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# Cos(A - B) Formula — Proof, Examples, Identity

### TL;DR

The cos(A - B) formula states that cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin Bcos(A−B)=cosAcosB+sinAsinB, the cosine difference identity. This article gives the formula, its unit-circle proof, why the sign is a plus (the opposite of cos(A+B)), six worked examples in degrees and radians, the most common sign mistake, and FAQs.

### The Identity That Builds Unfamiliar Angles From Familiar Ones

You know the cosine of 45° and the cosine of 30° cold — but the cosine of 15° isn't on any memorized table, and the **cos(A - B) formula** is exactly the tool that turns the two angles you know into the one you don't.

The **cos(A - B) formula** — also called the **cosine difference identity** — says:

;cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B;\boxed{;\cos(A - B) = \cos A \cos B + \sin A \sin B;};cos(A−B)=cosAcosB+sinAsinB;

It is one of the [sum and difference formulas](/content/math/trigonometry/sum-and-difference-identities/index.html) of trigonometry, and the plus sign between the two products is its signature — the detail students most often get backwards.

### What Is the Cos(A - B) Identity?

The cosine of a difference of two angles equals the product of their cosines **plus** the product of their sines. In symbols,  cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin Bcos(A−B)=cosAcosB+sinAsinB. It holds for all real angles AAA and BBB, in degrees or radians.

The point of the identity is decomposition: when an angle can be written as the difference of two angles you already know — 15°=45°−30°15° = 45° - 30°15°=45°−30°, or π12=π4−π6\dfrac{\pi}{12} = \dfrac{\pi}{4} - \dfrac{\pi}{6}12π​=4π​−6π​ — the formula gives its exact cosine without a calculator. It is the companion of cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin Bcos(A+B)=cosAcosB−sinAsinB, and the sign flip between the two is the whole story.

### How Is the Cos(A - B) Formula Proved?

The cleanest proof uses the unit circle and the fact that rotating two points together leaves the distance between them unchanged.

#### The unit-circle distance proof.
Put two points on the unit circle: P=(cos⁡A,sin⁡A)P = (\cos A, \sin A)P=(cosA,sinA) at angle AAA, and Q=(cos⁡B,sin⁡B)Q = (\cos B, \sin B)Q=(cosB,sinB) at angle BBB. The squared distance between them is:

∣PQ∣2=(cos⁡A−cos⁡B)2+(sin⁡A−sin⁡B)2.\|PQ\|^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2.

Expand and use sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1sin2θ+cos2θ=1 on each point:

∣PQ∣2=2−2(cos⁡Acos⁡B+sin⁡Asin⁡B).\|PQ\|^2 = 2 - 2(\cos A \cos B + \sin A \sin B).

Now rotate both points clockwise by BBB. Distance is preserved, and the points become P′=(cos⁡(A−B),sin⁡(A−B))P' = (\cos(A - B), \sin(A - B))P′=(cos(A−B),sin(A−B)) and Q′=(1,0)Q' = (1, 0)Q′=(1,0):

∣P′Q′∣2=(cos⁡(A−B)−1)2+sin⁡2(A−B)=2−2cos⁡(A−B).\|P'Q'\|^2 = (\cos(A - B) - 1)^2 + \sin^2(A - B) = 2 - 2\cos(A - B).

Set the two squared distances equal and cancel:

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B.\cos(A - B) = \cos A \cos B + \sin A \sin B.

#### A shortcut from cos(A + B).
If you already trust the sum formula, write cos⁡(A−B)=cos⁡(A+(−B))\cos(A - B) = \cos(A + (-B))cos(A−B)=cos(A+(−B)). Since cos⁡(−B)=cos⁡B\cos(-B) = \cos Bcos(−B)=cosB and sin⁡(−B)=−sin⁡B\sin(-B) = -\sin Bsin(−B)=−sinB:

cos⁡(A+(−B))=cos⁡Acos⁡B−sin⁡A(−sin⁡B)=cos⁡Acos⁡B+sin⁡Asin⁡B.\cos(A + (-B)) = \cos A \cos B - \sin A (-\sin B) = \cos A \cos B + \sin A \sin B.

The minus inside flips the sign of the sine term, turning the sum formula's −-− into the difference formula's +++.

### Double-Anchoring — Right Triangle and Unit Circle

The formula reads two ways, and seeing both anchors it.

