0
$\begingroup$

The trigonometric functions ($ \sin, \cos, \tan,...$) aren't invertible in $ R $ so we restrain it's domain to create inverses for them, the inverse trigonometric functions ($ \arcsin, \arccos, \arctan,...$). I'm not sure if it's a convention but all the books i've read restrain the sine function to the interval $ [-\pi/2, \pi/2] $ and the cosine function to the interval $ [0, \pi] $. My question is:

For what reasons mathematicians chose these intervals?

Thanks in advance

$\endgroup$
  • $\begingroup$ Like most conventions, it's done for the reson that it's convenient. If there was one way to do it which was objectively "better" than any alternative (mathematically speaking), we wouldn't call it "convention", we would call it "natural", "canonical", or something similar. $\endgroup$ – Arthur Mar 24 '18 at 14:53
3
$\begingroup$

Yes, it a convention. And a natural one. The longest possible length for such an interval os $\pi$ (after that, the functions aren't injective). And it's a natural choice to pick an interval starting at $0$ or containing it in its middle. The leaves us only with the choice of $\left[-\frac\pi2,\frac\pi2\right]$ for $\arcsin$ and $[0,\pi]$ for $\arccos$.

$\endgroup$
2
$\begingroup$

Well, we do need to restrain $\arcsin, \arccos$ to intervals on which $\sin, \cos$ are injective and upon which they achieve all possible values. As the maximums of these continuous functions are are $1$ and the minimums are $-1$ we must choose an intervals $[a,b]$ and $[c,d]$ where $\sin a = \pm 1; \sin b \mp 1$, $\cos a = \pm 1; \cos b = \mp 1$ and $b-a = d-c =$ half the period =$ \pi$.

Those are $[a = n\pi - \frac \pi 2, b=n\pi +\frac \pi 2]$ and $[c = m\pi, d= (m+1)\pi]$.

So why the convention that we let $n = m = 0$? Well, is that not the most natural/simple choice?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.