# Sin inequality proof , $0 \leq 2x/\pi \leq sin x$

1. Jan 21, 2017

### binbagsss

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Hi

How do I go about showing $0 \leq \frac{2x}{\pi} \leq sin x$?

for $0 \leq x \leq \pi /2$

I am completely stuck where to start.

Many thanks.

(I see it is a step in the proof of Jordan's lemma, but I'm not interested in this, and the proofs I find do not explain this actual step, ta).

Last edited: Jan 21, 2017
2. Jan 21, 2017

### Staff: Mentor

This isn't true in general. If $x > \pi/2$, the expression in the middle is larger than 1.

BTW, surround your TeX expressions with either $(inline) or  (standalone) at beginning and end. A single  character doesn't do anything. I edited your earlier post to fix this. 3. Jan 21, 2017 ### binbagsss edited to include range of x sorry I am aware of that for latex, running low on caffeine ! ta 4. Jan 21, 2017 ### Mark44 ### Staff: Mentor Try showing that$\sin x - \frac 2 {\pi} x \ge 0$on that interval, possibly using the Maclaurin expansion of$\sin x##.

5. Jan 21, 2017

### PeroK

Draw a graph! That's a good place to start.