Cos (a + b) Formula: Proof, Sign Rule, Examples

#Trigonometry
TL;DR
Cos (a + b) is the cosine addition formula, and it equals $\cos(a+b) = \cos a\cos b - \sin a\sin b$. You multiply the two cosines, multiply the two sines, then subtract the sine product from the cosine product. The one detail to guard is the sign: the sum formula carries a minus, while its partner $\cos(a-b)$ carries a plus.
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Bhanzu TeamLast updated on September 12, 202611 min read

What Is The Cos (a + b) Formula?

Cos (a + b) is the trigonometric identity for the cosine of a sum of two angles. Its value is:

$$\cos(a+b) = \cos a\cos b - \sin a\sin b$$

This is called the cosine addition formula, or a compound angle formula, because the angle $(a+b)$ is compounded from two separate angles $a$ and $b$. It lets you rewrite the cosine of an awkward angle using the sines and cosines of two friendlier angles you already know.

The angles work in both units. In degrees you might read it as $\cos(45^\circ + 30^\circ)$; in radians the same statement is $\cos\left(\frac{\pi}{4} + \frac{\pi}{6}\right)$. The formula does not care which unit you use, as long as $a$ and $b$ are measured the same way.

The rule in words: multiply cosine by cosine, multiply sine by sine, then subtract the second product from the first.

How Do You Prove The Cos (a + b) Formula?

The cleanest proof uses the unit circle and the distance formula. It first proves the difference formula $\cos(a-b)$, then turns it into the sum formula in one line. This is the standard rigorous route, and it makes the minus sign appear for a reason rather than by memory.

Place two points on the unit circle. Point $P$ sits at angle $a$, so $P = (\cos a, \sin a)$. Point $Q$ sits at angle $b$, so $Q = (\cos b, \sin b)$. The angle between them is $(a-b)$.

Write the squared distance $PQ^2$ with the distance formula:

$$PQ^2 = (\cos a - \cos b)^2 + (\sin a - \sin b)^2$$

Expand and collect terms:

$$PQ^2 = \cos^2 a - 2\cos a\cos b + \cos^2 b + \sin^2 a - 2\sin a\sin b + \sin^2 b$$

Group each angle's squares and use $\cos^2\theta + \sin^2\theta = 1$:

$$PQ^2 = (\cos^2 a + \sin^2 a) + (\cos^2 b + \sin^2 b) - 2(\cos a\cos b + \sin a\sin b)$$

$$PQ^2 = 2 - 2(\cos a\cos b + \sin a\sin b)$$

Now rotate the whole picture so the same angle $(a-b)$ sits from the positive $x$-axis. One point lands at $(1, 0)$ and the other at $(\cos(a-b), \sin(a-b))$. Rotation does not change the chord length, so:

$$PQ^2 = (\cos(a-b) - 1)^2 + \sin^2(a-b) = 2 - 2\cos(a-b)$$

Set the two expressions for $PQ^2$ equal:

$$2 - 2\cos(a-b) = 2 - 2(\cos a\cos b + \sin a\sin b)$$

$$\cos(a-b) = \cos a\cos b + \sin a\sin b$$

Finally, replace $b$ with $-b$. Since cosine is even, $\cos(-b) = \cos b$, and since sine is odd, $\sin(-b) = -\sin b$:

$$\cos(a+b) = \cos a\cos b + \sin a(-\sin b) = \cos a\cos b - \sin a\sin b$$

The minus sign is not a rule to memorise. It is the direct result of sine being an odd function.

How Do You Use Cos (a + b) To Find An Exact Value?

The formula turns a non-standard angle into two standard ones. The classic case is $75^\circ$, which is not on the special-angle table, but splits neatly as $45^\circ + 30^\circ$.

Example 1: Find $\cos 75^\circ$.

Write $75^\circ = 45^\circ + 30^\circ$ and apply the formula:

$$\cos 75^\circ = \cos(45^\circ + 30^\circ) = \cos 45^\circ\cos 30^\circ - \sin 45^\circ\sin 30^\circ$$

Substitute the special-angle values $\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, and $\sin 30^\circ = \frac{1}{2}$:

$$\cos 75^\circ = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$$

$$\cos 75^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$$

Final answer: $\cos 75^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4} \approx 0.2588$. A calculator confirms $\cos 75^\circ = 0.2588$ to four decimal places, so the exact surd form checks out.

Example 2: Find $\cos\dfrac{7\pi}{12}$ (that is, $105^\circ$).

In radians, $\frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$, which is $60^\circ + 45^\circ$:

$$\cos\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \cos\frac{\pi}{3}\cos\frac{\pi}{4} - \sin\frac{\pi}{3}\sin\frac{\pi}{4}$$

$$= \frac{1}{2}\cdot\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}$$

Final answer: $\cos\dfrac{7\pi}{12} = \dfrac{\sqrt{2} - \sqrt{6}}{4} \approx -0.2588$. The result is negative because $105^\circ$ lands in the second quadrant, where cosine is negative.

What Angles Can You Build With Cos (a + b)?

