# How do you write sin(arctan(a)) as an expression of "a" not involving any trigonometric functions or their inverses?

Jun 23, 2015

Let $\alpha = \arctan \left(a\right)$.

Then $- \frac{\pi}{2} < \alpha < \frac{\pi}{2}$, and $\cos \alpha > 0$.

So $\cos \alpha = \sqrt{1 - {\sin}^{2} \left(\alpha\right)}$

Hence $\sin \left(\alpha\right) = \frac{a}{\sqrt{1 + {a}^{2}}}$

#### Explanation:

Having let $\alpha = \arctan \left(a\right)$, we have:

$a = \tan \left(\alpha\right) = \sin \frac{\alpha}{\cos} \left(\alpha\right)$

$= \sin \frac{\alpha}{\sqrt{1 - {\sin}^{2} \left(\alpha\right)}}$

Multiply both ends by $\sqrt{1 - {\sin}^{2} \left(\alpha\right)}$ to get:

$a \sqrt{1 - {\sin}^{2} \left(\alpha\right)} = \sin \left(\alpha\right)$

Square both sides to get:

${\sin}^{2} \left(\alpha\right) = {a}^{2} \left(1 - {\sin}^{2} \left(\alpha\right)\right)$

$= {a}^{2} - {a}^{2} {\sin}^{2} \left(\alpha\right)$

Note that this will introduce a spurious solution to be eliminated later.

Add ${a}^{2} {\sin}^{2} \left(\alpha\right)$ to both ends to get:

$\left(1 + {a}^{2}\right) {\sin}^{2} \left(\alpha\right) = {a}^{2}$

Divide both sides by $\left(1 + {a}^{2}\right)$ to get:

${\sin}^{2} \left(\alpha\right) = {a}^{2} / \left(1 + {a}^{2}\right)$

So $\sin \left(\alpha\right) = \pm \frac{a}{\sqrt{1 + {a}^{2}}}$

Now remember that $- \frac{\pi}{2} < \alpha < \frac{\pi}{2}$, so $\cos \alpha > 0$.

Therefore $\sin \left(\alpha\right)$ must have the same sign as $\tan \left(\alpha\right) = a$

So the correct solution is $\sin \left(\alpha\right) = \frac{a}{\sqrt{1 + {a}^{2}}}$