Since I don't have the answer to this one, I want to make sure I've done this correctly.
Show that
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
Since
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
we have
$$\tan\left(\arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)\right) = x$$
Applying the Pythagorean theorem, we learn that $b = 1$, so we have $\tan(x/1) = x$.
It may be a bit short, but I really want to be sure I have this right before I continue on. 🙂
Best Answer
The connection is way more revealing in its simplicty if you just use trigonometry. For $x$ positive take a right-angled triangle with sides $\overline{AB} = 1$ and $\overline{BC} = x$. Then by definition $$\angle CAB = \arctan x.$$ The hypotenuse measures $\overline{AC} = \sqrt{1+x^2}$, by Pythagorean Theorem. So the same angle can be also defined as $$\angle CAB = \arcsin \left(\frac{x}{\sqrt{1+x^2}}\right).$$ For negative $x$ take $\overline{BC} = -x$ and recall the odd symmetry of both sine and tangent. As easy as that.