5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

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