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.

  • $\begingroup$ I suppose arcsin assumes the principal value $\endgroup$
    – Paladin
    Apr 15, 2017 at 3:06
  • 4
    $\begingroup$ This is a good question, and the answer is roughly the same as the answer to the question "why is $\sqrt{4}=2$ and not $-2$?" There are lots of angles whose sine is $-4/5$, but we want $\arcsin$ to be a function, so we pick a certain range of angles in which the solution is unique. For $\arcsin$, it is $-90^\circ$ to $90^\circ$, for $\arccos$ it is $0^\circ$ to $180^\circ$, and so on. $\endgroup$
    – user856
    Apr 15, 2017 at 3:14

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .