A-Level & IB Maths
A double angle formula rewrites a trig function of 2x in terms of x alone. There are three of them — one each for sin, cos and tan — and the cos one comes in three different forms, which is where almost everyone gets stuck.
Here are all of them first. Then this page shows where each one comes from, and the part most revision sites skip entirely: how to choose which form of cos 2x to use so you do not end up going round in circles.
The double angle formulae
sin 2x = 2 sin x cos x · cos 2x = cos²x − sin²x = 2cos²x − 1 = 1 − 2sin²x · tan 2x = 2 tan x ÷ (1 − tan²x)
Every double angle identity turns a function of 2x into something written only in terms of x. That matters because you cannot do much with sin 2x directly — you cannot differentiate it easily by parts, you cannot factorise it, and you cannot solve an equation that mixes sin 2x with sin x until they are in the same variable.
| Function | Double angle formula | Notes |
|---|---|---|
| sin 2x | 2 sin x cos x | Only one form. Nothing to choose. |
| cos 2x | cos²x − sin²x 2cos²x − 1 1 − 2sin²x | Three equal forms. Picking the right one is the whole skill. |
| tan 2x | 2 tan x1 − tan²x | Undefined wherever tan x = ±1, since the bottom becomes zero. |
You will see these called several things — the double angle formulas, a double angle identity, double angle trig identities, or trig double angle identities. They all mean the same three results above, so if you arrived here searching for one of those, you are in the right place.
The one sentence that explains all of them
Before the double angle results make any sense, you need the addition formulae — also called the compound angle formulae. These handle the sin, cos or tan of two angles added together or subtracted.
| Expression | Expands to |
|---|---|
| sin(A + B) | sin A cos B + cos A sin B |
| sin(A − B) | sin A cos B − cos A sin B |
| cos(A + B) | cos A cos B − sin A sin B |
| cos(A − B) | cos A cos B + sin A sin B |
| tan(A + B) | tan A + tan B1 − tan A tan B |
| tan(A − B) | tan A − tan B1 + tan A tan B |
The signs are the only thing people get wrong, and there are just two things to hold on to.
Watch the cos sign
cos(A + B) = cos A cos B − sin A sin B. The plus on the left becomes a minus on the right. Writing a plus here is the mistake that costs more marks in this topic than any other, because the working afterwards looks perfectly reasonable.
A quick way to check you have it the right way round: put A = B = 0°. Then cos(0 + 0) = cos 0 = 1, and the right-hand side gives (1)(1) − (0)(0) = 1. It matches. If you had written a plus you would still get 1 here, so instead try A = B = 90°: cos 180° = −1, and cos 90°cos 90° − sin 90°sin 90° = 0 − 1 = −1. Correct. With a plus you would have got +1, which is wrong.
Here is the step that makes the whole topic click. The double angle formulae are not separate results to learn — they are what you get when both angles in an addition formula are the same.
Write 2x as x + x and use the sin addition formula:
Both terms are the same thing, so they add together:
That is the sin 2x identity in full. It is the only one of the three with a single form.
Same idea with the cos addition formula, remembering the sign flips:
This is the base cos 2x identity, also written as the cosine double angle formula. The other two cos 2x identities come from it in the next section.
And the same again with the tan addition formula:
Three formulae, one idea: replace both A and B with x. If you can remember the addition formulae, you never need to memorise the double angle ones at all.
The derivation gave cos 2x = cos²x − sin²x. The other two forms come from substituting the Pythagorean identity into that one result:
sin²x + cos²x = 1
That identity can be rearranged two ways, and each one gives a different version of cos 2x.
Rearranging gives sin²x = 1 − cos²x. Substitute that in:
The other rearrangement gives cos²x = 1 − sin²x. Substitute that instead:
All three are the same number for every value of x. They are not alternatives that give different answers — they are the same result wearing different clothes. Which one you want depends entirely on what the rest of the question looks like.
This is the step that decides whether a question takes two lines or twenty. Most revision pages list the three forms and leave you to guess. The rule is actually simple, and it has nothing to do with the cos 2x term itself.
Look at the rest of the equation, not at cos 2x. Pick the form that makes everything match.
Scroll the diagram sideways →
Almost every double angle equation turns into a quadratic once there is only one trig function in it. If your equation has sin x in it and you pick the form with cos²x, you now have two different functions and no way forward. Pick the matching form and the equation collapses into something you already know how to solve.
Solve cos 2x + 3 sin x = 2 for 0° ≤ x ≤ 360°.
