Simplify $\sin \arctan x$, assuming $x>0$ Now, naturally I understand that $$\arctan x = \arcsin \frac{x}{\sqrt{x^2+1}}$$ and therefore I should have just $$\sin \arctan x = \frac{x}{\sqrt{x^2 + 1}}.$$ However, I am interested as to the effect of restraining the domain to $\{x &gt; 0\}$, as I believe the function may simplify further. Could someone point me in the right direction?
It cannot be simplified further. There is nothing simpler than  $ x/ \sqrt{ 1 -x^2} $ next in simplicity only to $ \tan^{-1}x $. Draw a triangle to look at the sides $ (\sqrt{1+x^2}, 1,x )$ for yourself. Domain is $ (-\infty, \infty) $.Choose only the positive part if you need only positive argument. Also there is atan2 function or equivalent to make a proper quadrant choice for numerical work for several CAS.
It cannot be simplified further. There is nothing simpler than  $ x/ \sqrt{ 1 -x^2} $ next in simplicity only to $ \tan^{-1}x $. Draw a triangle to look at the sides $ (\sqrt{1+x^2}, 1,x )$ for yourself. Domain is $ (-\infty, \infty) $.Choose only the positive part if you need only positive argument. Also there is atan2 function or equivalent to make a proper quadrant choice for numerical work for several CAS.