Question 767090
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15\cos A\ -\ 8\sin A\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos A\ =\ \frac{8}{15}\sin A]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2A\ =\ \frac{64}{225}\sin^2A]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2A\ +\ \frac{64}{225}\left(1\ -\ \cos^2A\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{289}{225}\cos^2A\ =\ \frac{64}{225}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2A\ =\ \frac{64}{289}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos A\ =\ \pm\frac{8}{17}]


But *[tex \LARGE 0\ <\ A\ <\ \frac{\pi}{2}], hence *[tex \LARGE \cos A\ >\ 0]


and therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos A\ =\ \frac{8}{17}]


Then using


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\varphi\ +\ \sin^2\varphi\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin A\ =\ \frac{15}{17]


From there


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin A\ +\ \cos A}{2\cos A\ -\ \sin A}]


is just arithmetic.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>