Addition formula for cosine

The addition formula for the cosine is: $$ \cos (\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $$

It’s incorrect to write:

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

A practical example

Let’s consider two angles, a=30° and b=60°.

$$ \cos a = \cos 30° = \frac{\sqrt{3}}{2} $$

$$ \cos b = \cos 60° = \frac{1}{2} $$

The cosine of a+b is not equal to the sum of the cosines of the two angles:

$$ \cos (a+b) \ne \cos(a) + \cos(b) $$

$$ \cos (30°+60°) \ne \cos(30°) + \cos(60°) = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1 + \sqrt{3}}{2} $$

Note: The cosine of 30°+60° is actually the cosine of 90°, and we know that the cosine of 90° is 0. Therefore, it’s a significant mistake to calculate the cosine of the sum of two angles, cos(a+b), as the sum of their individual cosines, cos(a)+cos(b).

Now, let’s calculate the cosine of a+b using the addition formula for cosine:

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

Substituting a=30° and b=60°, we get:

$$ \cos (30°+60°) = \cos 30° \cos 60° - \sin 30° \sin 60° $$

We already know the values: cos(30°) = √3/2 and cos(60°) = 1/2.

$$ \cos (30°+60°) = \frac{\sqrt{3}}{2} \cdot \frac{1}{2} - \sin 30° \sin 60° $$

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

The sine values for 30° and 60° are: sin(30°) = 1/2 and sin(60°) = √3/2.

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

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

$$ \cos (30°+60°) = 0 $$

Therefore, the cosine of 30°+60° is zero, and the result is correct.

The proof

The cosine of the sum of two angles a and b is:

$$ \cos(a+b) $$

This can be rewritten as:

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

Now, we can apply the cosine subtraction formula:

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

Note: The cosine subtraction formula is $$ \cos (a-b) = \cos a \cos b + \sin a \sin b $$ which I have already discussed in a previous note, where I also provide a detailed proof.

Cosine is an even function, so cos(-b)=cos(b).

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

Sine, on the other hand, is an odd function, so sin(-b)=-sin(b).

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

Thus, we derive the addition formula for cosine, as we wanted to prove:

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

And that’s the formula.