# Trig Question - $\arcsin(\sqrt{2}/2)$ and arc trig functions in general

I know $\arcsin(\sqrt{2}/2)$ is equal to $\pi/4$. However I don't understand why, I've done some searching on google about arc trig functions and I haven't found any webpages that explain it very well. Can any of you help?

Thanks!

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## 2 Answers

You have to understand how the inverse function is defined. If suppose $x = \sin\Bigl(\frac{\pi}{4}\Bigr)=\frac{1}{\sqrt{2}}$, then $\arcsin\Bigl(\frac{1}{\sqrt{2}}\Bigr) = \frac{\pi}{4}$. You can see more of this on

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The arc functions are sort of inverse functions for the trig functions. "Sort of" because the trig functions are periodic, so their inverse relations are not functions, but if we consider only a smaller part of the trig function, the inverse relation is a function. This is why the arc functions produce angles in a specific range. For sine, cosine, secant, and cosecant, the range of the corresponding arc function is the first quadrant angles ($0$ to $\frac\pi 2$) where the original function is positive, plus either the second or the fourth quadrant angles, whichever quadrant the original function is negative. For tangent and cotangent, the range is the first quadrant plus whichever of the second or fourth quadrants gives a contiguous range (because tangent and cotangent are discontinuous).

So, $\arcsin(\frac{\sqrt{2}}{2})$ is asking "For what angle $\theta$ (with $-\frac\pi 2\le\theta\le\frac\pi 2$) is $\sin\theta=\frac{\sqrt{2}}{2}$?"

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