Trigonometry/Addition Formula for Cosines

Cosine Formulae

We proved the sine addition formula; now we're going to prove the cosine addition formula.

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

Before we do that we will talk about subtraction formulae.

Template:ExampleRobox You do not need to learn or remember special subtraction formulas or 'angle difference' formulas for sine and cosine. You can work them out 'instantly' from the addition formulas for sine and cosine, using $\sin (-x)=-\sin (x)$ and $\cos (-x)=\cos (x)$ .

Let's put $(-\beta )$ in place of $\beta$ in the two addition formulas:

First the sine addition formula:

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

becomes:

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

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

Now for the cosine addition formula:

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

becomes:

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

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

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Combining all four formulas:

If we really want to we can write the four addition and 'angle difference' formulas in a more condensed notation like so:

$\cos (\alpha \pm \beta )=\cos (\alpha )\cos (\beta )\mp \sin (\alpha )\sin (\beta )$

$\sin (\alpha \pm \beta )=\sin (\alpha )\cos (\beta )\pm \cos (\alpha )\sin (\beta )$

If you like this style, use them. We'd recommend instead just learning the addition formulas and deriving the difference formulas from them when you need them. Template:Robox/Close

Now to prove:

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

as promised.

Proof

Video Link

There are many videos of the proof:

• Proof: $\cos (a+b)=\cos (a)\cos (b)-\sin (a)\sin (b)$

The Proof

We want to prove:

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

We will use the trick from the exercise on the previous page of setting $AB=1$ and exactly the same diagram as last time.

Because $\mathrm{△}ABE$ is a right angle triangle with hypotenuse 1 and angle $\beta$ , we have:

$\overline{BE}=\sin (\beta )$

$\overline{AE}=\cos (\beta )$

And because $\mathrm{△}ABC$ is a right angle triangle with hypotenuse 1 and angle $(\alpha +\beta )$ , we have:

$\overline{AC}=\cos (\alpha +\beta )$

Let's express $\overline{AF}$ and $\overline{DE}$ in terms of cos and sine of the angles. You'll need to look at the diagram to see which triangles we are using.

An expression for $\overline{AF}$ $\overline{AF}=\overline{AE}\cos (\alpha )$ so $\overline{AF}=\cos (\alpha )\cos (\beta )$

An expression for $\overline{DE}$ $\overline{DE}=\overline{BE}\sin (\alpha )$ so $\overline{DE}=\sin (\alpha )\sin (\beta )$

$\cos (\alpha +\beta )=\overline{AC}=\overline{AF}-\overline{CF}=\overline{AF}-\overline{DE}$

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

We're done!

Another Way

The proof looks mighty similar to the proof for $\sin (\alpha +\beta )$ .

We can in fact derive one from the other without using a diagram at all.

Template:ExampleRobox Starting from:

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

We use $\sin (x)=\cos (90^{\circ }-x)$ and $\cos (x)=\sin (90^{\circ }-x)$ and (substituting in several places):

$\cos (90^{\circ }-(\alpha +\beta ))=\cos (90^{\circ }-\alpha )\cos (\beta )+\sin (90^{\circ }-\alpha )\sin (\beta )$

Now we use $\cos (-x)=\cos (x)$ and $\sin (-x)=-\sin (x)$

$\cos (-(90^{\circ }-(\alpha +\beta )))=\cos (-(90^{\circ }-\alpha ))\cos (\beta )-\sin (-(90^{\circ }-\alpha ))\sin (\beta )$

$\cos (\alpha +\beta -90^{\circ })=\cos (\alpha -90^{\circ })\cos (\beta )-\sin (\alpha -90)\sin (\beta )$

This is true for all $\alpha ,\beta$ so if we put $\alpha =A+90^{\circ }$ and $\beta =B$ we get:

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

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Now it is your turn to practice deriving new formulas from old ones:

Starting from

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

Show

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

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A somewhat harder exercise:

Using

$\tan (x)=\frac{\sin (x)}{\cos (x)}$

and the addition formulae for sin and cos, show that

$\tan (A+B)=\frac{\tan (A)+\tan (B)}{1-\tan (A)\tan (B)}$

And now it is your turn to do the geometric proof of addition formulas.

Template:ExerciseRobox You might want to skip this exercise and come back to it later after you have used the cosine addition formula for a bit. It is a good exercise for getting to the stage where you are confident you can write a geometric proof of the formulas yourself.

Start from the diagram below:

Add labels to it, and write out a proof of

• Sine addition formula
• Cosine addition formula

based on the diagram and the letters you have chosen. Make sure you explain by chasing angles why the two angles labelled $\beta$ are the same. The labels given to the edge lengths are to help you. Your proof must spell out why those labels are correct, using the trig relations.

Compare the diagram with the one in the proof above. Just how different are they really?

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