IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read

Double Angle Formulae

Set B = A in the compound angle formulae and you get the double angle versions — single-angle expressions for sin 2θ, cos 2θ, and tan 2θ. They’re the workhorse for solving any equation with a mixed angle (like x and 2x) and for proving identities.

📘 What you need to know

sin 2θ = 2 sin θ cos θ.
cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ (three forms — pick the one that matches the rest of the equation).
tan 2θ = 2 tan θ / (1 − tan²θ).
The double angle formulas are derived from compound angles by setting B = A.
The angle in the formula is half of the angle in the term: sin 8θ = 2 sin 4θ cos 4θ.
Reverse direction: 2 sin 5θ cos 5θ = sin 10θ, and cos²3θ − sin²3θ = cos 6θ.
For solving equations with two different angles (e.g., 2x and x), apply the double angle formula to the bigger one — then everything is in the same angle.

The three formulae

sin double angle sin 2θ = 2 sin θ cos θ

cos double angle — three equivalent forms cos 2θ = cos²θ − sin²θ = 2 cos²θ − 1 = 1 − 2 sin²θ

tan double angle tan 2θ = 2 tan θ1 − tan²θ

The three versions of cos 2θ are all equivalent — they just suit different equations. If your equation has cos x linearly, use cos 2x = 2cos²x − 1. If it has sin x linearly, use cos 2x = 1 − 2sin²x. Pick the one that leaves a single trig function.

Rewriting expressions

Double angle formulas work in both directions. You can expand a single 2θ term into a product, or compress a product back into a 2θ term.

Form	Becomes	Why
sin 6θ	2 sin 3θ cos 3θ	angle in formula is half (3θ)
cos²5θ − sin²5θ	cos 10θ	angle in formula is double (10θ)
2 − 4 sin²3θ	2(1 − 2 sin²3θ) = 2 cos 6θ	factor out 2 first
4 sin 8θ cos 8θ	2 · 2 sin 8θ cos 8θ = 2 sin 16θ	pull out the 2 to match the formula

🧭 Recipe — solve an equation with a double angle

Spot the mismatch: are there two different angles (like 2x and x)? Or a 2θ and a single trig function?
Apply the right formula: replace the double-angle term with its single-angle expansion.
Choose wisely for cos 2θ: pick the form that matches the rest of the equation (linear sin → use 1 − 2sin²θ; linear cos → use 2cos²θ − 1).
Move everything to one side = 0, then factor out a common term or treat as a quadratic.
Solve each factor in the given interval; check no solutions are missed.

Worked examples

WE 1

Find sin 2θ and cos 2θ from sin θ

Given that sin θ = 4/5 and θ is acute, find the exact values of sin 2θ and cos 2θ.

Step 1: Find cos θ via Pythagoras (acute → positive) cos θ = √(1 − 16/25) = √(9/25) = 3/5 Step 2: sin 2θ = 2 sin θ cos θ = 2 · (4/5) · (3/5) = 24/25 Step 3: cos 2θ — use 1 − 2 sin²θ (sin θ already given) = 1 − 2(16/25) = 1 − 32/25 = −7/25 sin 2θ = 2425, cos 2θ = −725 classic 3-4-5 right triangle — check: 24² + 7² = 625 = 25² ✓

WE 2

Express as a single trig term

Express 8 sin 3x cos 3x in the form k sin nx, where k and n are constants to be found.

Step 1: Pull out a factor of 2 to match sin 2θ = 2 sin θ cos θ 8 sin 3x cos 3x = 4 · (2 sin 3x cos 3x) Step 2: Apply double angle (with θ = 3x → 2θ = 6x) = 4 · sin 6x k = 4, n = 6 → 4 sin 6x

WE 3

Solve cos 2x + cos x + 1 = 0

Solve the equation cos 2x + cos x + 1 = 0 for 0° ≤ x ≤ 360°.

Mixed 2x and x — use the cos 2x form that matches cos x linearly Step 1: Replace cos 2x with 2 cos²x − 1 2 cos²x − 1 + cos x + 1 = 0 2 cos²x + cos x = 0 Step 2: Factor cos x (2 cos x + 1) = 0 cos x = 0 or cos x = −1/2 Step 3: Solve each in [0°, 360°] cos x = 0 → x = 90°, 270° cos x = −1/2 → x = 120°, 240° x = 90°, 120°, 240°, 270° check x = 90°: cos 180° + cos 90° + 1 = −1 + 0 + 1 = 0 ✓

WE 4

Solve sin 2x = cos x in radians

Solve the equation sin 2x = cos x for 0 ≤ x ≤ 2π. Give answers in exact form.

Step 1: Replace sin 2x with 2 sin x cos x 2 sin x cos x = cos x Step 2: Move to one side, factor out cos x 2 sin x cos x − cos x = 0 cos x (2 sin x − 1) = 0 Step 3: Solve each factor cos x = 0 → x = π/2, 3π/2 sin x = 1/2 → x = π/6, 5π/6 x = π6, π2, 5π6, 3π2 never divide by cos x — you’d lose the cos x = 0 solutions

WE 5

Prove a double-angle identity

Prove that (sin θ + cos θ)² = 1 + sin 2θ.

Step 1: Expand the LHS (sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ Step 2: Group sin²θ + cos²θ = (sin²θ + cos²θ) + 2 sin θ cos θ = 1 + 2 sin θ cos θ Step 3: Apply 2 sin θ cos θ = sin 2θ = 1 + sin 2θ = RHS ✓ proved

WE 6

Solve cos 2x − 5 cos x + 3 = 0

Solve the equation cos 2x − 5 cos x + 3 = 0 for 0 ≤ x ≤ 2π. Give answers in exact form.

Linear cos x term → use cos 2x = 2 cos²x − 1 Step 1: Substitute 2 cos²x − 1 − 5 cos x + 3 = 0 2 cos²x − 5 cos x + 2 = 0 Step 2: Factor as a quadratic in cos x (2 cos x − 1)(cos x − 2) = 0 cos x = 1/2 or cos x = 2 (reject — outside [−1, 1]) Step 3: Solve cos x = 1/2 in [0, 2π] x = π/3 or x = 2π − π/3 = 5π/3 x = π3, 5π3 always reject cos x or sin x values outside [−1, 1]

💡 Top tips

Pick the right cos 2θ form. If the rest of the equation has sin x linearly, use 1 − 2sin²x. If it has cos x linearly, use 2cos²x − 1.
Look at half the angle. sin 8θ becomes 2 sin 4θ cos 4θ — the formula’s angle is half.
Look at double the angle when compressing. cos²5θ − sin²5θ becomes cos 10θ.
Don’t divide by cos x or sin x. Factor instead — division loses the cases where that term is zero.
Reject impossible values. After factoring a quadratic, drop any root with |sin| or |cos| > 1.

⚠ Common mistakes

Writing sin 2θ as 2 sin θ. It’s 2 sin θ cos θ — both factors matter.
Forgetting the factor of 2. 4 sin 5θ cos 5θ = 2 sin 10θ (pull out the 2 first).
Dividing by cos x in equations like sin 2x = cos x. You lose the cos x = 0 solutions.
Using the wrong cos 2θ form. Picking cos²θ − sin²θ when the equation has only cos x makes it harder to factor.
Mixing the angles. sin 2x ≠ 2 sin x — the 2 is on the angle, not a multiplier on sin x.

Next note: Relationship Between Trigonometric Ratios. How sin and cos relate via complementary angles (sin θ = cos(90° − θ)), and how to find every trig ratio (and double-angle ratios) when you’re given just one — using right triangles or identities, with quadrant signs to keep you honest.

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