Complicated inequality, how to prove? If $-\pi/2 <\theta<\pi/2$, how do I prove that  $\left|\sin (\theta) - \sum_{k=0}^n (-1)^k \frac{\theta^{2k+1}}{(2k+1)!}\right|  \leq \frac{|\theta|^{2n+2}}{(2n+2)!}.$
 A: The LHS is difference between $\sin{(\theta)}$ and part of the Maclaurin series for $\sin{(\theta)}$. In other words, the error after adding up $n$ terms.  Since the Maclaurin series is alternating, a bound on the error is the next term.
A: The Extended Mean Value Theorem states that if $f$, $f'$, ... , $f^{(N)}$ are continuous on a closed interval $[a,x]$, and $f^{(N+1)}$ exists on the open interval $(a,x)$, then there exists $\xi$ with $a<\xi <x$ such that 
$$f(x) = f(a)+ \sum_{n=1}^N \frac{f^{(n)}(a)}{n!}(x-a)^n +\frac{f^{(N+1)}(\xi)}{(N+1)!}(x-a)^{N+1}$$
The sine function meets all of these criteria on $(-\pi/2,\pi/2)$.  Thus, with $a=0$, we have with $N=2n+1$ odd,
$$\begin{align}
\sin \theta&=\sum_{k\ge1,\text{odd}}^{N} \frac{(-1)^k}{k!}(\theta)^k+\frac{(-1)^{(N-1)/2}\cos (\xi)}{(N+1)!}(\theta)^{(N+1)}\\
&=\sum_{k=1}^{n} \frac{(-1)^{2k+1}}{(2k+1)!}(\theta)^{2k+1}+\frac{(-1)^{n}\cos (\xi)}{(2n+2)!}(\theta)^{(2n+2)}
\end{align}$$
Subtracting the sum on the right-hand side from $\sin \theta$, taking absolute values, and using $|\cos x|\le 1$, we have the desired result!
A: Hint: series expansion of $\sin(\theta)$.
A: Use the Taylor series of $\sin{\theta}$, with either the Lagrange or the integral form of the remainder, and then estimate the remainder term.
A: Just notice that the Taylor series of $\sin \theta  $ is an alternating series which implies that the remainder
$$ |R_n | =\bigg| \sin\theta - S_n \bigg| \leq a_{n+1}$$  
$$ \implies \bigg| \sin\theta - \sum_{k=0}^n (-1)^k \frac{\theta^{2k+1}}{(2k+1)!} \bigg| \leq \frac{\theta^{2n+2}}{(2n+2)!} $$ 
