If you have "read up on Taylor polynomials", haven't you found the definitition? The "Taylor plolynomial about 0" (also known as the "MacLaurin" polynomial) for f(x) is $\displaystyle f(0)+ f'(0)x+ f''(0)/2 x^2+ ...+ f^(n)(0)/n! x^n)$. That is, each term is the kth derivative of f, evaluated at 0, divided by k!, times $\displaystyle x^k$. Can you differentiate $\displaystyle sin(\pi x/7)$ repeatedly? (Your formula is missing an "x". Without that, it is not a function.)
There is a forumula for the "error' that you could use to calculate what "n" you need to get four decimal places but the simplest thing to do is to just keep going until you get the same value, to four decimal places, twice in a row.
For example f(0)= sin(0)= 0, $\displaystyle f'(x)= \pi/7 cos(\pi x/7)$ so $\displaystyle f'(0)= \pi/7 cos(0)= \pi/7$. $\displaystyle f''= -\pi^2/49sin(\pi x/7)$ so $\displaystyle f''(0)= -\pi^2/49 sin(0)= 0$, $\displaystyle f'''(x)= -\pi^3/343 cos(\pi x/7)$ so [tex]f'''(0)= -\pi^3/343 cos(0)= -\pi^2/343.
That is, the 3rd degree Taylor polynomial about 0 is $\displaystyle (\pi/7)x- (\pi^3/343)x^3$.
(You should immediately notice pattern to the derivatives that will simplify this.)