## Addition Formulae Proof using Euler's Equation

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

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

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

Proof for sin and cos

Using Euler's Equation (a proof of which can be found here), $$e^{ix}=\cos x +i\sin x$$, let $$x=A+B$$ :

$e^{i(A+B)}=\cos (A+B) +i\sin (A+B)$

$e^{iA+iB}=\cos (A+B) +i\sin (A+B)$

$e^{iA} \times e^{iB}=\cos (A+B) +i\sin (A+B)$

Using Euler's Equation again gives us that $$e^{iA}=\cos A +i\sin A$$ and $$e^{iB}=\cos B +i\sin B$$. Substituting this in:

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

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

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

Comparing real and imaginary parts gives:

$①:\cos(A + B)=\cos A \cos B - \sin A \sin B$

$②:\sin(A + B)=\sin A \cos B + \sin B \cos A$

Now to prove the formula for $$A-B$$. If we substitute in $$-B$$ instead of $$B$$, we get:

$①:\cos(A -B)=\cos A \cos -B - \sin A \sin -B$

$②:\sin(A -B)=\sin A \cos -B + \sin -B \cos A$

Notice that sin is an odd function, i.e. $$\sin -x = -\sin x$$, whilst cos is an even function, i.e. $$\cos -x = \cos x$$. This means we can rewrite the above equations, giving:

$①:\cos(A -B)=\cos A \cos B + \sin A \sin B$

$②:\sin(A -B)=\sin A \cos B + \sin B \cos A$

$QED$

## Proof for tan

Using the results that we just proved:

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

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

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

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

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

$QED$