# How do you verify the identity (sinx+-siny)/(cosx+cosy)=tan((x+-y)/2)?

Aug 18, 2017

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

Consider the LHS (positive case):

$\frac{\sin x + \sin y}{\cos x + \cos y} = \frac{2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)}{2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)}$

$\text{ } = \frac{\sin \left(\frac{A + B}{2}\right)}{\cos \left(\frac{A + B}{2}\right)}$

$\text{ } = \tan \left(\frac{A + B}{2}\right)$

Consider the LHS (negative case):

$\frac{\sin x - \sin y}{\cos x + \cos y} = \frac{2 \cos \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)}{2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right)}$

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

$\text{ } = \tan \left(\frac{A - B}{2}\right)$

Hence, combining both cases, we have:

$\frac{\sin x \pm \sin y}{\cos x + \cos y} = \tan \left(\frac{A \pm B}{2}\right) \setminus \setminus \setminus \setminus$ QED