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

• Jul 17th 2010, 07:51 PM
ragnar
Prove x cosx < sinx on 0 < x < pi/2
I am asked to prove that $\displaystyle \frac{2}{\pi} < \frac{sin x}{x} < 1$ if $\displaystyle 0 < x < \frac{\pi}{2}$, where the definition of $\displaystyle sin$ is given by $\displaystyle sin(x) = \frac{1}{2}(e^{ix} + e^{-ix})$. 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 $\displaystyle \frac{sin x}{x}$ is always negative, which it is if I am to trust a graphing calculator.

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

So I need to show that $\displaystyle \frac{x cos x - sin x}{x^{2}}$ 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.
• Jul 17th 2010, 08:10 PM
ragnar
I suppose I should add that we have defined $\displaystyle \pi$ as the smallest positive number such that $\displaystyle cos(\frac{\pi}{2})$ = 0.
• Jul 17th 2010, 08:15 PM
Prove It
You should know that

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

So for $\displaystyle x > 0$, you that means

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

Now see what happens as $\displaystyle x \to 0$ and $\displaystyle x \to \frac{\pi}{2}$...
• Jul 17th 2010, 08:24 PM
Bruno J.
I would use that $\displaystyle \frac{\sin x}{x} = \cos(x/2)\cos(x/2^2)\cos(x/2^3)\dots$. If $\displaystyle 0 \leq x < y \leq \pi/2$, it is easy to see that $\displaystyle \cos(x/2) > \cos(y/2),\: \: \cos(x/2^2) >\cos(y/2^2)$, etc., so that $\displaystyle \frac{\sin x}{x} > \frac{\sin y}{y}$.
• Jul 17th 2010, 08:30 PM
Bruno J.
Another way using the same idea : taking the derivative, we have $\displaystyle \frac{d}{dx}\frac{\sin x}{x} = -\frac{\sin x}{x}\sum_{j=1}^\infty \frac{\tan x/2^j}{2^j}$, which is $\displaystyle \leq 0$ at least for $\displaystyle 0 \leq x \leq \pi$.

In fact, the derivative of $\displaystyle \frac{\sin x}{x}$ is not always negative.
• Jul 17th 2010, 08:40 PM
ragnar
Quote:

Originally Posted by Prove It
You should know that

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

So for $\displaystyle x > 0$, you that means

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

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

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.
• Jul 18th 2010, 02:38 AM
Use MVT on the interval $\displaystyle [0,x]$ where $\displaystyle 0 < x < \frac{\pi}{2}$
let $\displaystyle f(x)=sin(x)$ then: $\displaystyle sin(x)-sin(0) = f'(c)(x-0)$ where $\displaystyle 0 < c < x$.
So, $\displaystyle sin(x)=xf'(c)=xcos(c)$.
Now, note that the function cos(x) is decreasing on $\displaystyle [0,\frac{\pi}{2}]$ hence, $\displaystyle cos(c)>cos(x)$.
Therefore: $\displaystyle sin(x)=xcos(c)>xcos(x)$ if $\displaystyle 0 < x < \frac{\pi}{2}$