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:

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

\[=\frac{\sin A \cos B \pm \sin B \cos A}{\cos A \cos B \mp \sin A \sin B}\]

\[=\frac{\frac{\sin A \cos B}{\cos A \cos B} \pm \frac{\sin B \cos A}{\cos A \cos B}}{\frac{\cos A \cos B}{\cos A \cos B} \mp \frac{\sin A \sin B}{\cos A \cos B}}\]

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

\[=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\]

\[Q.E.D.\]