Had you chosen 2cos²x − 1 instead, you would have been left with cos²x and sin x together in one equation — two different functions, no factorising, and no route to an answer without starting again.
The double angle formula for tangent looks different from the other two because it is a fraction:
Unlike cos 2x, there is only one form, so there is nothing to choose. You will see it called the double angle formula for tangent, the double angle formula for tan, or the tangent double angle formula — all the same thing.
The denominator is 1 − tan²x, so the formula has no value whenever tan²x = 1, which means tan x = +1 or tan x = −1. Between 0° and 360° that happens at:
This is not a flaw in the formula. Check it against the original function: at x = 45°, tan 2x is tan 90°, which genuinely is undefined. The formula is simply telling you the truth about where the function does not exist.
An easy mark to lose
If a question asks you to solve an equation using tan 2x, check at the end whether any of your answers are at 45°, 135°, 225° or 315°. If one is, it has to be rejected — tan 2x does not exist there, so it cannot be a solution however neatly it came out of the algebra.
Because tan is sin over cos, you can always sidestep tan 2x entirely:
If a question is already written in sin and cos, this route often keeps everything in one place and avoids introducing tan at all.
There is one more thing these formulae are used for, and it catches people out because it runs the identity backwards. You cannot integrate sin²x or cos²x directly. But rearranging two of the cos 2x forms gives expressions you can integrate immediately.
Start from cos 2x = 1 − 2sin²x and make sin²x the subject:
And from cos 2x = 2cos²x − 1, make cos²x the subject:
The right-hand side of each is just a constant plus a cos term, and both of those integrate in one line. This is almost always what a question means when it says “using a suitable identity, find the integral of sin²x”.
Find ∫ sin²x dx.
Which direction you are running the identity
This is the natural next step after the addition formulae, and it is the part of the topic that tends to be buried at the back of a textbook. It is also where the addition formulae suddenly become useful rather than just something to memorise.
The problem it solves: an expression like 3 sin x + 4 cos x has two different trig terms, so you cannot solve it, cannot find its maximum, cannot do much at all. The harmonic form rewrites it as a single sine wave:
a sin x + b cos x = R sin(x + α)
Expand the right-hand side with the addition formula and compare it with the left:
Matching the sin x terms and the cos x terms on each side gives two equations:
Square both and add: a² + b² = R²(cos²α + sin²α) = R². Divide them instead: b ÷ a = tan α. So:
Which is exactly Pythagoras and SOH‑CAH‑TOA on a right-angled triangle with legs a and b:
Scroll the diagram sideways →
Write 3 sin x + 4 cos x in the form R sin(x + α), then state its maximum value.
Why harmonic form is worth the effort
Questions ask for different arrangements, and the sign inside the bracket follows the sign in the question. All four use the same R = √(a² + b²) and the same tan α = b/a.
| Starting expression | Rewrites as |
|---|---|
| a sin x + b cos x | R sin(x + α) |
| a sin x − b cos x | R sin(x − α) |
| a cos x + b sin x | R cos(x − α) |
| a cos x − b sin x | R cos(x + α) |
Notice the cos versions flip the sign. That is the same sign behaviour as the addition formulae, for the same reason — cos(A − B) is the one with the plus in it.
Find the exact value of sin 75° without a calculator.
If you need a reminder of the values used in step 3, our trig exact value chart has the full table and where each value comes from.
Solve sin 2x = sin x for 0° ≤ x ≤ 360°.
The mistake this question is testing
At step 2 it is tempting to divide both sides by sin x, leaving 2 cos x = 1. That gives x = 60° and 300° and looks like a finished answer — but you have just thrown away three solutions. Dividing by sin x is only valid when sin x is not zero, and the values where it is zero are themselves solutions. Always factorise instead of dividing. It is the single most common way to lose marks on this type of question.
Prove that (sin 2x) ÷ (1 + cos 2x) = tan x.
Step 2 is the whole question. Choosing 1 − 2sin²x instead would have given 1 + cos 2x = 2 − 2sin²x, which does not cancel with anything and leads nowhere.