#### From the unit circle.
cos⁡(A−B)\cos(A - B)cos(A−B) is the cosine of the angle _between_ the radii at AAA and BBB — the xxx-component of one point projected onto the other's direction. The dot product of the two unit vectors (cos⁡A,sin⁡A)(\cos A, \sin A)(cosA,sinA) and (cos⁡B,sin⁡B)(\cos B, \sin B)(cosB,sinB) is exactly cos⁡Acos⁡B+sin⁡Asin⁡B\cos A \cos B + \sin A \sin BcosAcosB+sinAsinB, and a dot product of unit vectors equals the cosine of the angle between them, A−BA - BA−B. The algebra and the geometry are the same statement.

#### From the right triangle.
For acute AAA and BBB with A>BA > BA>B, build adjacent right triangles sharing a side. Project the legs and the formula's two products — cos⁡Acos⁡B\cos A \cos BcosAcosB and sin⁡Asin⁡B\sin A \sin BsinAsinB — appear as the horizontal pieces that recombine into the adjacent side of the angle A−BA - BA−B. The triangle proof handles the intuitive acute case; the unit-circle proof handles the full real domain.

For a concrete anchor: at A=60°A = 60°A=60°, B=30°B = 30°B=30°, the formula gives cos⁡30°=cos⁡60°cos⁡30°+sin⁡60°sin⁡30°=12⋅32+32⋅12=32\cos 30° = \cos 60°\cos 30° + \sin 60°\sin 30° = \tfrac{1}{2}\cdot\tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2} = \tfrac{\sqrt{3}}{2}cos30°=cos60°cos30°+sin60°sin30°=21​⋅23​​+23​​⋅21​=23​​, which matches the known cos⁡30°\cos 30°cos30°.

### Examples of the Cos(A - B) Formula

#### Example 1
**Compute cos⁡15°\cos 15°cos15° exactly using cos⁡(45°−30°)\cos(45° - 30°)cos(45°−30°).**
Apply the difference formula:

cos⁡(45°−30°)=cos⁡45°cos⁡30°+sin⁡45°sin⁡30°.\cos(45° - 30°) = \cos 45° \cos 30° + \sin 45° \sin 30°.cos(45°−30°)=cos45°cos30°+sin45°sin30°.
Substitute cos⁡45°=sin⁡45°=22\cos 45° = \sin 45° = \dfrac{\sqrt{2}}{2}cos45°=sin45°=22​​, cos⁡30°=32\cos 30° = \dfrac{\sqrt{3}}{2}cos30°=23​​, sin⁡30°=12\sin 30° = \dfrac{1}{2}sin30°=21​:

cos⁡15°=22⋅32+22⋅12=6+24.\cos 15° = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}.cos15°=22​​⋅23​​+22​​⋅21​=46​+2​​.

In radians, 15°=π12=π4−π615° = \dfrac{\pi}{12} = \dfrac{\pi}{4} - \dfrac{\pi}{6}15°=12π​=4π​−6π​.

**Final answer:** cos⁡15°=6+24\cos 15° = \dfrac{\sqrt{6} + \sqrt{2}}{4}cos15°=46​+2​​.

#### Example 2
**Compute cos⁡(60°−30°)\cos(60° - 30°)cos(60°−30°).**
_Wrong attempt._ A student copies the sign from the angle operation: the angle has a minus, so they write the expansion with a minus too — cos⁡60°cos⁡30°−sin⁡60°sin⁡30°\cos 60° \cos 30° - \sin 60° \sin 30°cos60°cos30°−sin60°sin30° — and compute 12⋅32−32⋅12=0\tfrac{1}{2}\cdot\tfrac{\sqrt{3}}{2} - \tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2} = 021​⋅23​​−23​​⋅21​=0. They report cos⁡30°=0\cos 30° = 0cos30°=0.

_The break._ cos⁡30°\cos 30°cos30° is 32≈0.87\dfrac{\sqrt{3}}{2} \approx 0.8723​​≈0.87, not 000. The result 000 would mean 30° is a right angle, which it plainly is not. What went wrong: that minus sign belongs to the _sum_ formula, cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin Bcos(A+B)=cosAcosB−sinAsinB. The student accidentally computed cos⁡90°\cos 90°cos90°, which really is 000.

_Correct._ The cosine _difference_ formula carries a **plus**:

cos⁡(60°−30°)=cos⁡60°cos⁡30°+sin⁡60°sin⁡30°=34+34=32.\cos(60° - 30°) = \cos 60° \cos 30° + \sin 60° \sin 30° = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}.cos(60°−30°)=cos60°cos30°+sin60°sin30°=43​​+43​​=23​​.

