Why can we always draw right triangles for inverse trig functions? In trigonometry it's a basic technique to evaluate an expression like $$\tan\left(\arcsin\left(-\frac{4}{5}\right)\right)$$ by thinking of $\arcsin\left(-\frac{4}{5}\right)$ as an angle, setting  $\arcsin\left(-\frac{4}{5}\right) = \theta$, and drawing a right triangle to determine the solution. 
My question is why can we always draw a right triangle to represent these inverse trig expressions? I usually think of an expression like $\arcsin\left(-\frac{4}{5}\right)$ as "the angle whose sine is $-\frac{4}{5}$", but what if this angle is greater than $180$ degrees? What justifies drawing this angle in a right triangle if it is larger than the entire angle measure of the triangle? I appreciate any clarification. Thanks.
 A: Take a look at the graph of $\sin(x)$.  It isn't one-to-one (one way to tell is since it doesn't pass the Horizontal Line Test, its "inverse" wouldn't pass the Vertical Line Test, so its "inverse" isn't a function).  So we restrict the domain of $\sin(x)$ (and the other trig functions, too) to let us even get inverse sine to be a function.
We do that by restricting the domain of $\sin(x)$ to be from $-\pi/2$ to $\pi/2$.  Why that interval? It kind of makes sense to pick the "first" wave of a sine curve to be the restricted domain, so we could pick either $[-\pi/2, \pi/2]$ or $[\pi/2, 3\pi/2]$ and it kind of makes more sense to center it around $0$. So the mathematicians of the past defined arcsin as such.
So when we look at $\arcsin(-4/5)$, we want the angle in the first or fourth quadrant.  Since we're looking for an angle $\theta$ such that $\sin(\theta)=-4/5$, we want sine to be negative, which happens in the fourth quadrant, and then we build our right triangle like we usually would.