| The mistake | What goes wrong | Do this instead |
|---|---|---|
| Dividing by sin x or cos x | Silently deletes solutions. Dividing sin 2x = sin x by sin x loses three of the five answers, and the working still looks correct. | Move everything to one side and factorise. Then set each bracket to zero. |
| cos(A + B) with a plus sign | Writing cos A cos B + sin A sin B. Every line after it is wrong, but nothing looks odd. | Remember cos flips the sign. Test with A = B = 90°: the answer must be −1. |
| Picking the wrong cos 2x form | You end up with sin and cos in the same equation and no way to factorise, so the question stalls. | Match the form to the rest of the equation so only one trig function survives. |
| Thinking sin 2x = 2 sin x | Not true for any angle except multiples of 180°. At x = 30°: sin 60° = 0.866 but 2 sin 30° = 1. | The 2 is inside the function, not a multiplier. Always expand to 2 sin x cos x. |
| Keeping invalid tan 2x answers | Solutions at 45°, 135°, 225° or 315° come out of the algebra, but tan 2x does not exist there. | Check each answer against the original equation and reject any that make a denominator zero. |
| α in the wrong quadrant | tan α = b/a has two solutions in a full turn, and a calculator only gives you one of them. | Keep α acute and let the form (sin or cos, plus or minus) carry the sign instead. |
Work each one on paper first, then tap to check. They build up from a straight recall to a full exam-style question. Every answer below was checked symbolically, so if yours disagrees, look again at your working.
Straight recall, or derive from sin(x + x)
sin 2x = 2 sin x cos x
The rest of the equation has 5 cos x, so only cos appears
use cos 2x = 2cos²x − 1
That leaves 2cos²x + 5cos x − 3 = 0, a quadratic in cos x
Substitute into the tan double angle formula
tan 2x = 2(3) / (1 − 3²)
= 6 / (1 − 9)
= 6 / (−8)
tan 2x = −3/4
Cos flips the sign, so a minus outside becomes a plus inside
cos(A − B) = cos A cos B + sin A sin B
Split into two known angles: 15° = 45° − 30°
cos(45° − 30°) = cos45cos30 + sin45sin30
= (√2/2)(√3/2) + (√2/2)(1/2)
= √6/4 + √2/4
cos 15° = (√6 + √2) / 4
About 0.9659. Same value as sin 75°, which makes sense since 15 and 75 add to 90
Only cos elsewhere, so use 2cos²x − 1
2cos²x − 1 + cos x = 0
2cos²x + cos x − 1 = 0
Factorise the quadratic
(2 cos x − 1)(cos x + 1) = 0
cos x = 1/2 gives 60° and 300°; cos x = −1 gives 180°
x = 60°, 180°, 300°
Numerator has 1 − cos2x, so pick the form 1 − 2sin²x
1 − (1 − 2sin²x) = 2sin²x
Denominator expands as usual
sin 2x = 2 sin x cos x
Divide and cancel the 2 and one sin x
= 2sin²x / (2 sin x cos x)
= sin x / cos x
= tan x, as required
You cannot integrate cos²x directly, so run the identity backwards
cos²x = (1 + cos 2x) / 2
∫ (1 + cos 2x)/2 dx
= ½ (x + ½ sin 2x) + C
= x/2 + (sin 2x)/4 + C
R is the hypotenuse of a 5, 12 right triangle
R = √(5² + 12²) = √169 = 13
alpha comes from tan alpha = b/a
tan α = 12/5, so α = 67.4°
5 sin x + 12 cos x = 13 sin(x + 67.4°)
Maximum is R, since sine never exceeds 1
maximum = 13, at x = 22.6°
Expand sin 2x, then move everything to one side
2 sin x cos x = cos x
2 sin x cos x − cos x = 0
Factorise — do NOT divide by cos x
cos x (2 sin x − 1) = 0
cos x = 0 gives 90° and 270°; sin x = 1/2 gives 30° and 150°
x = 30°, 90°, 150°, 270°
Dividing by cos x would have lost 90° and 270°
A double angle formula rewrites a trig function of 2x in terms of x alone. There are three: sin 2x = 2 sin x cos x, cos 2x = cos²x − sin²x, and tan 2x = 2 tan x ÷ (1 − tan²x). They are not separate facts to learn — each one is an addition formula with both angles set to the same value.
sin 2x = 2 sin x cos x. cos 2x = cos²x − sin²x, which can also be written 2cos²x − 1 or 1 − 2sin²x. tan 2x = 2 tan x ÷ (1 − tan²x). Sin and tan have one form each; only cos has three.
Because the Pythagorean identity sin²x + cos²x = 1 can be substituted into cos²x − sin²x in two different ways. Replacing sin²x with 1 − cos²x gives 2cos²x − 1. Replacing cos²x with 1 − sin²x gives 1 − 2sin²x. All three forms are equal for every value of x — they are the same result written differently.
