Value of $\sin^{-1}(\sin 10)$

Question

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?

Answer 1

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

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

and,

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

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

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

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

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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:

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

Answer 2

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.