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

[binbagsss]

Homework Statement

Homework Equations

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).

 

[Mark44]

Homework Statement

Homework Equations

The Attempt at a Solution

Hi

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

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.

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).

 

[binbagsss]

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.

edited to include range of x

sorry I am aware of that for latex, running low on caffeine ! ta

 

[Mark44]

Try showing that $\sin x - \frac 2 {\pi} x \ge 0$ on that interval, possibly using the Maclaurin expansion of $\sin x$.

 

[PeroK]

I am completely stuck where to start.

Many thanks.

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