas
Two fractions to combine	common denominator
(a − b)(a + b) pattern	difference of squares: a² − b²

🧭 Recipe — prove a trig identity

Pick the messier side (more terms, more functions, fractions). Label it LHS or RHS.
Look at the target — what’s missing? What needs to disappear?
Apply identities to swap functions and combine fractions.
Simplify step by step until it matches the other side.
If stuck, switch sides and start again — both directions are valid.

Worked examples

WE 1

Pythagorean identity via difference of squares

Prove that (1 − sin θ)(1 + sin θ) = cos²θ.

Step 1: Expand LHS — difference of squares (1 − sin θ)(1 + sin θ) = 1 − sin²θ Step 2: Apply sin²θ + cos²θ = 1 → 1 − sin²θ = cos²θ = cos²θ = RHS ✓ proved

WE 2

tan θ + cot θ identity

Prove that tan θ + cot θ = 1sin θ cos θ.

Step 1: Rewrite tan and cot as sin/cos and cos/sin LHS = sin θ/cos θ + cos θ/sin θ Step 2: Common denominator sin θ cos θ = (sin²θ + cos²θ) / (sin θ cos θ) Step 3: Apply sin²θ + cos²θ = 1 = 1 / (sin θ cos θ) = RHS ✓ proved

WE 3

Double angle identity for tan

Prove that sin 2θ1 + cos 2θ = tan θ.

Step 1: Expand sin 2θ and choose the right cos 2θ form sin 2θ = 2 sin θ cos θ 1 + cos 2θ = 1 + (2 cos²θ − 1) = 2 cos²θ Step 2: Substitute LHS = (2 sin θ cos θ) / (2 cos²θ) Step 3: Cancel = sin θ / cos θ = tan θ = RHS ✓ proved choosing 2cos²θ − 1 makes the “+1” cancel beautifully

WE 4

sec² + cosec² identity

Prove that sec²θ + cosec²θ = sec²θ · cosec²θ.

Step 1: Rewrite each in sin/cos LHS = 1/cos²θ + 1/sin²θ Step 2: Common denominator sin²θ cos²θ = (sin²θ + cos²θ) / (sin²θ cos²θ) Step 3: sin²θ + cos²θ = 1 = 1 / (sin²θ cos²θ) Step 4: Match RHS — split the product = (1/cos²θ)(1/sin²θ) = sec²θ · cosec²θ = RHS ✓ proved

WE 5

Compound angle subtraction

Prove that sin(A + B) − sin(A − B) = 2 cos A sin B.

Step 1: Expand both terms with sin(A ± B) = sin A cos B ± cos A sin B sin(A + B) = sin A cos B + cos A sin B sin(A − B) = sin A cos B − cos A sin B Step 2: Subtract — sin A cos B terms cancel LHS = (sin A cos B + cos A sin B) − (sin A cos B − cos A sin B) = 2 cos A sin B = RHS ✓ proved

WE 6

Triple angle formula for cos

Prove that cos 3θ = 4 cos³θ − 3 cos θ.

Step 1: Write 3θ as 2θ + θ and apply compound angle cos 3θ = cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ Step 2: Substitute double angles cos 2θ = 2 cos²θ − 1 sin 2θ = 2 sin θ cos θ LHS = (2 cos²θ − 1) cos θ − (2 sin θ cos θ) sin θ = 2 cos³θ − cos θ − 2 sin²θ cos θ Step 3: Replace sin²θ with 1 − cos²θ = 2 cos³θ − cos θ − 2(1 − cos²θ) cos θ = 2 cos³θ − cos θ − 2 cos θ + 2 cos³θ = 4 cos³θ − 3 cos θ = RHS ✓ proved “target” was all in cos θ, so use the cos 2θ form that’s already in cos²θ, and swap sin² to 1 − cos²

💡 Top tips

Look at the target first. If RHS is in cos only, your LHS work has to remove every sin and tan.
Pick the right cos 2θ form: 2cos²θ − 1 if the target is in cos; 1 − 2sin²θ if the target is in sin.
If stuck, swap sides. Sometimes RHS → LHS is much easier.
Combine fractions early. A single fraction is easier to work with than two.
If totally stuck, rewrite everything in sin and cos. Brute-force, but it always works.

⚠ Common mistakes

Working both sides as an equation. You can’t multiply both sides by something — you’re proving they’re equal, not solving.
Wrong sign in the cos compound formula. cos(A + B) = cos A cos B − sin A sin B (sign flips!).
Forgetting the factor of 2 in sin 2θ = 2 sin θ cos θ.
Wrong cos 2θ choice — using cos²θ − sin²θ when target is in cos only makes it harder.
Stopping mid-proof. Keep going until the LHS literally matches the RHS — no “looks similar” allowed.

Next note: Strategy for Trigonometric Equations. A decision tree for picking the right approach — substitution, identities, factor or formula — based on what the equation looks like.

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