Cos A - Cos B : Difference-to-Product Formula, Proof, and Examples

What Is the Cos A − Cos B Formula?

The cos A − cos B formula is one of the difference-to-product identities. It states:

$$\cos A - \cos B = -2\sin\frac{A + B}{2}\sin\frac{A - B}{2}$$

The right side is a product of two sines, scaled by $-2$. The leading minus sign is part of the identity, not a slip - it is the single feature that separates a correct answer from a wrong one here. This identity belongs to the family of sum-to-product formulas and the wider trigonometric identities.

A reader question worth settling first: is cos A − cos B the same as cos(A − B)? No. $\cos(A - B)$ is a single cosine of one combined angle; $\cos A - \cos B$ is the difference of two separate cosines, and it expands into a product of sines.

The Difference That Factors

Two cosine waves subtracted from each other should be messy, yet the result collapses into a single clean product. That is the move behind noise-cancelling headphones and behind almost every trigonometric equation you will be asked to solve: a difference of cosines cannot be factored or set to zero as it stands, but as a product it factors instantly.

The cos A − cos B formula is what performs that collapse. It trades an expression you cannot solve for one you can, and the price is remembering a single leading minus sign.

How Do You Derive the Cos A − Cos B Formula?

The formula is not memorised, it is built. Introduce two helper angles $X$ and $Y$ so that

$$A = X + Y, \qquad B = X - Y, \qquad \text{which gives } X = \frac{A + B}{2}, \quad Y = \frac{A - B}{2}$$

Now expand each cosine with the compound-angle identities:

$$\cos A = \cos(X + Y) = \cos X \cos Y - \sin X \sin Y$$

$$\cos B = \cos(X - Y) = \cos X \cos Y + \sin X \sin Y$$

Subtract the second from the first. The $\cos X \cos Y$ terms cancel:

$$\cos A - \cos B = -2\sin X \sin Y$$

Substitute $X$ and $Y$ back:

$$\cos A - \cos B = -2\sin\frac{A + B}{2}\sin\frac{A - B}{2}$$

That minus sign comes directly from the $-\sin X \sin Y$ on top; it is structural, not decorative. The compound-angle step is the same one used in the cos(A − B) formula, and both sit inside the sum and difference identities.

Variable glossary. $A$ and $B$ are the two angles; $\frac{A + B}{2}$ is their average (half the sum) and $\frac{A - B}{2}$ is half their difference. The factor $-2$ is fixed by the derivation.

Examples of Cos A − Cos B

Example 1

Express $\cos 4x - \cos 2x$ as a product.

Set $A = 4x$, $B = 2x$, so $\frac{A + B}{2} = 3x$ and $\frac{A - B}{2} = x$.

$$\cos 4x - \cos 2x = -2\sin 3x \sin x$$

Example 2

Evaluate $\cos 75^\circ - \cos 15^\circ$.

Wrong attempt. A tempting shortcut is $\cos 75^\circ - \cos 15^\circ = \cos(75^\circ - 15^\circ) = \cos 60^\circ = \frac{1}{2}$.

Check it: $\cos 75^\circ \approx 0.259$ and $\cos 15^\circ \approx 0.966$, so the difference is about $-0.707$, which is negative - nowhere near $+\frac{1}{2}$. The difference of two cosines is not the cosine of the difference.

Correct. Apply the formula with $\frac{A + B}{2} = 45^\circ$ and $\frac{A - B}{2} = 30^\circ$:

$$\cos 75^\circ - \cos 15^\circ = -2\sin 45^\circ \sin 30^\circ = -2 \times \frac{\sqrt{2}}{2} \times \frac{1}{2} = -\frac{\sqrt{2}}{2}$$

That is about $-0.707$, matching the numerical check.

Example 3

Evaluate $\cos\frac{2\pi}{3} - \cos\frac{\pi}{3}$.

Here $\frac{A + B}{2} = \frac{1}{2}\left(\frac{2\pi}{3} + \frac{\pi}{3}\right) = \frac{\pi}{2}$ and $\frac{A - B}{2} = \frac{1}{2}\left(\frac{2\pi}{3} - \frac{\pi}{3}\right) = \frac{\pi}{6}$.

$$\cos\frac{2\pi}{3} - \cos\frac{\pi}{3} = -2\sin\frac{\pi}{2}\sin\frac{\pi}{6} = -2 \times 1 \times \frac{1}{2} = -1$$

A direct check agrees: $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\cos\frac{\pi}{3} = \frac{1}{2}$, whose difference is $-1$.

Example 4

Solve $\cos 3x - \cos x = 0$.

Factor the left side with $\frac{A + B}{2} = 2x$ and $\frac{A - B}{2} = x$:

$$\cos 3x - \cos x = -2\sin 2x \sin x = 0$$

A product is zero when a factor is zero, so $\sin 2x = 0$ or $\sin x = 0$. Both are covered by

$$x = \frac{n\pi}{2}, \quad n \in \mathbb{Z}$$

This is exactly why the product form matters: the original difference could not be set to zero, but the factored product can.

