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

The Product That Hides a Sum

An AM radio multiplies two cosine waves together, and the product secretly carries a sum inside it. That is not a metaphor: when a carrier wave $\cos A$ meets an audio signal $\cos B$, the product $\cos A \cos B$ splits into two separate frequencies, and pulling those frequencies apart is exactly how a receiver recovers the voice from the static.

That splitting is the whole point of the cos A cos B formula. A product of cosines is awkward to integrate and hard to read as a combination of frequencies; the formula trades it for a sum, which does both jobs cleanly.

What Is the Cos A Cos B Formula?

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

$$\cos A \cos B = \frac{1}{2}\left[\cos(A - B) + \cos(A + B)\right]$$

Cleared of the fraction, the same identity reads $2\cos A \cos B = \cos(A + B) + \cos(A - B)$, which is the form many textbooks list. Here $A$ and $B$ are any two angles, in degrees or radians, and the right side holds a sum of two cosines rather than a product.

A quick reader question worth settling now: is cos A cos B the same as cos(A + B)? No. $\cos(A + B)$ is a single cosine of a combined angle, while $\cos A \cos B$ is a product that expands into two cosines. This identity sits alongside the wider family of product-to-sum formulas and the broader trigonometric identities.

How Do You Derive the Cos A Cos B Formula?

The formula is not a rule to memorise; it falls straight out of the two cosine compound-angle identities. Start with them:

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

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

Add the two equations. The $\sin A \sin B$ terms have opposite signs, so they cancel:

$$\cos(A + B) + \cos(A - B) = 2\cos A \cos B$$

Now divide both sides by $2$:

$$\cos A \cos B = \frac{1}{2}\left[\cos(A - B) + \cos(A + B)\right]$$

That is the whole derivation. If the compound-angle step is unfamiliar, the cos(A − B) formula page builds it from the unit circle, and both live inside the sum and difference identities.

Variable glossary. $A$ and $B$ are the two angles being multiplied; $A - B$ is their difference and $A + B$ is their sum. The $\frac{1}{2}$ is not optional - it comes from dividing by the $2$ that appears when the two identities combine.

Examples of Cos A Cos B

Example 1

Express $\cos 9x \cos 7x$ as a sum.

Set $A = 9x$ and $B = 7x$, so $A - B = 2x$ and $A + B = 16x$.

$$\cos 9x \cos 7x = \frac{1}{2}\left[\cos 2x + \cos 16x\right]$$

Example 2

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

Wrong attempt. A tempting move is to treat the product as a single cosine: $\cos 75^\circ \cos 15^\circ = \cos(75^\circ + 15^\circ) = \cos 90^\circ = 0$.

Check that against the numbers: $\cos 75^\circ \approx 0.259$ and $\cos 15^\circ \approx 0.966$, so the product is about $0.25$, not $0$. Collapsing a product into the cosine of a sum is not a legal step.

Correct. Apply the formula with $A = 75^\circ$, $B = 15^\circ$, so $A - B = 60^\circ$ and $A + B = 90^\circ$:

$$\cos 75^\circ \cos 15^\circ = \frac{1}{2}\left[\cos 60^\circ + \cos 90^\circ\right] = \frac{1}{2}\left[\frac{1}{2} + 0\right] = \frac{1}{4}$$

That matches the $0.25$ from the numerical check.

Example 3

Evaluate $\cos\frac{5\pi}{12} \cos\frac{\pi}{12}$.

With $A = \frac{5\pi}{12}$ and $B = \frac{\pi}{12}$, the difference is $\frac{4\pi}{12} = \frac{\pi}{3}$ and the sum is $\frac{6\pi}{12} = \frac{\pi}{2}$.

$$\cos\frac{5\pi}{12} \cos\frac{\pi}{12} = \frac{1}{2}\left[\cos\frac{\pi}{3} + \cos\frac{\pi}{2}\right] = \frac{1}{2}\left[\frac{1}{2} + 0\right] = \frac{1}{4}$$

Example 4

Write $2\cos 5\theta \cos 3\theta$ as a sum.

Using the cleared form $2\cos A \cos B = \cos(A + B) + \cos(A - B)$ with $A = 5\theta$, $B = 3\theta$:

$$2\cos 5\theta \cos 3\theta = \cos 8\theta + \cos 2\theta$$

Example 5

Integrate $\displaystyle\int \cos 3x \cos x ,dx$.

