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

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:

$$\boxed{;\cos(A - B) = \cos A \cos B + \sin A \sin B;}$$

It is one of the sum and difference formulas 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, $\cos(A - B) = \cos A \cos B + \sin A \sin B$. It holds for all real angles $A$ and $B$, 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 — $15° = 45° - 30°$, or $\dfrac{\pi}{12} = \dfrac{\pi}{4} - \dfrac{\pi}{6}$ — the formula gives its exact cosine without a calculator. It is the companion of $\cos(A + B) = \cos A \cos B - \sin A \sin B$, 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: $P = (\cos A, \sin A)$ at angle $A$, and $Q = (\cos B, \sin B)$ at angle $B$. The squared distance between them is:

$$|PQ|^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2.$$

Expand and use $\sin^2\theta + \cos^2\theta = 1$ on each point:

$$|PQ|^2 = 2 - 2(\cos A \cos B + \sin A \sin B).$$

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

$$|P'Q'|^2 = (\cos(A - B) - 1)^2 + \sin^2(A - B) = 2 - 2\cos(A - B).$$

Set the two squared distances equal and cancel:

$$\cos(A - B) = \cos A \cos B + \sin A \sin B.$$

A shortcut from cos(A + B). If you already trust the sum formula, write $\cos(A - B) = \cos(A + (-B))$. Since $\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$:

$$\cos(A + (-B)) = \cos A \cos B - \sin A (-\sin B) = \cos A \cos B + \sin A \sin B.$$

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. $\cos(A - B)$ is the cosine of the angle between the radii at $A$ and $B$ — the $x$-component of one point projected onto the other's direction. The dot product of the two unit vectors $(\cos A, \sin A)$ and $(\cos B, \sin B)$ is exactly $\cos A \cos B + \sin A \sin B$, and a dot product of unit vectors equals the cosine of the angle between them, $A - B$. The algebra and the geometry are the same statement.

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

For a concrete anchor: at $A = 60°$, $B = 30°$, the formula gives $\cos 30° = \cos 60°\cos 30° + \sin 60°\sin 30° = \tfrac{1}{2}\cdot\tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2} = \tfrac{\sqrt{3}}{2}$, which matches the known $\cos 30°$.

Examples of the Cos(A - B) Formula

Example 1

Compute $\cos 15°$ exactly using $\cos(45° - 30°)$.

Apply the difference formula:

$$\cos(45° - 30°) = \cos 45° \cos 30° + \sin 45° \sin 30°.$$

Substitute $\cos 45° = \sin 45° = \dfrac{\sqrt{2}}{2}$, $\cos 30° = \dfrac{\sqrt{3}}{2}$, $\sin 30° = \dfrac{1}{2}$:

$$\cos 15° = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}.$$

In radians, $15° = \dfrac{\pi}{12} = \dfrac{\pi}{4} - \dfrac{\pi}{6}$.

Final answer: $\cos 15° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.

Example 2

Compute $\cos(60° - 30°)$.

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 — $\cos 60° \cos 30° - \sin 60° \sin 30°$ — and compute $\tfrac{1}{2}\cdot\tfrac{\sqrt{3}}{2} - \tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2} = 0$. They report $\cos 30° = 0$.

The break. $\cos 30°$ is $\dfrac{\sqrt{3}}{2} \approx 0.87$, not $0$. The result $0$ would mean $30°$ is a right angle, which it plainly is not. What went wrong: that minus sign belongs to the sum formula, $\cos(A + B) = \cos A \cos B - \sin A \sin B$. The student accidentally computed $\cos 90°$, which really is $0$.

Correct. The cosine difference formula carries a plus:

$$\cos(60° - 30°) = \cos 60° \cos 30° + \sin 60° \sin 30° = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}.$$

Final answer: $\cos 30° = \dfrac{\sqrt{3}}{2}$.

