Addition Formulae Proof using Euler's Equation

The addition formulae are:

\[\sin(A \pm B)=\sin A \cos B \pm \sin B \cos A\]

\[\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B\]

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

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:

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

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


Proof for tan

Using the results that we just proved:

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

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

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

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

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


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