Showing an inequality for $\sin1$ using Taylor's theorem

Question

The inequality I'm trying to show is in the question below:

I'm able to do the first part of the question but I can't get to the inequality. The way I tried it was by saying that $-1<\cos{c_n}<1$ so for some $m \in \mathbb N$ we can say that $$1-\frac{1}{3!}+\frac{1}{5!}-...+\frac{(-1)^{m-1}}{(2m-1)!}-\frac{1}{(2m+1)!}<\sin1<1-\frac{1}{3!}+\frac{1}{5!}-...+\frac{(-1)^{m-1}}{(2m-1)!}+\frac{1}{(2m+1)!}$$ which further gives $$1-\frac{1}{3!}+\frac{1}{5!}-...+\frac{(-1)^{n-1}(2m+1)(2m)-1}{(2m+1)!}<\sin1<1-\frac{1}{3!}+\frac{1}{5!}-...+\frac{(-1)^{n-1}(2m+1)(2m)+1}{(2m+1)!}$$ and that's as far as I've gotten. I'm not sure if I've did something wrong or if I'm supposed to somehow simplify what I've got further into the form shown in the question.

Answer 1

Hint: Put $n=2m$ and $n=2m-1$ and use $\cos(c_n) < 1$

Answer 2

Hint

Take $n=2m$ then $n=2m-1$ and use $\cos(x)\le1$ to get the two inequalities.