Direct integration of the product is awkward, so convert first. With $A = 3x$, $B = x$: $\cos 3x \cos x = \frac{1}{2}\left[\cos 2x + \cos 4x\right]$.

$$\int \cos 3x \cos x ,dx = \frac{1}{2}\int \left(\cos 2x + \cos 4x\right) dx$$

$$= \frac{1}{2}\left(\frac{\sin 2x}{2} + \frac{\sin 4x}{4}\right) + C = \frac{\sin 2x}{4} + \frac{\sin 4x}{8} + C$$

Example 6

Show that $\cos\theta \cos\theta = \frac{1}{2}\left(1 + \cos 2\theta\right)$.

Set $A = B = \theta$, so $A - B = 0$ and $A + B = 2\theta$. Since $\cos 0 = 1$:

$$\cos^2\theta = \frac{1}{2}\left[\cos 0 + \cos 2\theta\right] = \frac{1}{2}\left(1 + \cos 2\theta\right)$$

This is the power-reduction identity, and it drops out of the cos A cos B formula as the special case where the two angles are equal. The first-instinct error students make on this one is stopping at $\cos 0$ without writing its value, $1$; carrying the known value through is what closes the identity.

Where Is the Cos A Cos B Formula Used? - "Products become readable sums"

The formula exists to solve one recurring problem: a product of cosines is hard to work with, and a sum is easy. Two places make that concrete.

• Integration. A product like $\cos 3x \cos x$ has no clean antiderivative in product form, but rewritten as $\frac{1}{2}(\cos 2x + \cos 4x)$ it integrates term by term, as Example 5 showed. This is the standard first move for integrating products of trig functions.

• Frequency analysis. When two cosine signals multiply - a carrier and a message in amplitude modulation - the product form hides the actual frequencies present. The formula expands it into a sum of a difference-frequency and a sum-frequency, the two sidebands a radio receiver tunes between.

What competitor explainers usually skip is why the sine terms cancel and not the cosines: cosine is an even function, so $\cos(A - B)$ and $\cos(A + B)$ share the same $\cos A \cos B$ term, while the odd sine product flips sign between them and drops out. The difference-to-product companion, cos A − cos B, runs the same machinery in reverse.

Common Mistakes With Cos A Cos B

Mistake 1: Confusing the product formula with the sum formula

Where it slips in: Meeting $\cos A \cos B$ (a product) and reaching for the sum-to-product rule meant for $\cos A + \cos B$.

Don't do this: Applying $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ to a product.

The correct way: A product $\cos A \cos B$ uses the product-to-sum identity $\frac{1}{2}[\cos(A-B) + \cos(A+B)]$; a sum $\cos A + \cos B$ uses the sum-to-product identity. Read whether the two cosines are multiplied or added before choosing.

Mistake 2: Using a minus sign inside the bracket

Where it slips in: Blurring cos A cos B with the sin A sin B identity, which does carry a subtraction.

Don't do this: Writing $\cos A \cos B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$.

The correct way: Both terms are added: $\frac{1}{2}[\cos(A-B) + \cos(A+B)]$. The minus sign belongs to $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$, not to the cosine product. The habit that fixes this is deriving the formula from the two compound-angle identities rather than recalling the bracket cold.

Mistake 3: Dropping the factor of ½

Where it slips in: Rushing straight from $2\cos A \cos B = \cos(A+B) + \cos(A-B)$ to a value for $\cos A \cos B$ without halving.

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

The correct way: That right side equals $2\cos A \cos B$, so divide by $2$. The rusher who skips the halving doubles every answer.

Key Takeaways

• The cos A cos B formula is $\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)]$, a product-to-sum identity.

• It comes from adding the cosine sum and difference formulas, which cancels the $\sin A \sin B$ terms.

• Both cosines inside the bracket are added; the minus-sign version belongs to $\sin A \sin B$.

• Its main jobs are integrating products of cosines and separating a product of signals into its component frequencies.

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 tutoring with peers from 20+ countries.

A Practical Next Step

Practice these to solidify your understanding: express $\cos 8x \cos 2x$ as a sum, evaluate $\cos 105^\circ \cos 15^\circ$ using the formula, and integrate $\int \cos 5x \cos 3x ,dx$. If you get stuck on which angle is $A - B$ and which is $A + B$, come back to the derivation above. Want a live Bhanzu trainer to work through these with you? Book a free demo class.

Read More

• Trigonometric table — every standard-angle value the identities build on.

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

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

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

• Cos 2x — the double-angle identity behind Example 6's power reduction.