There was a question on my math test that asked for the principal value of $\arcsin(\sin(10)$. My friend says it is $10-3 \pi$ but I say it is $3 \pi -10$. However, taking into account the definition of the range of the inverse sin function, couldnt it be either? Since the answer $A$ has to be such that $\frac{\pi}{2}<A<-\frac{\pi}{2}$.
Can someone clarify this?
Note that the value $A$ should satisfy $-\frac{\pi}{2}\leq A\leq \frac{\pi}{2}$; so there are two problems with what your wrote at the end:
As Henning notes, the answer cannot be both, because sine gives different answers at those two. However, you can verify that numbers that satisfy $-\frac{\pi}{2}\leq A \lt 0$ will have $\sin(A)\lt 0$, and numbers $A$ that satisfy $0\lt A\leq \frac{\pi}{2}$ will have $\sin(A)\gt 0$. Since you are looking for an angle that has the same value as $\sin(10)$, and $\sin(10)\lt 0$, the answer has to satisfy $-\frac{\pi}{2}\leq A\lt 0$.
If you think about the cycles of $\sin x$, you have a full cycle between $0$ and $2\pi$, one between $2\pi$ and $4\pi$; $10$ lies in the latter; and in fact, $3\pi\lt 10 \lt 4\pi$, so $\sin(10)$ will be negative, hence $\arcsin(\sin 10)$ will be negative. If you think about the graph of sine, you should see that the value will be either $3\pi-10$ or $10-4\pi$, whichever lies in the correct interval, and so the answer is 3\pi-10$.
To three digits:
$\qquad \pi = 3.14$
$\qquad 0.5\pi = 1.57$
$\qquad 3\pi = 9.42$
$\qquad 3.5\pi = 10.99$
This is accurate enough to state that $3\pi < 10 < \dfrac{7\pi}{2}$
Which puts $10$ radians in the $3^{rd}$ quadrant.
Which means that the "reference angle is $10-3\pi$ and $\sin 10 < 0$.
It follows that $\arcsin(\sin 10) = -(10 - 3\pi) = 3\pi - 10$.
