I was playing with series with Wolfram Alpha online calculator when I've considered the plot of the function $$f(x)=\frac{(\sin x)^{\cos x}}{(\cos x)^{\sin x}},\tag{1}$$ over real numbers in $[0,\frac{\pi}{4}]$. And after I wondered next question.
Question. What about the asymptotic behaviour of $$\sum_{1\leq n\leq x}\frac{(\sin \left(\frac{\pi}{4n}\right))^{\cos \left(\frac{\pi}{4n}\right)}}{(\cos \left(\frac{\pi}{4n}\right))^{\sin \left(\frac{\pi}{4n}\right)}}\tag{2}$$ as $x\to\infty$? I am looking (as curiosity) a big/little oh statement, or well an asymptotic equivalence $\sim$ as $x\to\infty$. Since my series seems tedious and artificious only is required, if you want, that you explain a good strategy to get such asymptotic behaviour. Many thanks.
I know the Taylor series for the sine and cosine functions, and the Euler–Maclaurin formula, see this Wikipedia, but even this seems complicated to get. Finally one can write our sequence as
$$\sum_{1\leq n\leq x}\frac{\operatorname{exp}\left(\cos \left(\frac{\pi}{4n}\right)\log(\sin \left(\frac{\pi}{4n}\right))\right)}{\operatorname{exp}\left(\sin \left(\frac{\pi}{4n}\right)\log (\cos \left(\frac{\pi}{4n}\right))\right)}.\tag{3}$$
For small $t>0,$
$$\cos t < (\cos t)^{\sin t} < 1.$$
Thus as $t\to 0^+,$ $(\cos t)^{\sin t}\to 1.$ This takes care of the denominators of you terms.
As for the numerators, it seems likely that $(\sin t)^{\cos t}\sim t$ as $t\to 0^+.$ The intuition here being that $\sin t \sim t$ and $\cos t \sim 1.$ That's not a proof of course. However, $(\sin t)^{\cos t}\sim t$ does hold; see if you can prove it.
It follows that the terms of your series are asymptotic to $\dfrac{\pi}{4 n}.$ From this it follows that the sum of your terms from $1$ to $N$ is asymptotic to
$$\sum_{n=1}^{N} \frac{\pi}{4 n} = \frac{\pi}{4}\sum_{n=1}^{N} \frac{1}{n}\sim \frac{\pi}{4}\,\ln N.$$