IB Maths AA SL
Topic 3 — Trig Equations & Identities
Paper 1 & 2
~10 min read
Double Angle Formulae
The double angle formulas swap an expression in 2θ for one in θ — or the other way around. They’re the secret to almost every trig equation that mixes angles like 2x and x, and they unlock a lot of Paper 1 simplification questions too.
📘 What you need to know
- The sin double angle: sin 2θ = 2 sin θ cos θ — one formula, no choices.
- The cos double angle has three equivalent forms: cos 2θ = cos2θ − sin2θ = 2cos2θ − 1 = 1 − 2sin2θ.
- Pick the form of cos 2θ that matches the rest of your equation — if everything else is in sin, use the form that’s all in sin.
- The angle inside doubles or halves: sin 6θ = 2 sin 3θ cos 3θ.
- All three formulas are in your formula booklet.
What are the double angle formulas?
The double angle formulas connect a trig function of 2θ to functions of θ. There are two of them you need at SL — but the cos one comes in three equivalent dresses, so pay attention.
Don’t try to memorise these. They’re in the formula booklet — but you do need to know when to reach for each one. That’s where most of the marks live.
Why are there three forms of cos 2θ?
All three forms of cos 2θ are the same formula in disguise. They’re connected by the Pythagorean identity sin2θ + cos2θ = 1.
🤔 Where do the other two forms come from?
Start from the original: cos 2θ = cos2θ − sin2θ. Now use sin2θ = 1 − cos2θ to kill off sin2:
cos 2θ = cos2θ − (1 − cos2θ) = 2cos2θ − 1
Or use cos2θ = 1 − sin2θ to kill off cos2 instead:
cos 2θ = (1 − sin2θ) − sin2θ = 1 − 2sin2θ
Same identity, three faces — each useful in a different situation.
The three faces of cos 2θ — when to use each
Choosing the right form is the whole game. Here’s the rule of thumb: pick the form that only contains the trig ratio that matches the rest of your equation.
🧠Memory trick: “Match the rest of the equation”
Look at what’s not a 2θ term. If you see a stray sin θ, use the all-sin form (1 − 2sin2θ). If you see a stray cos θ, use the all-cos form (2cos2θ − 1). Pick whichever keeps everything in one language.
Reading the formulas in both directions
The formulas work both ways — left-to-right and right-to-left. Most exam questions use the right-to-left direction, where you spot a recognisable pattern and collapse it to a single term.
Forward direction — break a 2θ term into θ terms
sin 6θ
→
2 sin 3θ cos 3θ (angle halves)
cos 10x
→
cos2 5x − sin2 5x
2 cos 6θ
→
2 − 4sin2 3θ (using 1 − 2sin2)
Reverse direction — collapse a θ pattern into a 2θ term
2 sin 4x cos 4x
→
sin 8x (angle doubles)
cos2 5θ − sin2 5θ
→
cos 10θ
Whenever you see “2 sin (something) cos (something)” with the same “something” in both, your antennae should ping — that’s a hidden sin double angle ready to collapse.
Using double angle formulas to solve equations
The whole point of these formulas, in exam terms, is to turn an equation with mixed angles (some 2θ, some θ) into one with a single angle. Once everything is in θ, you can solve like normal.
📍The golden rule: same angle, same ratio
You can’t solve an equation while it has different angles in it (like 2x and x) or different ratios (like sin and cos and tan). Use the double angle formula to reduce to one angle. Then use the Pythagorean identity if you also need to reduce to one ratio.
Two situations you’ll see
- Different angles, same kind of trig. Example: sin 2θ = sin θ. Use sin 2θ = 2 sin θ cos θ, then factorise.
- Different angles, different ratios. Example: cos 2θ + 3 sin θ = 2. Use the form of cos 2θ that matches the lone sin term — here that’s 1 − 2sin2θ.
Worked examples
WE 1Rewrite a single term using the sin double angle
Write sin 8x in terms of sin 4x and cos 4x.
sin 8x = ?
8x is double 4x, so the sin double angle works with θ = 4x.
Use the identity sin 2θ = 2 sin θ cos θ
Let θ = 4x: sin 2(4x) = 2 sin 4x cos 4x
So: sin 8x = 2 sin 4x cos 4x
sin 8x = 2 sin 4x cos 4x
the angle inside is always half of the original
WE 2Collapse an expression into a single trig term
Show that cos2 5θ − sin2 5θ = cos 10θ.
cos2 5θ − sin2 5θ
This matches the cos double angle pattern cos2A − sin2A = cos 2A.
