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

#### ragnar

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.

#### 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.

#### Prove It

MHF Helper
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
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
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.

#### ragnar

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.

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 $$\displaystyle 0 < x < \frac{\pi}{2}$$