Why does $|\sin(\sqrt{4\pi^2n^2+x})| =|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|$ I need in a demonstration that $$|\sin(\sqrt{4\pi^2n^2+x})|
=|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|$$
but I don't understand why this equality is true. Can someone explain it to me?

From the angle-sum formulas $$\sin (A\pm B)=\sin A \cos B \pm \cos A \sin B$$ $$ \cos (A\pm B)=\cos A \cos B\mp\sin A\sin B$$ we obtain $$\sin C-\sin D=2\sin A\cos B$$ where $A=(C-D)/2$ and $B=(C+D)/2.$
If $y>0$ and $4\pi^2y^2+x>0,$  let $C=\sqrt {4\pi^2y^2+x}$ and $D=2\pi y.$ We have $$|A|=\frac {1}{2}|C-D|=\frac {1}{2}\frac {|C^2-D^2|}{C+D}=$$ $$=\frac {|x|}{2(C+D)}\le\frac {|x|}{2D}=\frac {|x|}{4\pi y}.$$ Therefore if $y>0$ and $4\pi^2y^2+x>0$ then $$|\sin \sqrt {(4\pi^2y^2+x)} -\sin 2\pi y\,|=|\sin  C-\sin D|=$$ $$=|2\sin A \cos B|\le|2\sin A|\le 2|A|\le \frac {|x|}{2\pi y}$$ which $\to 0$ as $y\to \infty.$
