MHB Confirming $\arcsin(-\sin(\frac{3\pi}{2}))$

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SUMMARY

The discussion confirms that $\arcsin(-\sin(-\frac{3\pi}{2}))$ simplifies to $\frac{\pi}{2}$. The reasoning involves recognizing that $-\frac{3\pi}{2}$ radians is equivalent to $\frac{\pi}{2}$ radians, which streamlines the calculation. The transformation steps include using the identity $\sin(\pi + \frac{\pi}{2})$ to arrive at $\sin(\frac{\pi}{2})$. This confirms that the final result is indeed $\frac{\pi}{2}$.

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What's $\arcsin(\sin(\frac{−3\pi}{2}))?$ I think it's equal to

$\arcsin(-\sin(\frac{3\pi}{2})) =\arcsin(-\sin(\pi+\frac{\pi}{2})) = \arcsin(\sin(\pi/2)) = \frac{\pi}{2}$

Is this correct?
 
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Guest said:
What's $\arcsin(\sin(\frac{−3\pi}{2}))?$ I think it's equal to

$\arcsin(-\sin(\frac{3\pi}{2})) =\arcsin(-\sin(\pi+\frac{\pi}{2})) = \arcsin(\sin(\pi/2)) = \frac{\pi}{2}$

Is this correct?

Yes it's correct, but if you realize that an angle of $\displaystyle \begin{align*} -\frac{3\,\pi}{2} \end{align*}$ radians is the same as an angle of $\displaystyle \begin{align*} \frac{\pi}{2} \end{align*}$ radians it saves about two lines of working...
 

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