0
$\begingroup$

For the past few hours, I have been trying to find out the value of the following term: $$\sin^{-1}(\sin 10) $$

I understand that the answer will not be $10$ as the range for a standard sine inverse function is $\big[-{\pi}/{2}, {\pi}/{2}\big]$ and $10^c$ is outside that range.

Now, I know that $1^c \approx 57^{\circ}$, with that in mind, we can state $10^c \approx 570^{\circ}$. We can write $570^{\circ} = 3\pi + {\pi}/{6}$.

Therefore, $$\sin^{-1}(\sin 10) = \sin^{-1}\bigg(\sin (3\pi + \dfrac{\pi}{6})\bigg) = \sin^{-1} \bigg( \sin(\pi + \dfrac{\pi}{6}) \bigg) = \sin^{-1} \big(-\frac{1}{2}\big) = - \dfrac{\pi}{6} $$

But the answer provided is $(3\pi - 10)$.

How and why? Which part of my process is wrong?

$\endgroup$
4
  • 1
    $\begingroup$ The value which you claim is approximate . $3π -10 $ would be the exact value . Observe that $3π - 10 < 0 $ $\endgroup$
    – user925963
    May 21 at 16:02
  • $\begingroup$ Yes, I do understand that $- \pi/6$ is an approximation, but how do I get to $3\pi -10$? $\endgroup$
    – Kiryu
    May 21 at 16:03
  • $\begingroup$ Because $\sin x$ has period $2\pi$ and \sin(\pi-x)=\sin x$. $\endgroup$
    – Bernard
    May 21 at 16:09
  • 1
    $\begingroup$ You already said that the Principal value branch is $[- \pi/2 , \pi/2 ] $so we need to " adjust " $10$ so that it may come under the required interval . You can try yourself that $3\pi - 10 $ would perfectly land in the interval . So that must be the answer. $\endgroup$
    – user925963
    May 21 at 16:10
4
$\begingroup$

We will be using the fact that for all real $x$,

$$\,\forall k\in\mathbb{Z}, \, \sin(2k\pi+x) = \sin(x) $$

and,

$$\sin(\pi-x) = \sin(x)$$


$$\sin(10) = \sin(-2\pi+10) = \sin(\pi-(-2\pi+10)) = \sin(3\pi-10)$$

And $3\pi-10 \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ therefore,

$$\sin^{-1}(\sin(10)) = \sin^{-1}(\sin(3\pi-10)) = 3\pi-10$$


Your answer gives you an approximation of the result because you assumed that $10 \, \mathrm{rad} \approx570° = 3\pi+\dfrac{\pi}{6}$ therefore:

$$\sin^{-1}(\sin(10)) \approx \sin^{-1}\left(\sin\left(3\pi+\dfrac{\pi}{6} \right) \right) = -\dfrac{\pi}{6}$$

Indeed you can check that:

$$\left|(3\pi -10)-\dfrac{\pi}{6}\right| = 10-\dfrac{19 \pi}{6} \approx 0.0516$$

$\endgroup$
1
$\begingroup$

Observe from the periodicity of the $\sin$ function that $\sin\theta=\sin x$ if and only if for each $k\in\mathbb Z,$ $$\theta=(-1)^k x+k\pi.$$

For the problem at hand, $x=10,$ and we wish to find the integer $k$ such that $\theta\in[-\frac\pi2,\frac\pi2].$ A quick test-and-check narrows down the possibilities to just $k=3.$ Thus, the required value is $$3\pi-10,$$ as given.

$\endgroup$

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.