Prove πsin2πx&gt;xtan⁡πx

πsin2πx&gt;xtan⁡πx Let u=πx then x&gt;2 means 0&lt;u&lt;π2. Then the inequality becomes usin2u&gt;1tan⁡u for 0&lt;u&lt;π2 and then utan⁡u&gt;sin2u for 0&lt;u&lt;π2 Since sin⁡u&lt;u&lt;tan⁡u for 0&lt;u&lt;π2 the inequality follows.

We want πsin2⁡πx&gt;xtan⁡πx When the denominators are non-zero, we can rewrite this as: πx&gt;sin⁡πxcos⁡πx Well, for all p≠0, |p|&gt;|sin⁡p|≥|sin⁡pcos⁡p| The expression is true for all x≠0 (and where defined).

πsin2⁡πx=πsin⁡πxsin⁡πx=1sin⁡πxcos⁡πxπcos⁡πxsin⁡πx=1tan⁡πxπcos⁡πxsin⁡πx =xtan⁡πxπxcos⁡πxsin⁡πx=xtan⁡πxπxcos⁡πxsin⁡πx πsin2⁡πx=xtan⁡πxπxcos⁡πxsin⁡πx Now, you need to estimate the second factor.

πsin2πx&gt;xtan⁡πx Let u=πx then x&gt;2 means 0&lt;u&lt;π2. Then the inequality becomes usin2u&gt;1tan⁡u for 0&lt;u&lt;π2 and then utan⁡u&gt;sin2u for 0&lt;u&lt;π2 Since sin⁡u&lt;u&lt;tan⁡u for 0&lt;u&lt;π2 the inequality follows.

We want πsin2⁡πx&gt;xtan⁡πx When the denominators are non-zero, we can rewrite this as: πx&gt;sin⁡πxcos⁡πx Well, for all p≠0, |p|&gt;|sin⁡p|≥|sin⁡pcos⁡p| The expression is true for all x≠0 (and where defined).

πsin2⁡πx=πsin⁡πxsin⁡πx=1sin⁡πxcos⁡πxπcos⁡πxsin⁡πx=1tan⁡πxπcos⁡πxsin⁡πx =xtan⁡πxπxcos⁡πxsin⁡πx=xtan⁡πxπxcos⁡πxsin⁡πx πsin2⁡πx=xtan⁡πxπxcos⁡πxsin⁡πx Now, you need to estimate the second factor.

Expert Answer to "Prove πsin2πx>xtan⁡πx"

Joan Thompson

Answered question

2022-01-17

Prove

\frac{π}{\sin^{2} \frac{π}{x}} > \frac{x}{\tan \frac{π}{x}}

Answer & Explanation

boronganfh

Beginner2022-01-18Added 33 answers

\frac{π}{\sin^{2} \frac{π}{x}} > \frac{x}{\tan \frac{π}{x}}

Let

u = \frac{π}{x}

then x>2 means

0 < u < \frac{π}{2}

. Then the inequality becomes

\frac{u}{\sin^{2} u} > \frac{1}{\tan u} for 0 < u < \frac{π}{2}

and then

u \tan u > \sin^{2} u for 0 < u < \frac{π}{2}

Since

\sin u < u < \tan u

for

0 < u < \frac{π}{2}

the inequality follows.

RizerMix

Expert2022-01-19Added 656 answers

We want
$\frac{π}{\sin^{2} \frac{π}{x}} > \frac{x}{\tan \frac{π}{x}}$
When the denominators are non-zero, we can rewrite this as:
$\frac{π}{x} > \sin \frac{π}{x} \cos \frac{π}{x}$
Well, for all $p \neq 0$ ,
$| p | > | \sin p | \geq | \sin p \cos p |$
The expression is true for all $x \neq 0$ (and where defined).

user_27qwe

Skilled2022-01-23Added 375 answers

\frac{π}{\sin^{2} \frac{π}{x}} = \frac{π}{\sin \frac{π}{x} \sin \frac{π}{x}} = \frac{1}{\frac{\sin \frac{π}{x}}{\cos \frac{π}{x}}} \frac{π}{\cos \frac{π}{x} \sin \frac{π}{x}} = \frac{1}{\tan \frac{π}{x}} \frac{π}{\cos \frac{π}{x} \sin \frac{π}{x}}

= \frac{x}{\tan \frac{π}{x}} \frac{π}{x \cos \frac{π}{x} \sin \frac{π}{x}} = \frac{x}{\tan \frac{π}{x}} \frac{\frac{π}{x}}{\cos \frac{π}{x} \sin \frac{π}{x}}

\frac{π}{\sin^{2} \frac{π}{x}} = \frac{x}{\tan \frac{π}{x}} \frac{\frac{π}{x}}{\cos \frac{π}{x} \sin \frac{π}{x}}

Now, you need to estimate the second factor.

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