**Final answer:** cos⁡30°=32\cos 30° = \dfrac{\sqrt{3}}{2}cos30°=23​​.

#### Example 3
**Find cos⁡75°\cos 75°cos75° as a difference, using cos⁡(105°−30°)\cos(105° - 30°)cos(105°−30°).**

cos⁡(105°−30°)=cos⁡105°cos⁡30°+sin⁡105°sin⁡30°.\cos(105° - 30°) = \cos 105° \cos 30° + \sin 105° \sin 30°.cos(105°−30°)=cos105°cos30°+sin105°sin30°.

With cos⁡105°=2−64\cos 105° = \dfrac{\sqrt{2}-\sqrt{6}}{4}cos105°=42​−6​​ and sin⁡105°=6+24\sin 105° = \dfrac{\sqrt{6}+\sqrt{2}}{4}sin105°=46​+2​​:

cos⁡75°=2−64⋅32+6+24⋅12=6−24.\cos 75° = \frac{\sqrt{2}-\sqrt{6}}{4}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{6}+\sqrt{2}}{4}\cdot\frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}.cos75°=42​−6​​⋅23​​+46​+2​​⋅21​=46​−2​​.

In radians, 75°=5π1275° = \dfrac{5\pi}{12}75°=125π​.

**Final answer:** cos⁡75°=6−24\cos 75° = \dfrac{\sqrt{6} - \sqrt{2}}{4}cos75°=46​−2​​.

#### Example 4
**Given cos⁡A=45\cos A = \dfrac{4}{5}cosA=54​, sin⁡A=35\sin A = \dfrac{3}{5}sinA=53​, cos⁡B=1213\cos B = \dfrac{12}{13}cosB=1312​, sin⁡B=513\sin B = \dfrac{5}{13}sinB=135​, find cos⁡(A−B)\cos(A - B)cos(A−B).**

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B=45⋅1213+35⋅513=48+1565=6365.\cos(A - B) = \cos A \cos B + \sin A \sin B = \frac{4}{5}\cdot\frac{12}{13} + \frac{3}{5}\cdot\frac{5}{13} = \frac{48 + 15}{65} = \frac{63}{65}.cos(A−B)=cosAcosB+sinAsinB=54​⋅1312​+53​⋅135​=6548+15​=6563​.

**Final answer:** 6365\dfrac{63}{65}6563​.

#### Example 5
**Simplify cos⁡(A−B)cos⁡B−sin⁡(A−B)sin⁡B\cos(A - B)\cos B - \sin(A - B)\sin Bcos(A−B)cosB−sin(A−B)sinB.**

This matches the _sum_ pattern cos⁡Xcos⁡Y−sin⁡Xsin⁡Y=cos⁡(X+Y)\cos X \cos Y - \sin X \sin Y = \cos(X + Y)cosXcosY−sinXsinY=cos(X+Y) with X=A−BX = A - BX=A−B and Y=BY = BY=B:

cos⁡((A−B)+B)=cos⁡A.\cos((A - B) + B) = \cos A.

The two difference and sum identities are inverse moves — applying one and then the other returns the original angle.

**Final answer:** cos⁡A\cos AcosA.

#### Example 6
**Use the formula to derive cos⁡2A\cos 2Acos2A from cos⁡(A−(−A))\cos(A - (-A))cos(A−(−A))... and explain why it gives 111 for the wrong reason, then the right identity.**

Setting B=−AB = -AB=−A in the difference formula gives cos⁡(A−(−A))=cos⁡(2A)\cos(A - (-A)) = \cos(2A)cos(A−(−A))=cos(2A)... but the formula expects a difference. Instead, treat it cleanly: cos⁡(A−A)=cos⁡0=1\cos(A - A) = \cos 0 = 1cos(A−A)=cos0=1. By the formula, cos⁡Acos⁡A+sin⁡Asin⁡A=cos⁡2A+sin⁡2A=1\cos A \cos A + \sin A \sin A = \cos^2 A + \sin^2 A = 1cosAcosA+sinAsinA=cos2A+sin2A=1. The identity recovers the Pythagorean relation as a special case.

