The double angle formulae are sin 2A = 2 sin A cos A, cos 2A = cos²A − sin²A (also written as 1 − 2 sin²A or 2 cos²A − 1) and tan 2A = 2 tan A / (1 − tan²A). They come from the addition formulae with B = A, and you use them to solve equations that mix x and 2x, prove identities, and integrate sin²x and cos²x.
The double angle formulae
- sin 2A = 2 sin A cos A
- cos 2A = cos²A − sin²A
- cos 2A = 1 − 2 sin²A
- cos 2A = 2 cos²A − 1
- tan 2A = 2 tan A / (1 − tan²A)
Rearranged (power-reduction) forms
- sin²A = (1 − cos 2A) / 2
- cos²A = (1 + cos 2A) / 2
They work in reverse too. 2 sin θ cos θ is always sin 2θ, whatever θ is, so 2 sin 3x cos 3x = sin 6x.
How are the double angle formulae derived?
They are the addition formulae with both angles equal. Put B = A into sin(A + B) = sin A cos B + cos A sin B and you get sin(A + A) = sin A cos A + cos A sin A, which is sin 2A = 2 sin A cos A. Doing the same with cos(A + B) = cos A cos B − sin A sin B gives cos 2A = cos²A − sin²A, and the tan addition formula gives tan 2A = 2 tan A / (1 − tan²A).
That means there is nothing new to memorise. If you can write down the addition formulae, you can derive all three double angle formulae in a few seconds. Learn to spot them in reverse as well: whenever you see 2 sin θ cos θ, you are looking at sin 2θ.
Which form of cos 2x should you use?
Start from cos 2A = cos²A − sin²A and use sin²A + cos²A = 1 to replace one of the two terms. That gives three equivalent versions:
- cos 2A = cos²A − sin²A, the raw form
- cos 2A = 1 − 2 sin²A, where everything becomes sine
- cos 2A = 2 cos²A − 1, where everything becomes cosine
All three are always true, so the question is never which one is correct. It is which one is useful. The rule is to look at the other terms in the equation and match the form to them. An equation with a stray sin x needs the sine form; one with a stray cos x needs the cosine form. Either way, the equation collapses into a quadratic in a single trig function, which you already know how to solve. Choose the wrong form and you are left with both sin x and cos x in the same equation, with no way forward.
How do you integrate sin²x and cos²x?
There is no rule for integrating a squared trig function as it stands. Instead, rearrange two of the cos 2A forms to make the squared term the subject:
- from cos 2A = 2 cos²A − 1, you get cos²A = (1 + cos 2A) / 2
- from cos 2A = 1 − 2 sin²A, you get sin²A = (1 − cos 2A) / 2
Now every term can be integrated. For example, ∫ sin²x dx = ∫ (½ − ½ cos 2x) dx = x/2 − (sin 2x)/4 + c.
These are called power-reduction forms because they turn a square into an expression in cos 2x with no powers. They are the reason double angle formulae turn up so often in integration questions. Keep the signs straight: cos²A takes the plus, sin²A takes the minus.
Solving double angle equations: factorise, never divide
After expanding sin 2x = 2 sin x cos x, you often end up with a common factor, as in 2 sin x cos x = sin x. Dividing both sides by sin x quietly throws away every solution where sin x = 0, and those solutions are worth marks. Instead, bring everything to one side and factorise: sin x(2 cos x − 1) = 0. Two brackets give two sets of solutions, and you need both.
Triple angle formulae and “prove, then hence solve” questions
There is no separate triple angle formula to learn. Write 3x as 2x + x, apply the addition formula, then use the double angle formulae. That route gives sin 3x = 3 sin x − 4 sin³x, worked through in Example 4 below.
The typical Pure 3 question has two parts. Part (a) asks you to prove an identity using the double angle formulae, and part (b) says “hence solve”. Always use your part (a) result in part (b). Starting again from scratch usually runs out of road, and the method marks are for using the result.
Worked examples
Example 1. Given that sin A = 3/5 and A is acute, find the exact value of sin 2A
- Find cos A. The 3, 4, 5 triangle gives cos A = 4/5, which is positive because A is acute.
- Use sin 2A = 2 sin A cos A.
- Substitute: sin 2A = 2 × 3/5 × 4/5 = 24/25.
Answer: sin 2A = 24/25.
Marks: M1 for finding cos A, A1 for the value.
Example 2. Solve cos 2x + 3 sin x = 2 for 0° ≤ x < 360°
- Look at the other terms. There is a 3 sin x, so use the all-sine form: cos 2x = 1 − 2 sin²x.
- Substitute: (1 − 2 sin²x) + 3 sin x = 2.
- Rearrange into a quadratic: 2 sin²x − 3 sin x + 1 = 0.
- Factorise: (2 sin x − 1)(sin x − 1) = 0, so sin x = ½ or sin x = 1.
