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
Compute $\cos 15°$ using $\cos(60° - 45°)$ and confirm it matches $\cos(45° - 30°)$.
Given $\cos A = \dfrac{8}{17}$, $\sin A = \dfrac{15}{17}$, $\cos B = \dfrac{3}{5}$, $\sin B = \dfrac{4}{5}$, find $\cos(A - B)$.
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.
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