What is a bound of 1sin⁡x−1x on (0,a]?Is 1sin⁡x−1x bounded in (0,a] , a&lt;π2?

For&nbsp;|t|≤π2&nbsp;we have,|1sin⁡t−1t|≤1−2πThat's all we have for it. l&nbsp;0≤t≤π2tan⁡tsin⁡t≥t2⇔−cos⁡tsin2t+1t2=(1sin⁡t−1t)′≥0as a result, the function,&nbsp;f:t↦1sin⁡t−1t&nbsp;is increasing on&nbsp;(0,π2)then for&nbsp;t∈(0,π2)f(0)≤f(t)≤f(π2)that is,0=f(0)=limh→0(1sin⁡h−1h)≤(1sin⁡t−1t)≤f(π2)=1−2πDue to the peculiar nature of the function, the conclusion is valid&nbsp;−t∈(0,π2)

Expert answers to "What is a bound of 1sin⁡x−1x on (0,a]?Is..."

Papierskiix5n

Answered question

2022-04-16

What is a bound of

\frac{1}{\sin x} - \frac{1}{x}

(0, a]

?
Is

\frac{1}{\sin x} - \frac{1}{x}

bounded in

(0, a]

a < \frac{π}{2}

Answer & Explanation

superpms01wks1

Beginner2022-04-17Added 13 answers

For $| t | \leq \frac{π}{2}$ we have,
$| \frac{1}{\sin t} - \frac{1}{t} | \leq 1 - \frac{2}{π}$
That's all we have for it. l $0 \leq t \leq \frac{π}{2}$
$\tan t \sin t \geq t^{2} \Leftrightarrow - \frac{\cos t}{\sin^{2} t} + \frac{1}{t^{2}} = {(\frac{1}{\sin t} - \frac{1}{t})}^{'} \geq 0$
as a result, the function, $f : t \mapsto \frac{1}{\sin t} - \frac{1}{t}$ is increasing on $(0, \frac{π}{2})$
then for $t \in (0, \frac{π}{2})$
$f (0) \leq f (t) \leq f (\frac{π}{2})$
that is,
$0 = f (0) = lim_{h \to 0} (\frac{1}{\sin h} - \frac{1}{h}) \leq (\frac{1}{\sin t} - \frac{1}{t}) \leq f (\frac{π}{2}) = 1 - \frac{2}{π}$
Due to the peculiar nature of the function, the conclusion is valid $- t \in (0, \frac{π}{2})$

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