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$$\arcsin\left(\sin\left(\frac{7\pi}{3}\right)\right)$$

I cannot use this formula, correct? $f(f^{-1}(x))=x$

The answer in the book is $\frac{\pi}{3}$

How do I approach solving a problem such as this?

The inverse sin function of $\sin\frac{7\pi}{3}$

Am I saying to myself there were 7 revolutions and $\frac{\pi}{3}$ a corresponding angle to $\frac{7\pi}{3}$

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    $\begingroup$ $7\pi/3=2\pi+\pi/3$ $\endgroup$ – Mark Viola Aug 13 '15 at 16:24
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You have to consider the restricted ranges of the inverse trig functions. Since $\sin\left(\frac{7\pi}{3}\right)=\frac{\sqrt{3}}{2}$, you want $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$, since the range of inverse sine is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).$

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  • $\begingroup$ I got it. Thanks for being so clear. So sin functions always have this range because they are being made to be one-to-one? $\endgroup$ – Sunny Aug 13 '15 at 16:35
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    $\begingroup$ Exactly - if we didn't limit the range so that it was one-to-one, the inverse wouldn't be a function. $\endgroup$ – Samir Khan Aug 13 '15 at 16:36
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Let $\sin^{-1}\sin\dfrac{7\pi}3=x$ where $-\dfrac\pi2\le x\le\dfrac\pi2$

$$\implies\sin x=\sin\dfrac{7\pi}3$$

$\implies x=n\pi+(-1)^n\dfrac{7\pi}3$ where $n$ is any integer

If $n$ is even $=2m$(say), $x=2m\pi+\dfrac{7\pi}3=\dfrac{(6m+7)\pi}3\implies -\dfrac\pi2\le\dfrac{(6m+7)\pi}3\le\dfrac\pi2$

$\iff-3\le12m+14\le3\implies-2<-\dfrac{17}{12}\le m\le-\dfrac{11}{12}<0\implies m=-1$

Check for odd $n=2m+1$(say)

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