Using two filters in series
In order for this answer to be compact I will define a new function.

$$
tri(x) = \text{abs}\Biggl(\biggl(\bigl(x-1\bigr) \mod 4\biggr)-2\Biggr)-1
$$

And it looks like this:



And you want to get this sine out of it, which looks like this:



In order to do that, you can just simply apply a LP (low pass) filter. Then the higher frequencies required to make the peak of the triangle wave will be removed, and you'll be left with your sine wave. 

But wait a second, there's one term that a LP filter passes through that is most certainly unwanted. That is the DC term, the 0 Hz. In the above image the DC term is 0 so everything works fine. But in reality there will be some offsets here and there, so in reality it might look like this:



With only a LP filter, you will get something similar to the sine wave above. A sine wave with a DC offset. Depending on how your instrument will be used, it might be wanted, but in most cases it is not wanted. 

This can be solved with a simple DC-blocking capacitor, it will work as a HP (high pass) filter. But if your frequency is very low, in the 50 Hz region, then that capacitor will have to be pretty big, or small with high gain. A LP filter followed by a HP filter will together work like a BP (band pass) filter. Iff the LP filter's cutoff frequency is higher than the HP filter's cutoff frequency. 

So if the end product is a BP filter, then you might as well use a BP filter instead of a HP filter. This will give you yet another LP filter inside the BP filter, so in some sense you will have 3rd order LP filter and a 1st order HP filter. That's better than a 2nd order LP filter and a 1st order HP filter. 



If I were making this for myself as a hobbyist,then I wouldn't care about 3rd order LP vs 2nd order LP. I would just use this circuit instead.