I was wondering how I could use the Maclaurin series for $\sin(x)$ to estimate $\sin(\frac{\pi}{36})$ to $5$dp?
I know the answer should be $0.08716$ to $5$dp, I tried to do this but using the first to terms of the expansion for $\sin(x)$ below: $$x-\frac{x^3}{3!}= \frac{\pi}{36}-\frac{\pi^3}{36^3 *3!}$$
Which I got to be $0.08716$ this seems right but according to my mark scheme, I should've done:
$$|\frac{x^7}{7!}|=\frac{|x^7|}{5040}$$
"which for $|x|\lt 0.3$ gives a bound of $4.3×10^{−8}$ on the error. Since we have $\frac{π}{36} =0.0872...$ we can therefore use $T_5$ to get the estimate"
I really don't understand why they used $T_5$ and what $x^7$ has to do with it?
If anyone can shed some light on this I would be very greatful!