**Final answer:** cos⁡(A−A)=cos⁡2A+sin⁡2A=1\cos(A - A) = \cos^2 A + \sin^2 A = 1cos(A−A)=cos2A+sin2A=1.

### Where the Cosine Difference Identity Earns Its Keep

The formula is the algebra behind comparing two directions or two phases.

- **Angle between vectors.** The cosine of the angle between two unit vectors is their dot product, cos⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)\cos A \cos B + \sin A \sin B = \cos(A - B)cosAcosB+sinAsinB=cos(A−B) — the foundation of lighting in 3D graphics, where the angle between a surface and a light source sets pixel brightness.

- **Wave interference and phase.** When two waves of the same frequency but different phases AAA and BBB overlap, the cosine of their phase difference cos⁡(A−B)\cos(A - B)cos(A−B) controls whether they reinforce or cancel — the core of interferometry and noise-cancelling audio.

- **GPS and signal correlation.** Matched-filter detection multiplies a received signal against a reference and sums; the cosine difference identity is what turns that product into a phase-alignment measure.

- **Astronomy.** The angular separation of two stars on the celestial sphere reduces to a cosine-difference expression in their coordinates.

Anywhere two angles need comparing rather than just measuring, this identity is the engine underneath.

### The Mathematicians Behind the Cos(A - B) Formula

**Claudius Ptolemy** (c. 100–170 CE, Greco-Egyptian) encoded the angle-difference relationship geometrically in the _Almagest_ through his chord theorem, which underlies every modern sum and difference identity. His chord tables powered astronomical prediction for over a millennium.

**Bhaskara II** (1114–1185, India) worked with the sine and cosine of angle sums and differences in the _Siddhanta-Shiromani_ (1150), treating them as practical tools for the astronomical calculations of his era.

### Where Cos(A - B) Goes Sideways

#### Mistake 1: Using a minus sign instead of a plus

**Where it slips in:** A reader copies the sign from the angle's subtraction into the expansion, writing cos⁡Acos⁡B−sin⁡Asin⁡B\cos A \cos B - \sin A \sin BcosAcosB−sinAsinB for cos⁡(A−B)\cos(A - B)cos(A−B).

**Don't do this:** Match the formula's sign to the sign between the angles.

**The correct way:** For cosine, the sign is _opposite_: cos⁡(A−B)\cos(A - B)cos(A−B) takes a **plus**, cos⁡(A+B)\cos(A + B)cos(A+B) takes a **minus**.

#### Mistake 2: Swapping the cosine and sine pairings

**Where it slips in:** A reader writes cos⁡Asin⁡B+sin⁡Acos⁡B\cos A \sin B + \sin A \cos BcosAsinB+sinAcosB — the _sine_ difference pattern — for the cosine formula.

**Don't do this:** Mix the cosine-product and sine-product terms, or borrow the sine formula's cross structure.

**The correct way:** Cosine pairs like with like: cos⁡Acos⁡B\cos A \cos BcosAcosB (both cosines) plus sin⁡Asin⁡B\sin A \sin BsinAsinB (both sines). The cross-paired form sin⁡Acos⁡B±cos⁡Asin⁡B\sin A \cos B \pm \cos A \sin BsinAcosB±cosAsinB belongs to sin⁡(A±B)\sin(A \pm B)sin(A±B) — a different identity entirely.

#### Mistake 3: Mixing degrees and radians in one evaluation

**Where it slips in:** A reader sets up cos⁡(π4−π6)\cos\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right)cos(4π​−6π​) but plugs in calculator values read in degree mode.

**Don't do this:** Switch angle units between substitution and evaluation.

**The correct way:** Hold one unit throughout. cos⁡(45°−30°)=cos⁡(π4−π6)\cos(45° - 30°) = \cos\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right)cos(45°−30°)=cos(4π​−6π​) — the same number in two notations — so state the unit beside the answer and match the calculator mode to it. The second-guesser who recomputes in the other unit "just to check" is the one who introduces the mismatch.

### Key Takeaways

- The **cos(A - B) formula** is cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin Bcos(A−B)=cosAcosB+sinAsinB — the cosine difference identity.

- The sign is a **plus**, the opposite of the minus in cos⁡(A+B)\cos(A + B)cos(A+B) — the detail most often reversed.

- The unit-circle distance proof is the standard derivation; the dot-product reading anchors it geometrically.

- It builds exact values for angles like 15° and 75° from familiar ones, and it measures the angle between two directions.

- The most common mistake is copying the angle's minus into the expansion instead of using the plus.