- sin x = ½ gives x = 30° and x = 150°.
- sin x = 1 gives just one solution, x = 90°, where the sine curve touches its maximum.
Answer: x = 30°, 90°, 150°.
Marks: M1 for choosing the sine form, M1 for forming the quadratic, A1 for all three solutions.
Example 3. Solve sin 2x = sin x for 0° ≤ x < 360°
- Expand the double angle: 2 sin x cos x = sin x.
- Bring everything to one side: 2 sin x cos x − sin x = 0.
- Factorise, without dividing by sin x: sin x(2 cos x − 1) = 0.
- sin x = 0 gives x = 0° and x = 180°.
- 2 cos x − 1 = 0 gives cos x = ½, so x = 60° and x = 300°.
Answer: x = 0°, 60°, 180°, 300°. Dividing by sin x at step 2 would have lost 0° and 180°.
Marks: M1 for expanding, M1 for factorising, A1 for all four solutions.
Example 4. Show that sin 3x ≡ 3 sin x − 4 sin³x
- Split the angle: sin 3x = sin(2x + x).
- Use the addition formula: sin(2x + x) = sin 2x cos x + cos 2x sin x.
- Use the double angle formulae. Replace sin 2x with 2 sin x cos x, and choose the all-sine form for cos 2x, 1 − 2 sin²x, because the result must be in terms of sin x only.
- This gives (2 sin x cos x) cos x + (1 − 2 sin²x) sin x = 2 sin x cos²x + sin x − 2 sin³x.
- Replace cos²x with 1 − sin²x: 2 sin x(1 − sin²x) + sin x − 2 sin³x = 2 sin x − 2 sin³x + sin x − 2 sin³x.
- Collect terms: 3 sin x − 4 sin³x, as required.
Marks: M1 for splitting 3x into 2x + x, M1 for using the double angle formulae, A1 for reaching the given result.
Common mistakes
- Dividing by sin x or cos x. This loses every solution where that function equals zero. Factorise instead.
- Choosing the wrong form of cos 2x. If you end up with sin x and cos x in the same equation, go back and pick the form that matches the other terms.
- Writing sin 2x as 2 sin x. Doubling the angle does not double the value: sin 2x = 2 sin x cos x.
- Swapping the signs in the power-reduction forms. cos²A = (1 + cos 2A) / 2 and sin²A = (1 − cos 2A) / 2. Getting these the wrong way round gives a wrong integral.
- Using the wrong range for 2x. When solving an equation in 2x for 0° ≤ x < 360°, find every value of 2x in 0° ≤ 2x < 720° before halving.
- Keeping an impossible root. If a quadratic gives something like sin x = 2 or cos x = −3, reject it and say why: sine and cosine always lie between −1 and 1.
- Ignoring part (a) in a “hence” question. The marks are for using the identity you have just proved.
How the double angle formulae are examined
In Edexcel International A-Level, the double angle formulae are Pure 3 content (paper WMA13), in the trigonometric addition formulae chapter. In Edexcel A-Level (9MA0) they are part of the Year 2 Pure content. Typical questions ask you to:
- find an exact value such as sin 2A or tan 2A from a given trig ratio
- show that an equation can be written as a quadratic in sin x or cos x, then hence solve it
- write sin²x or cos²x in the form a + b cos 2x, then hence integrate it
- prove an identity using the double angle formulae, then use it to solve an equation
Practise with the Pure 3 trigonometric addition formulae exam questions (with mark schemes), the Pure 3 trigonometric formulae worksheet, or generate unlimited questions in the A-Level trigonometry revision engine.
This topic builds on the addition formulae. It leads into the R sin(θ + α) form, further trig equations and integration of trig functions.
Frequently asked questions
What is the double angle formula for cos 2x?
There are three equivalent forms: cos 2x = cos²x − sin²x = 1 − 2 sin²x = 2 cos²x − 1. Use whichever one matches the other terms in your equation.
Are the double angle formulae in the formula booklet?
The Edexcel A-Level (9MA0) formula booklet gives the addition formulae but not the double angle formulae, so you need to learn them. If one slips your mind in the exam, set B = A in the matching addition formula and you can derive it in a line.
What is the difference between the addition formulae and the double angle formulae?
The addition formulae give sin, cos and tan of A + B and A − B. The double angle formulae are the special case where the two angles are equal, B = A.
How do you find tan 2x?
Use tan 2x = 2 tan x / (1 − tan²x). If you already know sin 2x and cos 2x, you can also divide them, because tan 2x = sin 2x / cos 2x.
Can the double angle formulae give half angle results?
Yes. Replace A with A/2 in the cos 2A forms: cos A = 1 − 2 sin²(A/2) = 2 cos²(A/2) − 1. Rearranging these gives sin²(A/2) and cos²(A/2) in terms of cos A.