## Geometric Proof of the Addition Formulae

The Addition Formulae are given by:

$\sin(A \pm B)=\sin A \cos B \pm \cos A \sin B$

$\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B$

$\tan(A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

## Proof for sin

Consider the following two triangles: We can draw in the height of these two triangles, like so: We can see by basic trigonometry that since the hypotenuse of the new big triangle (triangle xyz) is 1, the height must be $$\sin(A+B)$$ : The hypotenuse of the top triangle is 1, so by basic trigonometry, the base of the top triangle is $$\cos B$$ and the height is $$\sin B$$. This also means that the height of the bottom triangle is $$\sin A \cos B$$ (since the hypotenuse is $$\cos B$$ and the angle is $$A$$) : We can now extend the height of the bottom triangle to match the height of the whole construction, like so: Notice that we've formed a new triangle in the top right corner. Notice also that angle $$\alpha$$ is $$90-A$$ and therefore angle $$\beta$$ is $$A$$. This means that, again by basic trigonometry, the right hand side of the new triangle is equal to $$\cos A \sin B$$ (since the hypotenuse is $$\sin B$$ and the angle is $$A$$): Notice now that the line of length $$\sin(A+B)$$ is the same height as the lines of lengths $$\sin A \cos B$$ and $$\cos A \sin B$$ combined.

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

We can easily prove the formula for $$\sin(A-B)$$ by swapping $$B$$ for $$-B$$ in the above formula and noticing that $$\cos (-B)=\cos B$$ and $$\sin (-B)=- \sin B$$.  $$Q.E.D.$$

## Proof for cos

We know that $$\cos x = \sin \left(\frac{\pi}{2}-x\right)$$. Therefore:

$\cos (A \pm B) = \sin \left(\frac{\pi}{2}-(A \pm B)\right)$

$=\sin \left(\left(\frac{\pi}{2}-A\right) \pm B)\right)$

$=\sin \left(\frac{\pi}{2}-A\right) \cos B \pm \cos \left(\frac{\pi}{2}-A\right) \sin B$

$=\cos A \cos B \pm \sin A \sin B$

$Q.E.D.$

## Proof for tan

Using the results that we just proved:

$\tan (A \pm B)=\frac{\sin (A \pm B)}{\cos (A \pm B)}$

$=\frac{\sin A \cos B \pm \sin B \cos A}{\cos A \cos B \mp \sin A \sin B}$

$=\frac{\frac{\sin A \cos B}{\cos A \cos B} \pm \frac{\sin B \cos A}{\cos A \cos B}}{\frac{\cos A \cos B}{\cos A \cos B} \mp \frac{\sin A \sin B}{\cos A \cos B}}$

$=\frac{\frac{\sin A}{\cos A} \pm \frac{\sin B}{\cos B}}{1 \mp \frac{\sin A \sin B}{\cos A \cos B}}$

$=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

$Q.E.D.$