Use the identity cos2A − sin2A = cos 2A
Let A = 5θ: cos2 5θ − sin2 5θ = cos 2(5θ)
Simplify: = cos 10θ
cos2 5θ − sin2 5θ = cos 10θ ✓
when the angle inside doubles, you’ve collapsed correctly
WE 3Solve a trig equation using sin double angle
Without using a calculator, solve the equation sin 2θ = sin θ for 0° ≤ θ ≤ 360°.
sin 2θ = sin θ
Different angles (2θ and θ) — replace sin 2θ first.
Use the identity sin 2θ = 2 sin θ cos θ
Substitute: 2 sin θ cos θ = sin θ
Move to one side: 2 sin θ cos θ − sin θ = 0
Factor sin θ: sin θ (2 cos θ − 1) = 0
Set each factor = 0:
sin θ = 0 → θ = 0°, 180°, 360°
cos θ = ½ → θ = 60°, 300°
θ = 0°, 60°, 180°, 300°, 360°
never just divide both sides by sin θ — you’d lose the sin θ = 0 solutions!
WE 4Solve an equation mixing cos 2θ and sin θ
Solve cos 2θ + 3 sin θ = 2 for 0° ≤ θ ≤ 360°.
cos 2θ + 3 sin θ = 2
Mixed angles. Use the form of cos 2θ that matches the sin term, so we end up in sin only.
Use the identity cos 2θ = 1 − 2sin2 θ
Substitute: (1 − 2sin2 θ) + 3 sin θ = 2
Move to one side: −2sin2 θ + 3 sin θ − 1 = 0
× −1: 2sin2 θ − 3 sin θ + 1 = 0
Factor: (2 sin θ − 1)(sin θ − 1) = 0
sin θ = ½ → θ = 30°, 150°
sin θ = 1 → θ = 90°
θ = 30°, 90°, 150°
picking the right form of cos 2θ saves a whole step!
WE 5Find the exact values of sin 2θ and cos 2θ
Given that sin θ = 35 and that θ is acute, find the exact values of sin 2θ and cos 2θ.
sin θ = 35, θ acute
First find cos θ using the Pythagorean identity, then plug into the double angle formulas.
Find cos θ: cos2 θ = 1 − 925 = 1625
θ acute → cos θ > 0: cos θ = 45
Use sin 2θ = 2 sin θ cos θ:
sin 2θ = 2 · 35 · 45 = 2425
Use cos 2θ = cos2 θ − sin2 θ:
cos 2θ = 1625 − 925 = 725
sin 2θ = 2425, cos 2θ = 725
check: 1 − 2sin2 θ = 1 − 18/25 = 7/25 ✓ same answer, different form
💡 Top tips
- All three formulas are in the formula booklet — don’t waste time trying to remember them, just check before you start.
- Pick the matching form of cos 2θ. If the rest of the equation has only sin, use 1 − 2sin2θ. If only cos, use 2cos2θ − 1. Match the language and the equation collapses cleanly.
- When you see “2 sin (something) cos (something)” with the same “something” in both, that’s a sin 2θ in disguise — collapse it.
- When you see “cos2(something) − sin2(something)”, that’s a cos 2θ in disguise — same trick.
- Never divide both sides by sin θ or cos θ when solving — you’ll lose solutions where that ratio is zero. Always factor instead.
- The angle inside halves when you go forward (sin 6θ → 2 sin 3θ cos 3θ) and doubles when you go backward (2 sin 3θ cos 3θ → sin 6θ).
- For “find exact values” questions, draw a quick right-angled triangle to get the missing ratio first, then plug into the double angle formula.
⚠ Common mistakes
- Writing sin 2θ = 2 sin θ. No — there’s a cos θ in there too. The full formula is sin 2θ = 2 sin θ cos θ.
- Picking the wrong form of cos 2θ. Using cos2θ − sin2θ when the rest of the equation is in sin only forces an extra step. Choose 1 − 2sin2θ instead.
- Dividing by sin θ to “simplify”. Equations like 2 sin θ cos θ = sin θ should be factored, not divided. Dividing throws away the sin θ = 0 solutions.
- Forgetting the angle inside halves. sin 6θ becomes 2 sin 3θ cos 3θ, not 2 sin 6θ cos 6θ. The angle inside is half of the outside.
- Sign errors when distributing. When substituting cos 2θ = 1 − 2sin2θ, bracket it: (1 − 2sin2θ). Otherwise the next step will go sideways.
- Stopping after finding one solution. Once you’ve factored, set each factor = 0 separately, and find every angle in the given range — including 0°, 180°, 360° if applicable.
- Mixing up cos 2θ with (cos θ)2. cos 2θ means cos of (2θ); cos2θ means (cos θ)2. They are completely different things.
Once double angles feel natural, you’ll start spotting them everywhere — in integration, in modelling, and in the upcoming “Quadratic Trig Equations” note. Practice the spotting more than the formula itself.
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