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

Jun 2010
205
16
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.
 
Jun 2010
205
16
I suppose I should add that we have defined \(\displaystyle \pi\) as the smallest positive number such that \(\displaystyle cos(\frac{\pi}{2})\) = 0.
 

Prove It

MHF Helper
Aug 2008
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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 \(\displaystyle x \to \frac{\pi}{2}\)...
 

Bruno J.

MHF Hall of Honor
Jun 2009
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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 \(\displaystyle \frac{\sin x}{x} > \frac{\sin y}{y}\).
 

Bruno J.

MHF Hall of Honor
Jun 2009
1,266
498
Canada
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.
 
Jun 2010
205
16
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.
 
Feb 2010
100
27
Lebanon - Beirut
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 \(\displaystyle 0 < x < \frac{\pi}{2}\)
 
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