unborn needle
Does this inequality hold for all $b\geq a$?
$\frac{a-b}{2} \leq $sin\frac{a-b}{2}cos\frac{a+b}{2}$ \leq \frac{b-a}{2}$
Please check if my proof below is correct
$sin(\frac{a-b}{2})cos(\frac{a+b}{2})$
equals to $\frac{sina-sinb}{2}$, using Lagrange Mean value theorem: suppose f(x) = sin(x) continuous on [a,b] and differentiable on (a,b), then there exists c in (a,b) such that $sina-sinb = (a-b)cosc \to |sina-sinb| = |a-b||cosc| \leq |a-b| \to \frac{|sina-sinb|}{2} \leq \frac{|a-b|}{2}$, with $b\geq a$, remove the abs -> $\frac{a-b}{2} \leq \frac{sina-sinb}{2} \leq \frac{b-a}{2} \qed$
neat comet
plush dirge
vague scroll