Any angle that splits into a sum of two special angles is fair game. The table pairs each target angle with a split you can feed into the formula, in both degrees and radians.

Table: Angles you can evaluate with the cosine addition formula, and their exact values.

Target angle

Split as

Radians

Exact $\cos$ value

Decimal

$75^\circ$

$45^\circ + 30^\circ$

$\frac{\pi}{4}+\frac{\pi}{6}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

$0.2588$

$90^\circ$

$45^\circ + 45^\circ$

$\frac{\pi}{4}+\frac{\pi}{4}$

$0$

$0.0000$

$105^\circ$

$60^\circ + 45^\circ$

$\frac{\pi}{3}+\frac{\pi}{4}$

$\frac{\sqrt{2}-\sqrt{6}}{4}$

$-0.2588$

$120^\circ$

$60^\circ + 60^\circ$

$\frac{\pi}{3}+\frac{\pi}{3}$

$-\frac{1}{2}$

$-0.5000$

The special-angle inputs come straight from the trigonometric table, and the fastest values to memorise are cos 45 degrees and cos 30 degrees, which build most of the useful splits.

What Is The Difference Between Cos (a + b) And Cos (a − b)?

These two identities look almost identical, and the only difference is one sign. That single sign is the most common place students lose a mark.

Table: The sum and difference cosine formulas side by side.

Formula

Expansion

Sign

$\cos(a+b)$

$\cos a\cos b - \sin a\sin b$

minus

$\cos(a-b)$

$\cos a\cos b + \sin a\sin b$

plus

The trap: a plus inside the bracket produces a minus in the expansion, and a minus inside produces a plus. The signs are crossed. A quick memory hook is that cosine disagrees with the sign in the bracket, while sine agrees with it.

For the full treatment of the partner identity, see the cos (a − b) formula, and for the matching sine version, the sin (a + b) formula. Both live inside the wider family of sum and difference identities.

Why Does Cos (a + b) Use A Minus Sign?

The minus is not arbitrary, and the unit circle explains it in one picture.

  • Adding a positive angle $b$ to $a$ rotates the point further anticlockwise, which pushes its $x$-coordinate (the cosine) down toward zero and beyond. The subtraction encodes that downward push.

  • Sine is an odd function, so $\sin(-b) = -\sin b$. When you rewrite $\cos(a+b)$ as $\cos(a-(-b))$ using the difference formula, that odd-function flip is exactly what converts the plus into a minus.

  • Cosine is an even function, so $\cos(-b) = \cos b$, and the cosine-times-cosine term keeps its sign. Only the sine-times-sine term flips.

So the minus sign is a direct fingerprint of sine's odd symmetry. If sine were even, the sum and difference formulas would be identical, and they are not.

Who Discovered The Cos (a + b) Formula?

Angle-addition results are among the oldest tools in trigonometry, worked out long before modern sine and cosine notation existed. Ancient astronomers needed them to predict the positions of the Sun, Moon, and planets, and the "chord" of an angle was the quantity they tabulated instead of the sine.

Two other figures shaped the same idea:

  • Hipparchus of Nicaea (c. 190–120 BCE, Greece) built the first known table of chords, the direct ancestor of Ptolemy's work and of the addition theorems.

  • Aryabhata (476–550 CE, India) tabulated the half-chord, the jya, which became our sine, and Indian astronomers used sum-and-difference relations to refine it. Later, Madhava of Sangamagrama (c. 1340–1425, India) pushed these ideas toward the power series that now define sine and cosine.

Where Is Cos (a + b) Used In The Real World?

The cosine addition formula is the mathematics of combining two oscillations, which is why it appears far outside the classroom.

  • Sound and music: when two tones overlap, their combined loudness depends on the sum of their phase angles, and the addition formula predicts the beats and interference you hear.

  • Alternating current: electrical engineers add voltage and current waves that are out of phase, and the power delivered depends on the cosine of the combined phase angle.

  • GPS and navigation: position fixes and signal timing rely on adding angular phases, where sum-and-difference identities keep the arithmetic exact.

  • Computer graphics: rotating an object by angle $a$ and then by angle $b$ multiplies to a single rotation by $a+b$, and the entries of that rotation use exactly $\cos(a+b)$ and $\sin(a+b)$.

  • Astronomy: the original use, combining angular positions to predict where a body will appear, still runs on the same identity Ptolemy wrote down.

One formula for adding two angles quietly runs sound systems, power grids, satellites, and game engines. The same rule that finds $\cos 75^\circ$ on paper is the rule that blends two waves in the physical world.

What Are The Most Common Mistakes With Cos (a + b)?

These four errors account for most lost marks on compound-angle questions, verified against The Math Doctors and Paul's Online Math Notes trig-error guides.

Splitting the cosine across the plus sign.

Where it slips in:

A student writes $\cos(a+b)$ as $\cos a + \cos b$, treating cosine as if it distributes over addition.

Don't do this:

Do not break a function apart over a sum. $\cos(a+b)$ is not $\cos a + \cos b$.

The correct way:

Use the full identity: $\cos(a+b) = \cos a\cos b - \sin a\sin b$. As a numeric check, $\cos(30^\circ+60^\circ) = \cos 90^\circ = 0$, while $\cos 30^\circ + \cos 60^\circ \approx 1.37$, so the two are clearly different.

