Show $\frac{\sin(x)}{x}>\cos(x)$ for $0I'm trying to show the inequality $$\frac{\sin(x)}{x}>\cos(x)$$ by for $0<x<\pi$ using the Mean Value Theorem, but I don't know how to start. I can show that $\sin(x)<x$, but I can't see how I can use it. I just need some help to get started.
 A: This is not true in general.
For an interval of clear counterexamples, consider that for $x\in(\frac32\pi,2\pi)$ we have
$$ \frac{\sin x}{x} < 0 < \cos x$$

Update after the question was amended to specify $0<x<\pi$:
The mean value theorem says that $\frac{\sin x}{x} = \sin'(\alpha)$ for some $\alpha\in(0,x)$. We have $\sin'(\alpha)=\cos\alpha$ so what you need to show is merely that $\cos \alpha > \cos x$. Hopefully you already know that the cosine decreases monotonically between $0$ and $\pi$...
A: Consider the function $f(x)=\sin(x)$.  Fix a point $y$ between $0$ and $\pi$.  By applying the mean value theorem to the points $x=0$ and $x=y$, you know that for some point $z$ between $0$ and $y$,
$$
\cos(z)=f'(z)=\frac{f(y)-f(0)}{y-0}=\frac{\sin(y)}{y}.
$$
Since $\cos(x)$ is a decreasing function on the interval $0$ to $\pi$ and $y>z$, it follows that $\cos(y)<\cos(z)$.  Therefore,
$$
\cos(y)<\cos(z)<\frac{\sin(y)}{y}.
$$ 
A: Since $\cos$ is strictly decreasing on $[0,\pi]$ one has
$${\sin x\over x}=\int_0^1\cos(t\>x)\>dt>\cos x\qquad(0<x\leq\pi)\ .$$
A: Wait, if you already know how to show $\sin(x) > x$, then you just need to observe that $\cos(x) \leq 1$ for all $x$ to get what you want.
