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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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4 Answers 4

up vote 3 down vote accepted

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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Since both left and right sides are smooth and monotonous on (0,π] you could prove the inequality for π and for x->0+. If the inequality holds on this two points than it will hold for every value between them.

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Rearrange this as x<= tan (x), and think about the visual representation of tangent on the unit circle. Since it is exactly the same for (0, pi/2] and [pi/2, pi], it is clear that you only have to prove this for (0, pi/2]. Now lets split this up into 2 different functions, a=x and b=tan x. Now let's take the derivative of each: da/dx=1, and db/dx=(sec x)^2. So a increases at a constant rate of 1. If you graph sec^2 x, you will see that it starts at 1 and increases on the interval (0, pi/2]. Therefore, b increases faster than a, therefore a<=b, so x<=tan x. Hope this helps!

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