Computing $\arctan x$ in terms of a certain collection of other functions

Question

I know that $$\tan(x) = \frac{\sin(x)}{\cos(x)}.$$ Does this relationship hold in the inverse in any form? For example:

• atan(x) = asin(x) / acos(x), or
• atan(x) = acos(x) / asin(x), or
• atan(x) = asin(x) acos(x),
• or something else similar (you get the idea).

Context: There's a game that involves trigonometry and offers a limited in-game scripting environment, which I'm trying to use to semi-automate the math. I have $\sin()$, $\cos()$, $\text{asin}()$, $\text{acos}()$, $\log()$, and $\text{sqrt}()$ functions (and some others, but nothing more mathematically complex), but no $\tan()$ or $\text{atan}()$, and $\text{atan}()$ is the one I need. So, I'm trying to compose it, but I have an interested layman's understanding of the relationships between the various quantities (that is to say: almost none, and certainly nothing specific enough to be helpful).

Halp? (Correct answers will be rewarded with screenshots of the devastation I inflict in-game with the help of my trig script.)

Answer 1

The indicated relationships cannot hold, more or less just reasons that the domains of the involved functions aren't compatible.

On the other hand, if we denote by $\theta$ the measure of one acute vertex of a right triangle, and $x$ the length of the leg opposite the labeled angle, and then declare the length of the leg adjacent to the angle to be $1$, by definition we can write $\theta$ as both $$\theta = \arctan x \qquad \text{and} \qquad \theta = \arcsin \frac{x}{\sqrt{1 + x^2}},$$ giving a formula for $\arctan$ in terms of $\arcsin$ and elementary operations: $$\arctan x = \arcsin \frac{x}{\sqrt{1 + x^2}}.$$

One could just as well write $\arctan x$ in terms of, say, $\arccos x$.

(One can also easily write $\arctan$ in terms of the complex logarithm function, but presumably this isn't what you meant by $\log$.)

Answer 2

$$ \arctan{x} = \arcsin{\left(\frac{x}{\sqrt{1+x^2}}\right)} $$ Proof: there is a unique $y$ with $-\pi/2 < y <\pi/2 $ and $x=\tan{y}$. Then $$ \sin{y} = \frac{\tan{y}}{\sec{y}} = \frac{\tan{y}}{\sqrt{1+\tan^2{y}}} = \frac{x}{\sqrt{1+x^2}}, $$ and take $\arcsin$ of both sides.