What Is The Cos A+Cos B Formula?
The Cos A+Cos B formula is the sum-to-product identity that turns the sum of two cosines into a single product:
$$\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
Read it slowly. The left side adds two cosines. The right side multiplies two cosines, but the angles change: the new angles are the half-sum $\frac{A+B}{2}$ and the half-difference $\frac{A-B}{2}$, and a factor of $2$ sits out front. This is one of the four sum-to-product identities, and it belongs to the wider family of trigonometric identities.
Why bother rewriting a sum as a product? Because products are far easier to work with in three places: evaluating awkward angle sums by hand, solving equations where a sum of cosines must equal zero (a product is zero when either factor is zero), and integrating, where a product of cosines has a clean antiderivative and a bare sum does not.
The angles $A$ and $B$ can be in degrees or radians, as long as both use the same unit. For $\cos 75^\circ + \cos 15^\circ$ the angles are $\frac{5\pi}{12}$ and $\frac{\pi}{12}$ in radian form; if you ever need a refresher on that conversion, see what is a radian.
How Do You Derive The Cos A+Cos B Formula?
The derivation starts from the two compound-angle expansions for cosine, the ones you already meet in the sum and difference identities. Write them with placeholder angles $X$ and $Y$:
$$\cos(X + Y) = \cos X \cos Y - \sin X \sin Y$$
$$\cos(X - Y) = \cos X \cos Y + \sin X \sin Y$$
Now add the two equations. The $\sin X \sin Y$ terms cancel, since one is negative and the other positive:
$$\cos(X + Y) + \cos(X - Y) = 2\cos X \cos Y$$
That is already a sum-to-product statement, just written with the "wrong" letters. To match the identity we want, rename the two combined angles:
$$A = X + Y, \qquad B = X - Y$$
Solving this small pair for $X$ and $Y$ gives the half-sum and half-difference:
$$X = \frac{A+B}{2}, \qquad Y = \frac{A-B}{2}$$
Substitute those back into $\cos(X+Y) + \cos(X-Y) = 2\cos X \cos Y$, and $\cos(X+Y)$ becomes $\cos A$, $\cos(X-Y)$ becomes $\cos B$:
$$\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
The whole identity is just the two cosine expansions added together, then relabelled. The halving of the arguments is not a trick to memorise, it is a direct consequence of solving $A = X+Y$ and $B = X-Y$ for the original angles.
How Do You Use Cos A+Cos B?
The method is always the same three moves: compute the half-sum, compute the half-difference, then evaluate the two cosines and multiply by $2$.
Example 1: Evaluate $\cos 75^\circ + \cos 15^\circ$.
Here $A = 75^\circ$ and $B = 15^\circ$. First the two new angles:
$$\frac{A+B}{2} = \frac{75^\circ + 15^\circ}{2} = 45^\circ, \qquad \frac{A-B}{2} = \frac{75^\circ - 15^\circ}{2} = 30^\circ$$
Substitute into the identity:
$$\cos 75^\circ + \cos 15^\circ = 2\cos 45^\circ \cos 30^\circ$$
Both are special angles. Using cos 45 degrees $= \frac{\sqrt{2}}{2}$ and cos 30 degrees $= \frac{\sqrt{3}}{2}$:
$$2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2}$$
Final answer: $\cos 75^\circ + \cos 15^\circ = \dfrac{\sqrt{6}}{2} \approx 1.2247$. A calculator confirms it: $\cos 75^\circ \approx 0.2588$ and $\cos 15^\circ \approx 0.9659$, and their sum is $1.2247$.
Example 2: Simplify $\cos 7x + \cos 3x$.
The identity works on symbolic angles just as well as numbers. With $A = 7x$ and $B = 3x$:
$$\frac{A+B}{2} = \frac{7x + 3x}{2} = 5x, \qquad \frac{A-B}{2} = \frac{7x - 3x}{2} = 2x$$
$$\cos 7x + \cos 3x = 2\cos 5x \cos 2x$$
This second form is what makes the identity useful in calculus and in solving equations. If a problem asks you to solve $\cos 7x + \cos 3x = 0$, the product form says $2\cos 5x \cos 2x = 0$, so either $\cos 5x = 0$ or $\cos 2x = 0$, and each is a standard equation to finish.