Using the wrong sign.

Where it slips in:

A student copies the plus from inside the bracket straight into the expansion, writing $\cos a\cos b + \sin a\sin b$ for the sum.

Don't do this:

Do not match the expansion sign to the bracket sign. They are crossed for cosine.

The correct way:

Remember cosine disagrees with the bracket: a plus inside gives a minus outside. The sum uses $-\sin a\sin b$; only the difference $\cos(a-b)$ uses $+\sin a\sin b$.

Mixing degrees and radians.

Where it slips in:

A student writes $a$ in degrees and $b$ in radians, or leaves a calculator in the wrong mode, and gets a value that is far off.

Don't do this:

Do not combine two angles measured in different units, and do not trust a calculator result without checking its angle mode.

The correct way:

Convert both angles to the same unit first. If you need radians, see what is a radian, and set the calculator to match before evaluating.

Choosing a split that is not a special angle.

Where it slips in:

A student splits $75^\circ$ as $70^\circ + 5^\circ$, then cannot find exact values for either piece.

Don't do this:

Do not split into angles you do not know exactly.

The correct way:

Split only into special angles from the table, such as $30^\circ$, $45^\circ$, and $60^\circ$. For $75^\circ$, use $45^\circ + 30^\circ$; for $105^\circ$, use $60^\circ + 45^\circ$.

Practice Problems On Cos (a + b)

Try each with the formula $\cos(a+b) = \cos a\cos b - \sin a\sin b$, then check the answer.

  1. Expand $\cos(x + 60^\circ)$.
    (Answer: $\frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x$.)

  2. Find $\cos 90^\circ$ using the split $45^\circ + 45^\circ$.
    (Answer: $\left(\frac{\sqrt{2}}{2}\right)^2 - \left(\frac{\sqrt{2}}{2}\right)^2 = 0$.)

  3. Find the exact value of $\cos 105^\circ$.
    (Answer: $\frac{\sqrt{2}-\sqrt{6}}{4} \approx -0.2588$.)

  4. If $\cos a = \frac{3}{5}$, $\sin a = \frac{4}{5}$, $\cos b = \frac{5}{13}$, $\sin b = \frac{12}{13}$, find $\cos(a+b)$.
    (Answer: $\frac{3}{5}\cdot\frac{5}{13} - \frac{4}{5}\cdot\frac{12}{13} = \frac{15-48}{65} = -\frac{33}{65}$.)

  5. Show that $\cos(a+b) + \cos(a-b) = 2\cos a\cos b$.
    (Answer: the $\sin a\sin b$ terms cancel, leaving $2\cos a\cos b$.)

  6. Use $\cos(45^\circ + 45^\circ)$ to confirm $\cos 90^\circ = 0$.
    (Answer: matches Problem 2, confirming the identity.)

Where Should You Go Next After Cos (a + b)?

The cosine addition formula opens onto the whole family of compound-angle work, and a few natural doors follow from here.

  1. Sum and difference identities. The complete set for sine, cosine, and tangent, so you can expand any compound angle.

  2. Double angle formula. Set $b = a$ in the sum formula to derive $\cos 2a$ in one step.

  3. Trigonometric identities. The broader toolkit, including the cofunction identities and the basics in sin cos tan.

If your child is building these foundations, a live Bhanzu trainer teaches the addition formulas starting from the unit circle, so the minus sign is understood rather than memorised, in the Bhanzu trigonometry program.

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Frequently Asked Questions

What is the Cos (a + b) formula?
It is the cosine addition formula, $\cos(a+b) = \cos a\cos b - \sin a\sin b$. You multiply the cosines, multiply the sines, and subtract the sine product from the cosine product.
Why does Cos (a + b) have a minus sign?
Because sine is an odd function. Rewriting the sum through the difference formula flips $\sin(-b)$ to $-\sin b$, which turns the plus into a minus. Cosine is even, so the cosine term keeps its sign.
What is the difference between cos(a + b) and cos(a − b)?
The sign of the sine term. The sum $\cos(a+b) = \cos a\cos b - \sin a\sin b$ uses a minus, while the difference $\cos(a-b) = \cos a\cos b + \sin a\sin b$ uses a plus.
How do you use Cos (a + b) to find cos 75 degrees?
Split $75^\circ$ into $45^\circ + 30^\circ$ and apply the formula. The result is $\cos 75^\circ = \frac{\sqrt{6}-\sqrt{2}}{4} \approx 0.2588$.
Does the cos(a + b) formula work in radians?
Yes. The identity holds in any consistent unit. For example, $\cos\left(\frac{\pi}{4}+\frac{\pi}{6}\right)$ gives the same value as $\cos 75^\circ$, as long as both angles use the same unit.
Is cos(a + b) the same as cos a + cos b?
No. Cosine does not distribute over addition. As a check, $\cos(30^\circ+60^\circ)=0$, but $\cos 30^\circ + \cos 60^\circ \approx 1.37$, so the two expressions are not equal.
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