Trouble showing $a_n=\frac{\sin \frac{n\pi}{3n+1}}{\sqrt{n+3}}$ is monotonically decreasing I've tried reducing the expression to something obviously true, but I couldn't get anywhere. Also tried induction but I got stuck at the inductive step and failed to obtain anything useful. I've also tried to prove it by contradiction, still to no avail. 
Also tried to show that $a_n/a_{n+1} > 1$, again without success because I don't know how to deal with the sines that appear in the expression, and there I ran out of ideas.
EDIT: I could perhaps consider the continuos case of a function given with the same expression, by testing it's derivative, however that's not really what I'm trying to do here unless someone knows a more rigorous line of reasoning to use this idea somehow.
 A: The fraction $\frac{n\pi}{3n+1}$ increases strictly with $n$. Therefore, by Aristarchus's inequality,
$$
\frac{\sin\left(\frac{(n+1)\pi}{3n+4}\right)}{\sin\left(\frac{n\pi}{3n+1}\right)} < \frac{(n+1)(3n+1)}{n(3n+4)} = 1 + \frac1{n(3n+4)}.
$$
Write $t = \frac1{n(3n+4)} < 1$. For $n \geqslant 2$,
$$
\left(1 + \frac1{n(3n+4)}\right)^2 = 1 + 2t + t^2 < 1 + 3t < 1 + \frac1{n(n+1)} < 1 + \frac1{n+3} = \frac{n+4}{n+3}.
$$
Hence:
$$
\frac{a_{n+1}}{a_n} = \frac{\sin\left(\frac{(n+1)\pi}{3n+4}\right)}{\sin\left(\frac{n\pi}{3n+1}\right)} \cdot \frac{\sqrt{n+3}}{\sqrt{n+4}} < 1 \quad (n \geqslant 2).
$$
We also have
$$
a_1 = \frac{1}{2\sqrt2} > 0.35 > \frac{\sin\left(\frac{2\pi}{7}\right)}{\sqrt5} = a_2,
$$
so the inequality does actually hold for all $n \geqslant 1$.
For a slightly tidier finish, one could argue thus:
$$
2\sin^2\left(\frac{2\pi}{7}\right) = 1 - \cos\left(\frac{4\pi}{7}\right) = 1 + \sin\left(\frac{\pi}{14}\right) < 1 + \frac{\pi}{14} < \frac{5}{4}.
$$
A: For $n\gt 0$, $sin\frac{\pi}{4}\lt sin\frac{n\pi}{3n+1}\lt sin\frac{\pi}{3}$.  But $\frac{1}{\sqrt{n+3}}$ decreases faster than $sin\frac{n\pi}{3n+1}$ increases as $n$ increases.
To quantify my answer, I estimated the differences in question .   $sin\frac{(n+1)\pi}{3n+4}-sin\frac{n\pi}{3n+1}\lt \frac{1}{(3n+1)(3n+4)}$  Meanwhile $\frac{1}{\sqrt{n+3}}-\frac{1}{\sqrt{n+4}}=\frac{1}{\sqrt{n+3}}(1-\sqrt\frac{n+3}{n+4})\approx \frac{1}{2\sqrt{n+3}(n+4)}$.
A: Beside the solutions already given in answers, considering
$$a_n=\frac{\sin \left(\frac{n\pi  }{3 n+1}\right)}{\sqrt{n+3}}$$ you have
$$\frac{a_{n+1}}{a_n}=\frac{\sin \left(\frac{(n+1)\pi  }{3 n+4}\right)}{\sqrt{n+4}}\frac{\sqrt{n+3}}{\sin \left(\frac{n\pi  }{3 n+1}\right)}$$ and you could use Taylor series for large values of $n$; this would give
$$\frac{a_{n+1}}{a_n}=1-\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$
Similarly, for large $n$
$$a_n=\frac{\sqrt{3}}{2}  \frac 1 {\sqrt n}+O\left(\frac{1}{n^{3/2}}\right)$$
