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

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?

• The value which you claim is approximate . $3π -10$ would be the exact value . Observe that $3π - 10 < 0$
– user925963
May 21 at 16:02
• Yes, I do understand that $- \pi/6$ is an approximation, but how do I get to $3\pi -10$? May 21 at 16:03
• Because $\sin x$ has period $2\pi$ and \sin(\pi-x)=\sin x$. May 21 at 16:09 • You already said that the Principal value branch is$[- \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.
– user925963
May 21 at 16:10

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:

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

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.