Question #db20d

Apr 14, 2017

$\cos x = \pm \sqrt{\frac{{m}^{2} - 1}{{n}^{2} - 1}}$

Explanation:

We have equivalently

$\left\{\begin{matrix}\sin \frac{x}{\sqrt{1 - {\sin}^{2} x}} = n \sin \frac{y}{\sqrt{1 - {\sin}^{2} y}} \\ \sin y = \frac{1}{m} \sin x\end{matrix}\right.$

or

$\sin \frac{x}{\sqrt{1 - {\sin}^{2} x}} = \left(\frac{n}{m}\right) \sin \frac{x}{\sqrt{1 - \frac{1}{m} ^ 2 {\sin}^{2} x}}$

or

$\sqrt{1 - \frac{1}{m} ^ 2 {\sin}^{2} x} = \left(\frac{n}{m}\right) \sqrt{1 - {\sin}^{2} x}$

and squaring both sides

$1 - \frac{1}{m} ^ 2 {\sin}^{2} x = {\left(\frac{n}{m}\right)}^{2} \left(1 - {\sin}^{2} x\right)$

solving for ${\sin}^{2} x$ we have

${\sin}^{2} x = \frac{{n}^{2} - {m}^{2}}{{n}^{2} - 1} = 1 - {\cos}^{2} x$

and finally

${\cos}^{2} x = \frac{{m}^{2} - 1}{{n}^{2} - 1}$

$\cos x = \pm \sqrt{\frac{{m}^{2} - 1}{{n}^{2} - 1}}$

Apr 15, 2017

Given

$\tan x = n \tan y \ldots \ldots . . \left[1\right]$

$\sin x = n \sin y \ldots \ldots . . \left[2\right]$

Dividing  by 

$\cos x = \frac{m}{n} \cos y \ldots \ldots . . \left[3\right]$

$\implies \cos y = \frac{n}{m} \cos x$

Squaring 

${\sin}^{2} x = {m}^{2} {\sin}^{2} y$

$\implies 1 - {\cos}^{2} x = {m}^{2} \left(1 - {\cos}^{2} y\right)$

$\implies 1 - {\cos}^{2} x = {m}^{2} \left(1 - {n}^{2} / {m}^{2} {\cos}^{2} x\right)$

$\implies 1 - {\cos}^{2} x = {m}^{2} - {n}^{2} {\cos}^{2} x$

$\implies {n}^{2} {\cos}^{2} x - {\cos}^{2} x = {m}^{2} - 1$

$\implies \left({n}^{2} - 1\right) {\cos}^{2} x = {m}^{2} - 1$

$\implies {\cos}^{2} x = \frac{{m}^{2} - 1}{{n}^{2} - 1}$

$\implies \cos x = \pm \sqrt{\frac{{m}^{2} - 1}{{n}^{2} - 1}}$