Prove x cosx < sinx on 0 < x < pi/2

**ragnar** · July 17th 2010, 07:51 PM

I am asked to prove that $\frac{2}{\pi} < \frac{sin x}{x} < 1$ if $0 < x < \frac{\pi}{2}$ , where the definition of $sin$ is given by $sin(x) = \frac{1}{2}(e^{ix} + e^{-ix})$ . I think I have a bit of the problem figured out. I reason that I can solve the whole problem if I know the derivative of $\frac{sin x}{x}$ is always negative, which it is if I am to trust a graphing calculator.

Once that's obtained, I think the rest follows from: $\displaystyle \lim_{x \rightarrow 0^{+}} \frac{\sin x}{x} = \lim_{x \rightarrow 0^{+}} cos(x) = 1$ , and $\frac{sin(\frac{\pi}{2}}{\frac{\pi}{2}} = \frac{2}{\pi}$ .

So I need to show that $\frac{x cos x - sin x}{x^{2}}$ is negative within these bounds. But I don't see how to do so, or if I need to re-consider how I'm approaching the task.

**ragnar** · July 17th 2010, 08:10 PM

I suppose I should add that we have defined $\pi$ as the smallest positive number such that $cos(\frac{\pi}{2})$ = 0.

**Prove It** · July 17th 2010, 08:15 PM

You should know that

$-1 \leq \sin{x} \leq 1$ for all $x$ .

So for $x > 0$ , you that means

$-\frac{1}{x} \leq \frac{\sin{x}}{x} \leq \frac{1}{x}$ .

Now see what happens as $x \to 0$ and $x \to \frac{\pi}{2}$ ...

**Bruno J.** · July 17th 2010, 08:24 PM

I would use that $\frac{\sin x}{x} = \cos(x/2)\cos(x/2^2)\cos(x/2^3)\dots$ . If $0 \leq x < y \leq \pi/2$ , it is easy to see that $\cos(x/2) > \cos(y/2),\: \: \cos(x/2^2) >\cos(y/2^2)$ , etc., so that $\frac{\sin x}{x} > \frac{\sin y}{y}$ .

**Bruno J.** · July 17th 2010, 08:30 PM

Another way using the same idea : taking the derivative, we have $\frac{d}{dx}\frac{\sin x}{x} = -\frac{\sin x}{x}\sum_{j=1}^\infty \frac{\tan x/2^j}{2^j}$ , which is $\leq 0$ at least for $0 \leq x \leq \pi$ .

In fact, the derivative of $\frac{\sin x}{x}$ is not always negative.

**ragnar** · July 17th 2010, 08:40 PM

Originally Posted by Prove It

You should know that

$-1 \leq \sin{x} \leq 1$ for all $x$ .

So for $x > 0$ , you that means

$-\frac{1}{x} \leq \frac{\sin{x}}{x} \leq \frac{1}{x}$ .

Now see what happens as $x \to 0$ and $x \to \frac{\pi}{2}$ ...

I can't say that I follow. You're seeing what the function does at infinity and at +/- pi/2. But if I don't know about the monotonicity of the function in the interval I'm considering, it seems like the function could be absolutely anything in-between these points.

**mohammadfawaz** · July 18th 2010, 02:38 AM

Use MVT on the interval $[0,x]$ where $0 < x < \frac{\pi}{2}$
let $f(x)=sin(x)$ then: $sin(x)-sin(0) = f'(c)(x-0)$ where $0 < c < x$ .
So, $sin(x)=xf'(c)=xcos(c)$ .
Now, note that the function cos(x) is decreasing on $[0,\frac{\pi}{2}]$ hence, $cos(c)>cos(x)$ .
Therefore: $sin(x)=xcos(c)>xcos(x)$ if $0 < x < \frac{\pi}{2}$

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