If sinx+siny=a and cosx+cosy=b how do you find x,y ?

2 Answers
Aug 30, 2016

Given

$\sin x + \sin y = a \ldots \ldots \left(1\right)$

$\cos x + \cos y = b \ldots . . \left(2\right)$

Squaring and adding (1) and (2)

${\cos}^{2} x + {\sin}^{2} x + {\cos}^{2} y + {\sin}^{2} y + 2 \cos x \cos y + 2 \sin x \sin y = {a}^{2} + {b}^{2}$

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

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

$\implies x - y = {\cos}^{-} 1 \left(\frac{{a}^{2} + {b}^{2} - 2}{2}\right) \ldots \ldots \left(3\right)$

Dividing (1) by (2)

$\frac{\sin x + \sin y}{\cos x + \cos y} = \frac{a}{b}$

=>(2sin((x+y)/2)cos((x-y)/2))/ (2cos((x+y)/2)cos((x-y)/2))=a/b

$\implies \tan \left(\frac{x + y}{2}\right) = \frac{a}{b}$

$\implies \frac{x + y}{2} = {\tan}^{-} 1 \left(\frac{a}{b}\right)$

$\implies \left(x + y\right) = 2 {\tan}^{-} 1 \left(\frac{a}{b}\right) \ldots \ldots . \left(4\right)$

Adding (3) & (4)

$\implies 2 x = {\cos}^{-} 1 \left(\frac{{a}^{2} + {b}^{2} - 2}{2}\right) + 2 {\tan}^{-} 1 \left(\frac{a}{b}\right)$

$\implies x$
$= {\tan}^{-} 1 \left(\frac{a}{b}\right) + \frac{1}{2} {\cos}^{-} 1 \left(\frac{{a}^{2} + {b}^{2} - 2}{2}\right)$

Subtracting (3) from (4)

$\implies 2 y$
$= 2 {\tan}^{-} 1 \left(\frac{a}{b}\right) - {\cos}^{-} 1 \left(\frac{{a}^{2} + {b}^{2} - 2}{2}\right)$

$\implies y$
$= {\tan}^{-} 1 \left(\frac{a}{b}\right) - \frac{1}{2} {\cos}^{-} 1 \left(\frac{{a}^{2} + {b}^{2} - 2}{2}\right)$

Aug 30, 2016

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

Explanation:

Calling $u = \sin x$ and $v = \sin y$ we have

{ (u + v = a), (sqrt(1-u^2)+sqrt(1-v^2) = b) :}

or

{ (u + v = a), (sqrt(1-(a-v)^2)+sqrt(1-v^2) = b) :}

squaring the second equation

$1 - {a}^{2} - {v}^{2} - 2 a v = {b}^{2} + 1 - {v}^{2} - 2 b \sqrt{1 - {v}^{2}}$

or

$2 b \sqrt{1 - {v}^{2}} = {b}^{2} + {a}^{2} - 2 a v$

squaring again

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

and

$4 \left({a}^{2} + {b}^{2}\right) {v}^{2} - 4 \left({a}^{2} + {b}^{2}\right) a v + {\left({b}^{2} + {a}^{2}\right)}^{2} - 4 {b}^{2} = 0$

Solving for $v$ we obtain

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

and

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

Finally

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