Example 3

Find $\cos 75°$ as a difference, using $\cos(105° - 30°)$.

$$\cos(105° - 30°) = \cos 105° \cos 30° + \sin 105° \sin 30°.$$

With $\cos 105° = \dfrac{\sqrt{2}-\sqrt{6}}{4}$ and $\sin 105° = \dfrac{\sqrt{6}+\sqrt{2}}{4}$:

$$\cos 75° = \frac{\sqrt{2}-\sqrt{6}}{4}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{6}+\sqrt{2}}{4}\cdot\frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}.$$

In radians, $75° = \dfrac{5\pi}{12}$.

Final answer: $\cos 75° = \dfrac{\sqrt{6} - \sqrt{2}}{4}$.

Example 4

Given $\cos A = \dfrac{4}{5}$, $\sin A = \dfrac{3}{5}$, $\cos B = \dfrac{12}{13}$, $\sin B = \dfrac{5}{13}$, find $\cos(A - B)$.

$$\cos(A - B) = \cos A \cos B + \sin A \sin B = \frac{4}{5}\cdot\frac{12}{13} + \frac{3}{5}\cdot\frac{5}{13} = \frac{48 + 15}{65} = \frac{63}{65}.$$

Final answer: $\dfrac{63}{65}$.

Example 5

Simplify $\cos(A - B)\cos B - \sin(A - B)\sin B$.

This matches the sum pattern $\cos X \cos Y - \sin X \sin Y = \cos(X + Y)$ with $X = A - B$ and $Y = B$:

$$\cos((A - B) + B) = \cos A.$$

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

Final answer: $\cos A$.

Example 6

Use the formula to derive $\cos 2A$ from $\cos(A - (-A))$... and explain why it gives $1$ for the wrong reason, then the right identity.

Setting $B = -A$ in the difference formula gives $\cos(A - (-A)) = \cos(2A)$... but the formula expects a difference. Instead, treat it cleanly: $\cos(A - A) = \cos 0 = 1$. By the formula, $\cos A \cos A + \sin A \sin A = \cos^2 A + \sin^2 A = 1$. The identity recovers the Pythagorean relation as a special case.

Final answer: $\cos(A - A) = \cos^2 A + \sin^2 A = 1$.

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, $\cos A \cos B + \sin A \sin B = \cos(A - B)$ — 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 $A$ and $B$ overlap, the cosine of their phase difference $\cos(A - B)$ 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 $\cos A \cos B - \sin A \sin B$ for $\cos(A - B)$.

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

The correct way: For cosine, the sign is opposite: $\cos(A - B)$ takes a plus, $\cos(A + B)$ takes a minus.

Mistake 2: Swapping the cosine and sine pairings

Where it slips in: A reader writes $\cos A \sin B + \sin A \cos B$ — 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: $\cos A \cos B$ (both cosines) plus $\sin A \sin B$ (both sines). The cross-paired form $\sin A \cos B \pm \cos A \sin B$ belongs to $\sin(A \pm B)$ — a different identity entirely.

Mistake 3: Mixing degrees and radians in one evaluation

Where it slips in: A reader sets up $\cos\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right)$ 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. $\cos(45° - 30°) = \cos\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right)$ — 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 $\cos(A - B) = \cos A \cos B + \sin A \sin B$ — the cosine difference identity.

• The sign is a plus, the opposite of the minus in $\cos(A + B)$ — 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.

Practice These Three

1. Compute $\cos 15°$ using $\cos(60° - 45°)$ and confirm it matches $\cos(45° - 30°)$.

2. Given $\cos A = \dfrac{8}{17}$, $\sin A = \dfrac{15}{17}$, $\cos B = \dfrac{3}{5}$, $\sin B = \dfrac{4}{5}$, find $\cos(A - B)$.

3. Show that $\cos(A - B) - \cos(A + B) = 2\sin A \sin B$.

If Problem 1 gives $\dfrac{\sqrt{6} + \sqrt{2}}{4}$ both ways, the two decompositions of $15°$ agree — a built-in check. Want a live Bhanzu trainer to walk your child through the compound-angle identities and the Class 11 trigonometry chapter? Book a free demo class — online globally.