# Proof that $\cos x \leq \frac{\sin x}{x}, x \in [0,\pi ]$

As the title implies I am trying to prove the inequality $$\cos x \leq \frac{\sin x}{x}, x \in [0,\pi ]$$ However, I am not sure how to approach it. Any help would be greatly appreciated.

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Rewrite as $\tan x\ge x$. Note equality at $x=0$, and then differentiate. That takes care of $x\lt\pi/2$. For $\pi/2\le x\le\pi$, $\cos x\le0\le\sin x/x$.

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Geometrically, it's clear, at least for $x\lt\pi/2$. See upload.wikimedia.org/wikipedia/commons/thumb/4/45/… –  lhf Feb 1 '12 at 11:53

Visually, $\cos (x)$ is the slope of the tangent line to the graph of the sinus at $x$, while $\sin (x) / x$ is the slope of the line which links the point $(0,0)$ and the point $(x, \sin (x))$. The inequality comes from the concavity of the sinus on $[0, \pi]$. Now, to translate it into sweet equations, write:

$$\sin (x) = \int_0^x \cos (t) dt = x \cdot \frac{\sin (x)}{x},$$

and use the fact that $\cos (t) \geq \cos (x)$ for all $0 \leq t \leq x \leq \pi$ (this is where we use the concavity of the sinus: its derivative - the cosinus - is decreasing on this interval).

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