Two ways to show that $\sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ Show that: $\large \sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ on: $0<x<\frac {\pi}2$
I tried to solve it in two ways and got a little stuck:
One way is to use Cauchy's MVT, define $f,g$ such that $f(x)=\sin x -x +\frac {x^3}{3!}$ and $g(x)=\frac {x^5}{5!}$ so $\frac {f(x)}{g(x)}<1$ and both are continuous and differentiable on  $x\in (0,\frac{\pi}2)$ so 
$$\large \frac {f(x)}{g(x)}=\frac{f(x)-f(0)}{g(x)-g(0)}=\frac {f'(c)}{g'(c)}\Rightarrow
\frac{\sin x -x +\frac {x^3}{3!}} {\frac {x^5}{5!}}=\frac {\cos y -1 +\frac{y^2}{2}}{\frac{y^4}{24}}$$
Such that $y\in (0,x)$.
Now what ? I need to show that $\frac {f'(y)}{g'(y)}<1$ but I tried to input $\frac{\pi} 2$ as a value to $f'/g'$ but it was larger than 1.

The other way: With Taylor expansion. 
$\large \sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}+R_{5,0}(x)\Rightarrow\\ 
\large \sin x-x+\frac {x^3}{3!}-\frac {x^5}{5!}=R_{5,0}(x)$
Now we're left with showing that $R_{5,0}(x)<0$
So $\large R_{5,0}(x)=\frac{-\sin(c)c^7}{7!}$ for $c\in (0,x)$, now can I say that because it's negative and $(\sin(c)c^7>0)$ for all $c\in(0,x)$,  $R_{5,0}(x)<0$ ?
Also, would it be correct to say that: $\large R_{5,0}(x)=\frac{cos(c)c^6}{6!}$?
 A: Use increasing functions. 
For example
$$ 1- \frac{x^2}{2} <\cos x$$
take 
$$f(x)=\cos x +\frac{x^2}{2}-1$$ then 
$$f^{\prime}=-\sin x + x  >0$$ so $f$ is increasing, and $f(0) <f(x)$ gives the desired inequality. Next take $x-\frac{x^3}{3} < \sin x$ then go back to the corresponding $cos$ expansion and finally your ready to prove your inequality. Each step relies on the last.
A: Repeat use of the Cauchy Mean Value theorem $3$ times we have to prove:
$\dfrac{sinx}{6x} < 1$. But since $x > 0$, we need to prove: $h(x) = sinx - 6x < 0$.
$h'(x) = cosx - 6 < 0$, so $h(x) < h(0) = 0$. Done.
A: Using the Mean value theorem is the right idea. You need the MVT to prove the following useful lemma: 

If $f'(x)>g'(x)$ for all $a\ge x>0$ and $f(0)=g(0)$, then $f(x)>g(x)$ for all $a\ge x>0$. 

Letting $g(x)=\sin(x)$ and $f(x)=x-\frac{x^3}{3!} +\frac{x^5}{5!}$ and then iterating this lemma will give you your answer. 
Proof of the lemma:

Suppose we are given $f$ and $g$ such that $f'(x)>g'(x)$ for all $a\ge x>0$ and $f(0)=g(0)$. Let $h(x)=f(x)-g(x)$, then $h(0)=0$ and $h'(x)>0$. By MVT there is an $x_0\in [0,x]$, for any $0<x\le a$, such that 
  $$h'(x_0)=\frac{f(x)-g(x)-(f(0)-g(0))}{x}=\frac{f(x)-g(x)}{x}>0.$$
  Since $x>0$, we know $f(x)-g(x)>0$ for any $x\in [0,a]$.  

A: Observe first that $f_0(x):=1-\cos x>0$ for $0<x\leqslant\pi/2$. Then 
$$f_1(x):=\int_0^xf_0(t)\,\mathrm dt=x-\sin x>0$$
for $0<x\leqslant\pi/2$. Continue in this way, with
$$f_{n+1}(x):=\int_0^xf_n(t)\,\mathrm dt>0$$
for $0<x\leqslant\pi/2$ and $n=0,1,...$, and we arrive at your result at the stage $n=4$.
It is easy to generalize this procedure to show by induction that
$$(-1)^n\left(\sum_{k=0}^n(-1)^k\frac{x^{2k}}{(2k)!}-\cos x\right)>0$$
and
$$(-1)^n\left(\sum_{k=0}^n(-1)^k\frac{x^{2k+1}}{(2k+1)!}-\sin x\right)>0$$
for $0<x\leqslant\pi/2$ and $n=0,1,...$.
