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?
$$\tan\left(2\cos^{-1}\left(\dfrac{x}{5}\right)\right)$$
 A: You will need two things. One is the identity for $tan(2x)$ in terms of $tan(x)$. You can look this up the purpose is to get rid of the $2$ in your problem.
The other thing is to know what $\tan(\cos^{-1}(x/5))$ is.
You can figure out by realizing that the angle $\cos^{-1}(x/5)$ corresponds to a right triangle with an adjacent side $x$ and a hypotenuse $5$. With this information you should be able to solve the right triangle completely and determine what the tangent of that angle is. 
The key to handling inverse trig functions is to realize that they represent some angle in a right triangle. For instance $\theta = \tan^{-1}(x+2)$ is an angle whose opposite side is $(x+2)$ and whose adjacent side is $1$. The hypotenuse of this hypothetical triangle is $\sqrt{(x+2)^2+1}$. Therefore we can conclude that $\sin(\tan^{-1}(x+2)) = \frac{x+2}{\sqrt{(x+2)^2+1}}$
A: 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?
