Trigonometry/Addition Formula for Cosines

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Cosine Formulas[edit]

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

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

Before we do that we will talk about subtraction formulas.

Subtraction formulas

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 \displaystyle \sin(-x) =-\sin x and \displaystyle \cos(-x) =\cos x.

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

First the sine addition formula:

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


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

Now for the cosine addition formula:

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


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

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:

\displaystyle\cos(\alpha\pm \beta)=\cos\alpha\cos\beta \mp \sin\alpha\sin\beta
\displaystyle\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.

Now to prove:

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

as promised.


Video Link[edit]

There is a video of the proof which may be easier to follow at the Khan Academy:

The Proof[edit]


We want to prove:

\displaystyle\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 \displaystyle AB=1 and exactly the same diagram as last time.

Because \triangle ABE is a right angle triangle with hypotenuse 1 and angle \displaystyle \beta, we have:

\displaystyle \overline{BE}=\sin\beta
\displaystyle \overline{AE}=\cos\beta

And because \triangle ABC is a right angle triangle with hypotenuse 1 and angle \displaystyle (\alpha + \beta), we have:

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

Let's express \displaystyle \overline{AF} and \displaystyle \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 \displaystyle \overline{AF}

\displaystyle \overline{AF}=\overline{AE}\cos\alpha


\displaystyle \overline{AF}=\cos\beta\cos\alpha

An expression for \displaystyle \overline{DE}

\displaystyle \overline{DE}=\overline{BE}\sin\alpha


\displaystyle \overline{DE}=\sin\beta\sin\alpha

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

We're done!

Another Way[edit]

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

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

Worked Example: Cosine Addition Formula from Sine Addition Formula

Starting from:

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

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

\displaystyle\cos( 90^\circ - (\alpha +\beta)) = \cos( 90^\circ- \alpha) \cos \beta + \sin( 90^\circ- \alpha) \sin \beta

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

\displaystyle\cos( -(90^\circ - (\alpha +\beta))) = \cos( -(90^\circ- \alpha)) \cos \beta - \sin(-( 90^\circ- \alpha)) \sin \beta
\displaystyle\cos( \alpha +\beta -90^\circ) = \cos( \alpha - 90^\circ) \cos \beta - \sin( \alpha -90) \sin \beta

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

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

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

Exercise: Sine Addition Formula from Cosine Addition Formula

Starting from

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


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

A somewhat harder exercise:

Exercise: Tangent Addition Formula


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

and the addition formulae for sin and cos, show that

\displaystyle \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.

Exercise: Using a different Diagram for the proof

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:

An alternative diagram

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?