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.

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 $\pi$...
