Taylor series convergence for sin x a. $\forall x\in(0,\pi/2),\quad x-\frac{x^3}{3!}<\sin x<x-\frac{x^3}{3!}+\frac{x^5}{5!},$
b. $\forall x\in(0,\pi/2),\quad x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots-\frac{x^{4k-1}}{(4k-1)!}<\sin x<x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots+\frac{x^{4k+1}}{(4k+1)!},$
c. $\forall x\in(0,\pi/2),\quad x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots = \sin x.$
Hello!
I have been trying to prove these, but only had luck with the first two. I do get that you have to use part b. to solve c., but I have no idea how to connect them.
 A: For a) and by the Maclaurin formula for the sine function there's $\theta\in(0,1)$ such that
$$\sin x=x-\frac{x^3}{3!}+\frac{x^4}{4!}\sin^{(4)}(\theta x)=x-\frac{x^3}{3!}+\frac{x^4}{4!}\sin(\theta x)>x-\frac{x^3}{3!},\; \forall x\in(0,\frac\pi2)$$
and
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^6}{6!}\sin^{(6)}(\theta x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^6}{6!}\sin(\theta x)\\<x-\frac{x^3}{3!}+\frac{x^5}{5!},\; \forall x\in(0,\frac\pi2)$$
so we deduce the result. Can you generalize this to find b)?
A: We know that $\sin'(x)=\cos(x)$, $\sin''(x)=-\sin(x)$, $\sin^{[3]}(x)=-\cos(x)$ etc. Notice that the derivatives are periodic, alternating between $\sin$ & $\cos$ & changing sign every $2^{nd}$ term.
So in general, $\sin^{[n]}(x)=\begin{cases}\sin(x), & n=0+4k\\
\cos(x), & n=1+4k\\
-\sin(x), & n=2+4k\\
-\cos(x), & n=3+4k\\
\end{cases}\quad$
for integer $k$.
If we center the series at $a=0$, we have $\sin^{[n]}(0)=\begin{cases}0, & n\text{ is even}\\
1, & n=1+4k\\
-1, & n=3+4k\\
\end{cases}$
Sub. this into the definition for the Taylor series & get odd terms with coefficients of $\frac{(-1)^{n}}{n!}$. This should look familiar to $c$. Hope I helped.
