Cosine Addition Formula

The cosine addition formula allows you to calculate the cosine of the sum of two angles using the sine and cosine of the individual angles:

$$ \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta $$

This is one of the most important identities in trigonometry. It appears in geometry, calculus, physics, engineering, and many other areas of mathematics.

A common mistake is to think that:

$$ \cos(\alpha+\beta)=\cos\alpha+\cos\beta $$

However, this relationship is not true. The cosine of a sum cannot be obtained by simply adding the cosine values of the two angles.

An Example

Let's see why with a simple example.

Consider the angles a = 30° and b = 60°.

Their cosine values are:

$$ \cos30^\circ=\frac{\sqrt{3}}{2} $$

$$ \cos60^\circ=\frac{1}{2} $$

If we incorrectly add these values, we get:

$$ \cos30^\circ+\cos60^\circ=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{1+\sqrt{3}}{2} $$

But the angle sum is:

$$ 30^\circ+60^\circ=90^\circ $$

Therefore:

$$ \cos(30^\circ+60^\circ)=\cos90^\circ $$

and we know that:

$$ \cos90^\circ=0 $$

The two results are clearly different.

Note: This example illustrates an important property of trigonometric functions. In general, the value of a trigonometric function evaluated at a sum of angles is not equal to the sum of the function values evaluated separately.

To find the correct result, we must use the cosine addition formula:

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

Substituting a = 30° and b = 60° gives:

$$ \cos(30^\circ+60^\circ)=\cos30^\circ\cos60^\circ-\sin30^\circ\sin60^\circ $$

Using the exact trigonometric values

$$ \cos30^\circ=\frac{\sqrt{3}}{2}, \qquad \cos60^\circ=\frac{1}{2} $$

we obtain:

$$ \cos(30^\circ+60^\circ)=\frac{\sqrt{3}}{2}\cdot\frac{1}{2}-\sin30^\circ\sin60^\circ $$

$$ \cos(30^\circ+60^\circ)=\frac{\sqrt{3}}{4}-\sin30^\circ\sin60^\circ $$

The corresponding sine values are:

$$ \sin30^\circ=\frac{1}{2}, \qquad \sin60^\circ=\frac{\sqrt{3}}{2} $$

Substituting them into the formula gives:

$$ \cos(30^\circ+60^\circ)=\frac{\sqrt{3}}{4}-\frac{1}{2}\cdot\frac{\sqrt{3}}{2} $$

$$ \cos(30^\circ+60^\circ)=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4} $$

$$ \cos(30^\circ+60^\circ)=0 $$

This matches the known value of cos(90°), confirming that the addition formula produces the correct result.

Proof of the Cosine Addition Formula

We now prove the identity:

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

Start by rewriting the sum as a difference:

$$ \cos(a+b)=\cos[a-(-b)] $$

Next, apply the cosine subtraction formula:

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

Note: The cosine subtraction formula states that

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

Its proof is discussed in detail in the corresponding article.

Since cosine is an even function,

$$ \cos(-b)=\cos(b) $$

Substituting this property gives:

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

Since sine is an odd function,

$$ \sin(-b)=-\sin(b) $$

Therefore:

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

Simplifying the expression yields:

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

which is exactly the cosine addition formula.

The proof is complete.