<s>[INST] Simplifying an inverse trig function? I am trying to figure out how to simplify this expression but I am not quite sure how these inverses work. What sort of approach should I take for this equation?
$$\tan\left(2\cos^{-1}\left(\dfrac{x}{5}\right)\right)$$
 [/INST] As @Spencer notes, there exists a triangle with hypotenuse 5 and adjacent $x$. Pythagoras tells us that we must therefore have an opposite of $\sqrt{5^2 - x^2} = \sqrt{25-x^2}$ 
If 
$$\cos(\theta) = x/5$$ 
then 
$$\tan(\theta) = \frac{\sqrt{25-x^2}}{x}$$
Given 
$$\tan(2\theta) = \frac{2\tan(\theta)}{1 - tan²(\theta)}$$
We now have
$$
\begin{array}{ll}
\tan\left(2\cos^{-1}\left(\frac{x}{5}\right)\right) &= \frac{2\frac{\sqrt{25-x^2}}{x}}{1 - \left(\frac{\sqrt{25-x^2}}{x}\right)^2} \\
&=2\frac{\frac{\sqrt{25-x^2}}{x}}{1 - \frac{25-x^2}{x^2}} \\
&=2x\frac{\sqrt{25-x^2}}{x^2 - (25-x^2)} \\
&=2x\frac{\sqrt{25-x^2}}{2x^2 - 25}
\end{array}
$$
Is that simpler?
</s>