# Trigonometric inequality

1. Nov 18, 2014

### DavideGenoa

I read that, for $\delta>0$, if $\delta<z\leq\pi$, then $\sin\frac{z}{2}\geq\frac{2\delta}{\pi}$.
I cannot prove it. I know that $\forall x\in\mathbb{R}\quad|\sin x|\leq |x|$, but that does not seem useful here...
Thank you so much for any help!

2. Nov 18, 2014

### Staff: Mentor

Is this a homework assignment? for what course?

If so then you need to use the homework template so we know what you know and what relevant theorems or formulas you've learned that would apply.

delta and z are elements of Reals too?

3. Nov 18, 2014

### DavideGenoa

Thank you so much for your comment! Oh, sorry, I meant $z$ real.
It is a statement I find inside Kolmogorov and Fomin's proof of Fejér's theorem here at p. 417.
By drawing some graphs I have convinced myself that there is an error in Kolmogorov-Fomin's and have logged in to amend it. Nevertheless, for $x=z/2\in[0,\pi/2]$, the convexity of the cosine in $(0,\pi)$ guarantee that the line $y_A+\frac{y_A- 1}{x_A-\pi/2}(x-x_A)$ "lies below" the sinusoid for any $(x_A,\sin(x_A))$ with $x_A\in(0,\pi/2)$ and the continuity of the function $\mathbb{R}^2\to\mathbb{R}$, $(x_A,y_A)\mapsto\sin x -(y_A+\frac{y_A- 1}{x_A-\pi/2}(x-x_A))$ in $(0,0)$ guarantee that it also does for $(x_A,y_A)=(0,0)$.
Therefore $\forall z\in\mathbb{R}(0<\delta<z\leq\pi\Rightarrow\sin\frac{z}{2}\geq\frac{\delta}{\pi})$.
For the target of Kolmogorov and Fomin's proof that is enough to prove the boundedness of the Fejér kernel, which is what the inequality is used for.

Last edited: Nov 18, 2014