Look at the rest of the equation, not at the cos 2x. If only sin x appears elsewhere, use 1 − 2sin²x. If only cos x appears, use 2cos²x − 1. If you need to factorise, use cos²x − sin²x, which is a difference of two squares. The goal is always to end up with just one trig function so the equation becomes a quadratic.
tan 2x = 2 tan x ÷ (1 − tan²x). It comes from the tan addition formula with A and B both set to x. It is undefined wherever tan²x = 1, which means at 45°, 135°, 225° and 315°, because the denominator becomes zero there.
sin 2x = 2 sin x cos x. It comes from sin(x + x) = sin x cos x + cos x sin x, where the two terms are identical and so add together. There is only one form, so there is nothing to choose between.
cos 2x = cos²x − sin²x, and equally 2cos²x − 1 or 1 − 2sin²x. The first comes from cos(x + x) = cos x cos x − sin x sin x. The other two come from substituting sin²x + cos²x = 1 in each of the two possible directions.
No. This is one of the most common errors in the topic. The 2 is inside the function, not a multiplier outside it. At x = 30°, sin 2x means sin 60° = 0.866, while 2 sin x means 2 × 0.5 = 1. They only happen to agree where sin x is zero or cos x is 1.
You need to be able to produce them, but memorising is not the only route. Every one is an addition formula with A = B, and the addition formulae are printed in most formula booklets. Deriving sin 2x from sin(x + x) takes about fifteen seconds, which is a useful safety net if your memory goes under pressure.
They expand the sin, cos or tan of two angles combined. sin(A ± B) = sin A cos B ± cos A sin B. cos(A ± B) = cos A cos B ∓ sin A sin B, where the sign flips. tan(A ± B) = (tan A ± tan B) ÷ (1 ∓ tan A tan B). They are also called the compound angle formulae.
It is a memory aid for the cos addition formula: cos(A + B) = cos A cos B − sin A sin B. The terms appear in exactly that order — cos times cos, then sin times sin. Crucially the sign between them flips, so a plus on the left becomes a minus on the right.
It rewrites a sin x + b cos x as a single wave, R sin(x + α), where R = √(a² + b²) and tan α = b ÷ a. It is also called harmonic form or the wave function. It matters because two trig terms added together cannot easily be solved or maximised, but one can.
Expand R sin(x + α) with the addition formula to get R sin x cos α + R cos x sin α, then compare it with a sin x + b cos x. Matching terms gives a = R cos α and b = R sin α. Squaring and adding gives R = √(a² + b²); dividing gives tan α = b ÷ a. It is Pythagoras and SOH-CAH-TOA on a triangle with legs a and b.
Write it in harmonic form R sin(x + α). Since a sine wave never goes above 1 or below −1, the maximum is R and the minimum is −R, where R = √(a² + b²). For 5 sin x + 12 cos x, R = √169 = 13, so the maximum is 13. No differentiation is needed.
Compound angle formulae, also called addition formulae, handle two different angles: sin(A + B), cos(A − B) and so on. Double angle formulae are the special case where both angles are the same, so A + B becomes x + x = 2x. Every double angle result is a compound angle result with A = B.
Wherever the denominator 1 − tan²x equals zero, which means tan x = +1 or tan x = −1. Between 0° and 360° that is at 45°, 135°, 225° and 315°. This is not a flaw in the formula: at x = 45°, tan 2x is tan 90°, which genuinely does not exist. If a solution lands on one of these angles it must be rejected.
Run the double angle identity backwards. Rearranging cos 2x = 1 − 2sin²x gives sin²x = (1 − cos 2x) ÷ 2, and rearranging cos 2x = 2cos²x − 1 gives cos²x = (1 + cos 2x) ÷ 2. Both right-hand sides are a constant plus a cos term, which integrate in one line. So ∫ sin²x dx = x/2 − (sin 2x)/4 + C.
1 + tan²x = sec²x. It comes from dividing sin²x + cos²x = 1 through by cos²x. It is worth knowing alongside the double angle formulae because questions often mix them, and it is the quickest way to convert between tan and sec in an equation.
Double angle questions rarely say “use the double angle formula”. They say solve, prove, integrate or find the maximum — and the formula is the move that unlocks it. What examiners are really testing is whether you can spot which identity turns an unworkable expression into a workable one.
Everything above, in seven lines
If you are still building the foundations underneath this, start with the trig exact value chart — several of the questions above need those values, and the addition formulae are much easier once the basic chart is automatic.
Trig identities are unforgiving — a plus where a minus should be, or dividing where you should factorise, and correct method still scores zero. Our tutors teach IB and A‑level maths and work through your own past papers, so the slips we fix are the ones you are actually making.
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