Example 5

Simplify $\dfrac{\cos x - \cos 5x}{\cos 2x - \cos 4x}$.

Convert numerator and denominator separately, using $\sin(-\theta) = -\sin\theta$ to clear the negative half-differences.

$$\cos x - \cos 5x = -2\sin 3x \sin(-2x) = 2\sin 3x \sin 2x$$

$$\cos 2x - \cos 4x = -2\sin 3x \sin(-x) = 2\sin 3x \sin x$$

$$\frac{\cos x - \cos 5x}{\cos 2x - \cos 4x} = \frac{2\sin 3x \sin 2x}{2\sin 3x \sin x} = \frac{\sin 2x}{\sin x} = 2\cos x$$

The last step uses $\sin 2x = 2\sin x \cos x$.

Example 6

Write $\cos 40^\circ - \cos 80^\circ$ as a product and state its sign.

With $\frac{A + B}{2} = 60^\circ$ and $\frac{A - B}{2} = -20^\circ$:

$$\cos 40^\circ - \cos 80^\circ = -2\sin 60^\circ \sin(-20^\circ) = 2\sin 60^\circ \sin 20^\circ = \sqrt{3},\sin 20^\circ$$

Both sine factors end up positive, so the result is positive - sensible, since $\cos 40^\circ > \cos 80^\circ$. The first-instinct error here is dropping the sign on $\sin(-20^\circ)$; carrying $\sin(-\theta) = -\sin\theta$ through is what flips the leading minus into a plus.

Where Is the Cos A − Cos B Formula Used? - "A difference you can finally solve"

The formula earns its place by turning an unsolvable-looking difference into a factorable product. Two settings make that concrete.

• Solving trigonometric equations. An equation like $\cos 3x - \cos x = 0$ has no obvious next step as a difference, but factored into $-2\sin 2x \sin x$ it splits into two simple equations, as Example 4 showed. This is the standard technique whenever a trig equation pits one cosine against another.

• Simplifying ratios and expressions. A quotient of two cosine differences, like the one in Example 5, cancels cleanly once each part is a product - the shared $\sin 3x$ factor divides out and a page of algebra becomes $2\cos x$.

What most explainers skip is why the cosine terms cancel while the sine terms survive: cosine is even, so $\cos X \cos Y$ is identical in both expansions and subtracts to zero, while the odd sine product keeps its sign and doubles. The product-to-sum companion, cos A cos B, runs the same idea in the opposite direction.

Common Mistakes With Cos A − Cos B

Mistake 1: Dropping the leading minus sign

Where it slips in: Recalling the bracket as $2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$ without the $-$ in front.

Don't do this: $\cos A - \cos B = 2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$.

The correct way: The identity is $-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$. The minus is what makes the sign of the answer come out right, as Example 2's negative result showed. The habit that fixes this is deriving from the two expansions, where the minus appears on its own.

Mistake 2: Treating cos A − cos B as cos(A − B)

Where it slips in: Collapsing the difference of two cosines into a single cosine of the difference.

Don't do this: $\cos A - \cos B = \cos(A - B)$.

The correct way: These are unrelated. $\cos 75^\circ - \cos 15^\circ = -\frac{\sqrt{2}}{2}$, while $\cos(75^\circ - 15^\circ) = \cos 60^\circ = \frac{1}{2}$. The rusher who reads "minus" as "combine the angles" gets the wrong value and the wrong sign at once.

Mistake 3: Mishandling the sign when A is less than B

Where it slips in: When $\frac{A - B}{2}$ comes out negative, as in Example 6.

Don't do this: Writing $\sin\frac{A-B}{2}$ as positive and ignoring the negative argument.

The correct way: Use $\sin(-\theta) = -\sin\theta$ to rewrite the factor, then track the sign through.

Key Takeaways

• The cos A − cos B formula is $\cos A - \cos B = -2\sin\frac{A + B}{2}\sin\frac{A - B}{2}$, a difference-to-product identity.

• It comes from subtracting the two cosine compound-angle expansions, which cancels the cosine terms and leaves the minus sign.

• The leading $-2$ is essential; dropping it reverses the sign of the answer.

• Its main job is factoring a difference of cosines so trigonometric equations can be solved and ratios simplified.

To take the cos A − cos B formula further with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor for identity practice, or live math classes online with peers from 20+ countries.

A Practical Next Step

Practice these to solidify your understanding: express $\cos 6x - \cos 2x$ as a product, evaluate $\cos 105^\circ - \cos 15^\circ$ using the formula, and solve $\cos 5x - \cos x = 0$. If you get stuck on the sign, come back to the derivation and watch where the minus appears. Want a live Bhanzu trainer to work through these with you? Book a free demo class.

Read More

• Sin A + sin B formula — the sum-to-product identity for sine, built the same way.

• Cosine function — the even function whose symmetry makes the cosine terms cancel.

• Trigonometry formulas — the full identity toolkit in one reference.

• 2 sin a cos a formula — the double-angle identity used to finish Example 5.

• Cos 2x — the double-angle cosine identity that pairs with these.