Use the taylor polynomials p1,p3,p5,.... about 0, successively, to evaluate sin(pi/7) to four decimal places, showing all your working.
Any help please?!
Can you tell us where your difficulty begins?
If you don't know what the Taylor (Maclaurin) series for sin(x) is, it shouldn't be hard to find. It's on wikipedia for exampe,
Taylor series - Wikipedia, the free encyclopedia
I suppose you'd have to express pi/7 as a decimal to at least four places.
Perhaps your problem is that you don't know how many terms you need?
$\displaystyle \sin(x)=P_{2n+1}(x)+R_{2n+2}(x)$ where $\displaystyle P_{2n+1}(x)=x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + \frac{(-1)^n}{(2n+1)!} x^{2n+1}$ and
$\displaystyle |R_{2n+2}(x)|\le\frac{1}{(2n+3)!}|x|^{2n+3}$ (*)
This follows from the Taylor series for sin(x) and the Taylor's theorem with the remainder in the Lagrange form. Since you need four correct decimal places, you need $\displaystyle |R_{2n+2}(x)|<5\cdot10^{-5}$. Find the smallest n satisfying this inequality (just calculate the right-hand side of (*) for n = 1, 2, ... and $\displaystyle x = \pi/7$ until it becomes strictly smaller than $\displaystyle 5\cdot10^{-5}$), compute $\displaystyle P_{2n+1}$ and round it to four decimal digits.