Simplifying $\sin\left(\arctan\frac x a\right)$ without geometry

Question

I want to simplify the expression $$\sin\left(\arctan\frac x a\right)$$

I can do this with geometry. In a triangle $ABC$, let $AB=x$, $BC=a$ and $\angle B=90^\circ$. Then, $\arctan\frac xa=\angle C$ and $$\sin\left(\arctan\frac x a\right)=\frac{AB}{AC}=\frac{x}{\sqrt{a^2+x^2}}$$

But how to simplify this without geometry; i.e, using trigonometric formulas and algebra?

I tried using the formula: $$\sin x=\frac{2\tan\frac x2}{1+\tan^2\frac x2}$$ and got the expression $$\sin\left(\arctan\frac x a\right)=\frac{2\tan\left(\dfrac {\arctan\frac xa}2\right)}{1+\tan^2\left(\dfrac {\arctan\frac xa}2\right)}$$ But this doesn't seem to work.

Answer 1

By the Pythagorean identities, each trig function can be expressed in terms of the others. Here you can use the identity

$$\sin \theta =\pm \frac{\tan \theta}{\sqrt{ 1+\tan^2 \theta}}, $$

and the result follows (note the $\pm$).

Answer 2

Draw a right-angled triangle with the non-hypotenuse edges having length x and a. The angle opposite x is arctan of x/a. Now, you want to compute sine of this angle. You should know what to do!

Answer 3

Easier with a trig drawing that is the basis of the Pythagorean triangle/thm.

Let $ t= \dfrac{x}{a}= \dfrac{t}{1}=\tan \theta $

Read off directly from the drawing

$$\sin \theta =\pm \frac{ t}{\sqrt{ 1+t^2}} $$

Due to the radical sign we needed to insert $\pm$in front of the fraction.

This ambiguity meant that our $\theta $ was either in the first or in the third quadrants.