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In radians, $$\arcsin\left(\dfrac{\frac1e^{\frac1\pi}+\frac1\pi^{\frac1e}}{\pi^{\frac1\pi}+e^{\frac1e}}\right)=0.500350052703\ldots\approx 0.5.$$

The Maclaurin expansion of $\arcsin(x)$ is $$x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} + \cdots$$ and the calculations get really complicated once we do the substitution of $x$.

Is there a faster way to prove this, or is it merely a coincidence?

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  • $\begingroup$ You consider $0.50035...$ "so close" from $0.5$ ! There are myriads of such "so closednesses" ! For example $\sqrt{470}+\sqrt{300}=38.999991...$ is very very close to $39$ according to your criteria... The question "Why that" ? is devoided of interest, especially when there is no special closedness. $\endgroup$
    – Jean Marie
    Dec 17 '17 at 20:53
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Don't know if this is the answer you're looking for.

Well, $e \approx 3 \approx \pi$.

So $$\frac{\frac{1}{e^{\frac{1}{\pi}}} + \frac{1}{\pi^{\frac{1}{e}}}}{\pi^{\frac{1}{\pi}} + e^{\frac{1}{e}}} \approx \frac{\frac{1}{3^{\frac{1}{3}}} + \frac{1}{3^{\frac{1}{3}}}}{3^{\frac{1}{3}} + 3^{\frac{1}{3}}} = \frac{\frac{2}{3^{\frac{1}{3}}}}{2 \cdot 3^{\frac{1}{3}}} = \frac{2}{2 \cdot \left(3^{\frac{1}{3}}\right)^2} = 3^{-\frac{2}{3}} \approx 0.480749856769\dots$$

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  • $\begingroup$ I like your answer beginning by "Well, $e \approx 3 \approx \pi$". I didn't perceive at first the offbeat humor herein. $\endgroup$
    – Jean Marie
    Dec 20 '17 at 23:40

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