a question about arcsin

[bbm25]

 

[ksdhart2]

If I'm understanding you right, you've noticed that:

$\displaystyle \sin \left(-\frac{\pi}{3} \right) = -\frac{\sqrt{3}}{2}$ and $\displaystyle \sin \left(\frac{5\pi}{3} \right) = -\frac{\sqrt{3}}{2}$

But are struggling to reconcile this information with what your calculator tells you:

$\displaystyle \sin^{-1} \left( -\frac{\sqrt{3}}{2} \right) = -\frac{\pi}{3}$

Assuming this is correct, the trick here lies in the fact that we define $\displaystyle \sin^{-1}(x) = \arcsin(x)$ as the inverse function of sine. This means that for every $x$ in the domain, there must be exactly one output. To ensure this, we restrict the domain of inverse sine to $\displaystyle [-1, 1]$ which also necessarily restricts the range to $[-\frac{\pi}{2}, \frac{\pi}{2}]$

Hence, the inverse sine function maps $-\frac{\sqrt{3}}{2}$ uniquely to $-\frac{\pi}{3}$, rather than any of the infinitely many other $y$ such that $\sin(y) = -\frac{\sqrt{3}}{2}$

 

[Dr.Peterson]

To state it just a little differently, we define the arcsine function by "arcsin(x) is the angle [MATH]\theta[/MATH] in the interval [MATH][-\pi/2\le \theta \le\pi/2][/MATH] whose sine is x". It is not just any such angle.

We have to make such a restriction so that it will be a function; other intervals could be used, but this is what we agree to use.

 

[Steven G]

View attachment 11450

arcsin -sqrt(3)/2 really has no meaning. It seems to me to say arcsin minus[sqrt(3)/2]. Since arcsin has no meaning you can't subtract sqrt(3)/2 from it. You meant to write arcsin(-sqrt(3)/2) which is read as arcsin of (-sqrt(3)/2)

 

[HallsofIvy]

It's not that "arcsin has no meaning", it is that "arcsin" is NOT a number so that you cannot subtract a number from it.