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.

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)?
