Another question I have unfortunately drawn a blank on..
Any help is greatly appreciated!
$\displaystyle \sin x(\cos y+2\sin y)=\cos x(2\cos y-\sin y)\Leftrightarrow$
$\displaystyle \Leftrightarrow\sin x\cos y+2\sin x\sin y=2\cos x\cos y-\sin y\cos x\Leftrightarrow$
$\displaystyle \Leftrightarrow \sin x\cos y+\sin y\cos x=2(\cos x\cos y-\sin x\sin y)\Leftrightarrow$
$\displaystyle \Leftrightarrow\sin(x+y)=2\cos(x+y)\Leftrightarrow \tan(x+y)=2$
$\displaystyle \sinx\cosy+2 \sinx\siny=2 \cosx\siny-\cosx\siny$
$\displaystyle \sinx\cosy+\cosx\siny=2(\cosx\cosy-\sinx\siny)$
$\displaystyle \sin(x+y)=2cos(x+y)$
$\displaystyle \frac{\sin(x+y)}{cos(x+y)}=2$
$\displaystyle \tan(x+y)=2$
Sorry about some confusion .. Can someone pls check my latex .