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If you put $0.5^{0.5}$ into a calculator you will see that: $$0.5^{0.5}\approx0.707106781187$$ And, if you also put $\sin(\frac{\pi}{4})$ into that same calculator you will get: $$\sin(\frac{\pi}{4})\approx0.707106781187$$ Is there any specific reason why $0.5^{0.5}=\sin(\frac{\pi}{4})$, or is it just a coincidence?

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    $\begingroup$ you do know that $0.5^{0.5}=1/\sqrt2$, don't you? $\endgroup$ – Lord Shark the Unknown Dec 23 '17 at 20:16
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    $\begingroup$ @LordSharktheUnknown Well, I think if OP made that connection, they wouldn't be asking this question. That sounds like the most rational conclusion, don't you think? And if you agree (and I hope you do), then your kind-of-rude comment is unwarranted. $\endgroup$ – layman Dec 23 '17 at 20:18
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    $\begingroup$ @Lord Shark the Unknown I think the OP knows that, but doesn't know why $\sin\pi/4$ has the same value. $\endgroup$ – Professor Vector Dec 23 '17 at 20:19
  • $\begingroup$ Actually I think when one sees $0.5^{0.5}$ one can forget that what that means is $\frac 12^{\frac 12} = \frac 1\sqrt{2}$. and if one isn't thinking might find that surprising. (And then one would kick oneself for not realizing why that should have been obvious.) $\endgroup$ – fleablood Dec 23 '17 at 20:31
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$$0.5^{0.5}=\sqrt{0.5}=\frac1{\sqrt2}$$ and $$\sin\left(\frac\pi4\right)=\frac1{\sqrt2}$$

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    $\begingroup$ Wow for some reason I expected something morec complicated lol. I wonder what other values of the sine have this property.. $\endgroup$ – Plungerdz Dec 23 '17 at 20:17
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    $\begingroup$ Have what property? Every real value, $x$ between $-1 < x < 1$ is the sin of some number. So there will be some $m$ so that $\sin m = \frac 1{\sqrt[n]{n}}$. Note $\sqrt{2}= \sqrt[4]{4}$ so $.25^{.25} = \sin \frac \pi 4$. $\endgroup$ – fleablood Dec 23 '17 at 20:26
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That's true also for $\cos\left(\frac\pi4\right)$ $$\sin\left(\frac\pi4\right)=\cos\left(\frac\pi4\right)=0.5^{0.5}=\frac1{\sqrt2}=\frac{\sqrt2}{2}$$

Here is a picture of what is going on geometrically:

enter image description here

Since

$$-1\le \sin x \leq 1$$

for each $\theta$ such that $$e^{-\frac 1e} <\sin \theta < 1$$

you can find two values of $x$ such that

$$x^x=\sin \theta$$

and exactly one for $\sin \theta = e^{-\frac 1e}$ and $<\sin \theta = 1$

enter image description here

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