# Given {(sinx+siny=a),(cosx+cosy=b):} calculate sin((x-y)/2)= ?

Dec 21, 2016

$\sin \left(u + v\right) + \sin \left(u - v\right) = 2 \sin \left(u\right) \cos \left(v\right) = a$
$\cos \left(u + v\right) + \cos \left(u - v\right) = 2 \cos \left(u\right) \cos \left(v\right) = b$

so

$4 {\sin}^{2} \left(u\right) {\cos}^{2} \left(v\right) = {a}^{2}$
$4 {\cos}^{2} \left(u\right) {\cos}^{2} \left(v\right) = {b}^{2}$

$4 {\cos}^{2} \left(v\right) = {a}^{2} + {b}^{2}$ or

$4 \left(1 - {\sin}^{2} \left(v\right)\right) = {a}^{2} + {b}^{2}$ or

$\sin \left(v\right) = \pm \frac{1}{2} \sqrt{4 - \left({a}^{2} + {b}^{2}\right)}$

but $u + v = x$ and $u - v = y$ so $v = \frac{x - y}{2}$ and finally

$\sin \left(\frac{x - y}{2}\right) = \pm \frac{1}{2} \sqrt{4 - \left({a}^{2} + {b}^{2}\right)}$

Dec 24, 2016

Given

$\sin x + \sin y = a \ldots \ldots . . \left[1\right]$

$\cos x + \cos y = b \ldots \ldots . \left[2\right]$

Squaring and adding [1] and [2] we get

${\left(\sin x + \sin y\right)}^{2} + {\left(\cos x + \cos y\right)}^{2} = {a}^{2} + {b}^{2}$

$\implies {\sin}^{2} x + {\sin}^{2} y + {\cos}^{2} x + {\cos}^{2} y + 2 \left(\cos x \cos y + \sin x \sin y\right) = {a}^{2} + {b}^{2}$

$\implies 2 + 2 \left(\cos \left(x - y\right)\right) = {a}^{2} + {b}^{2}$

$\implies 4 - 2 + 2 \left(\cos \left(x - y\right)\right) = {a}^{2} + {b}^{2}$

$\implies 4 - 2 \left[1 - \cos \left(x - y\right)\right] = {a}^{2} + {b}^{2}$

$\implies 4 - 2 \cdot 2 {\sin}^{2} \left(\frac{x - y}{2}\right) = {a}^{2} + {b}^{2}$

$\implies {\sin}^{2} \left(\frac{x - y}{2}\right) = \frac{4 - {a}^{2} - {b}^{2}}{4}$

$\implies \sin \left(\frac{x - y}{2}\right) = \pm \sqrt{\frac{4 - {a}^{2} - {b}^{2}}{4}} = \pm \frac{\sqrt{4 - {a}^{2} - {b}^{2}}}{2}$