What Are The Related Sum-To-Product Formulas?
Cos A+Cos B has three siblings. Together the four cover every sum or difference of two sines or two cosines, and they share the same $\frac{A+B}{2}$, $\frac{A-B}{2}$ pattern. The differences are which function appears (cosine or sine) and whether a minus sign creeps in.
Table: The four sum-to-product identities.
Sum or difference | Product form |
|---|---|
$\cos A + \cos B$ | $2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ |
$\cos A - \cos B$ | $-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ |
$\sin A + \sin B$ | $2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ |
$\sin A - \sin B$ | $2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ |
Two features are worth noticing. The cosine-difference row is the odd one out: it turns into two sines and picks up a leading minus sign, which is the single most common source of errors on this topic. Bhanzu has dedicated pages for the ones you will reach for most, including cos A minus B formula and sin A plus sin B. For the full set in one place, see sum to product formulas.
Why Does The Cos A+Cos B Formula Work?
The algebra is settled above, but there are two deeper reasons the identity looks the way it does.
The cancellation is the whole point. When you add $\cos(X+Y)$ and $\cos(X-Y)$, the parts that depend on the sign of $Y$ are the $\sin X \sin Y$ terms, and they are equal and opposite, so they vanish. What survives is symmetric in $Y$, which is exactly why the answer is a product of two even (cosine) factors.
The half-angles are an average and a spread. The quantity $\frac{A+B}{2}$ is the average of the two angles, and $\frac{A-B}{2}$ is half the gap between them. So the identity says: two cosines add up to twice the cosine of their average, scaled by the cosine of how far apart they are. When $A$ and $B$ are close, $\frac{A-B}{2}$ is small, $\cos\frac{A-B}{2}$ is near $1$, and the sum is almost $2\cos(\text{average})$. When they are far apart, that second factor shrinks the result. This is precisely the behaviour behind acoustic beats.
That second reading is why the formula is not just bookkeeping. It connects a symbolic sum to a physical picture: the average angle sets the fast wiggle, and the difference angle sets the slow envelope. For the underlying right-triangle and unit-circle meaning of the cosines themselves, sin cos tan is the place to start.
Who Discovered The Cos A+Cos B Formula?
Long before logarithms, astronomers faced a brutal problem: multiplying the huge numbers in their star tables by hand, over and over, with every product risking a fresh arithmetic slip. Someone noticed that identities like this one could turn a hard multiplication into an easy addition, and a whole technique grew up around that idea.
Two more names shaped this story:
Claudius Ptolemy (c. 100 to 170 CE, Alexandria) built the chord tables of the Almagest using sum-and-difference relationships for chords, the ancient ancestors of these identities.
John Napier (1550 to 1617, Scotland) invented logarithms in 1614, which did the same job (turning multiplication into addition) more directly, and prosthaphaeresis faded from use soon after.
Where Is Cos A+Cos B Used In The Real World?
Adding two cosines is not a paper exercise. It is what physically happens whenever two waves of similar frequency overlap.
Acoustic beats. Two notes at close frequencies add exactly like $\cos A + \cos B$. The average frequency is the pitch you hear, and the slow $\cos\frac{A-B}{2}$ envelope is the throb, which is how musicians tune by ear until the beats disappear.
Wave interference. Light, water, and radio waves that meet combine as sums of cosines. The product form predicts where they reinforce (bright or loud) and where they cancel (dark or silent).
Signal processing and Fourier analysis. Complex signals are built from many cosines, and sum-to-product identities let engineers factor, filter, and compress them. Amplitude modulation in AM radio is a direct application.
Alternating current. When AC sources of nearly equal frequency combine on a shared line, the resulting voltage swells and fades on the same beat pattern this identity describes.
Seismology and medical imaging. Overlapping reflected waves in earthquake data and ultrasound are separated using the same sum-to-product structure.
One identity, learned for a trigonometry exam, quietly runs music tuning, radio, and imaging. The math class and the physics of waves are the same subject wearing different clothes.
What Are The Most Common Mistakes With Cos A+Cos B?
These four errors account for most lost marks on sum-to-product problems, and each traces back to a worked-example step where students slip.
Using $A+B$ and $A-B$ instead of the halved angles.
Where it slips in:
A student writes $\cos A + \cos B = 2\cos(A+B)\cos(A-B)$, dropping the division by $2$ inside each cosine.
Don't do this:
Do not put the full sum and difference inside the cosines. The arguments are halved.
The correct way:
Compute $\frac{A+B}{2}$ and $\frac{A-B}{2}$ first, as separate numbers, before touching the cosines. For $\cos 75^\circ + \cos 15^\circ$ that means $45^\circ$ and $30^\circ$, not $90^\circ$ and $60^\circ$.
Confusing $\cos A + \cos B$ with $\cos A - \cos B$.
Where it slips in:
A student reuses the cosine-sum result for a cosine-difference, keeping cosines and forgetting the minus sign.
Don't do this:
Do not assume the difference formula also gives cosines. It does not.
The correct way:
Remember the family pattern: $\cos A + \cos B$ gives two cosines, but $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ flips to sines and carries a leading minus.
Mixing degrees and radians inside the halves.
Where it slips in:
A student halves a degree angle but leaves a calculator in radian mode, or vice versa, so $\cos 45^\circ$ is evaluated as $\cos 45$ radians.
Don't do this:
Do not switch units partway through, and do not trust a calculator without checking its mode.
The correct way:
Keep $A$ and $B$ in one unit throughout. Halving $90^\circ$ gives $45^\circ$; halving $\frac{\pi}{2}$ gives $\frac{\pi}{4}$. The two are the same angle, so the answer must match either way.
Forgetting the leading factor of $2$.
Where it slips in:
A student reaches $\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ and stops, leaving the answer at half its true size.
Don't do this:
Do not drop the $2$ out front. It comes from adding two identical $\cos X \cos Y$ terms in the derivation.
The correct way:
Write the $2$ before you write anything else, so it cannot be lost. Every sum-to-product identity has a leading $2$; the sign is what changes between them.
Practice Problems On Cos A+Cos B
Evaluate or simplify each using the identity. Answers follow each line, verified numerically.
Evaluate $\cos 105^\circ + \cos 15^\circ$.
(Answer: $2\cos 60^\circ \cos 45^\circ = 2 \cdot \tfrac{1}{2} \cdot \tfrac{\sqrt{2}}{2} = \tfrac{\sqrt{2}}{2} \approx 0.7071$.)Evaluate $\cos 90^\circ + \cos 30^\circ$.
(Answer: $2\cos 60^\circ \cos 30^\circ = 2 \cdot \tfrac{1}{2} \cdot \tfrac{\sqrt{3}}{2} = \tfrac{\sqrt{3}}{2} \approx 0.8660$.)Evaluate $\cos 120^\circ + \cos 60^\circ$.
(Answer: $2\cos 90^\circ \cos 30^\circ = 2 \cdot 0 \cdot \tfrac{\sqrt{3}}{2} = 0$.)Simplify $\cos 80^\circ + \cos 40^\circ$ to a single cosine.
(Answer: $2\cos 60^\circ \cos 20^\circ = \cos 20^\circ \approx 0.9397$.)Write $\cos 3A + \cos A$ as a product.
(Answer: $2\cos 2A \cos A$.)Solve $\cos 5x + \cos x = 0$ for the product form.
(Answer: $2\cos 3x \cos 2x = 0$, so $\cos 3x = 0$ or $\cos 2x = 0$.)
Where Should You Go Next After Cos A+Cos B?
Cos A+Cos B is one door into the sum-to-product family, and a few natural next steps open from here.
Sum to product formulas. See all four identities together and practise choosing the right one.
Sum and difference identities. The $\cos(A \pm B)$ expansions this derivation was built from.
Trigonometric table. The special-angle values ($30^\circ$, $45^\circ$, $60^\circ$) you plug into every worked example.
If your child is building these foundations, a live Bhanzu trainer teaches the sum-to-product identities starting from the "why" (the wave picture and the derivation, not the memorised formula) in the Bhanzu trigonometry program.
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