Question

How can I show that the function f(x)=(1/x)-(cos(x)/sin(x)), ∀x∈R and x≠nπ,n∈Z, is strictly increasing on each interval (nπ,(n+1)π. if the inequality |sin(x)|<|x| is used it have to be proven as well.

Answer 1

\[f(x)=\frac{1}{x}-\cot x\]???

Answer 2

so we must prove\[f'(x)=-\frac{1}{x^2}+1+\cot^2 x>0\] for \(n\pi<x<(n+1)\pi\)

Answer 3

Yes, just posted it in the other form becuase they wrote it that way, but again yes.

Answer 4

\(f'(x)=-\frac1{x^2}+\frac1{\sin^2x}\)
Using your inequality: \(\frac1{\sin^2x}=\frac1{|\sin x|^2}>\frac1{x^2}\) or \(\frac1{\sin^2x}-\frac1{x^2}>0\) on each interval (nπ,(n+1)π)
So, \(f'(x)=-\frac1{x^2}+\frac1{\sin^2x}>0\). It means that your function is strictly increasing on each interval (nπ,(n+1)π).

Answer 5

How do you get f'(x) to that? Can follow you in the first part, but the second not quite.

Answer 6

Can you find the derivative of your function?

Answer 7

Ok now i got it. Sorry just had to try it by hand.

Answer 8

Are you satisfied with the result or you have any questions?

Answer 9

I think so the reason emm.. can you elaborate the inequality?

Answer 10

Yes. What about a geometric proof?

Answer 11

Would be fine.

Answer 12

There is the circumference or the radius = 1. Area of the sector will be \(\frac x2\) and the area of the triangle will be \(\frac {\sin x} 2\). It is obvious that the area of the sector is bigger than the area of the triangle. So \(\frac{\sin x}2<\frac x 2\) or \(\sin x< x\).

Answer 13

I see, and the inequality you show in the caculation is just a rewriteing of this inequality?

Answer 14

Yes. But I proved it for xє[0,п/2]. You can get your inequality from mine.

Answer 15

No i ment this one (see the attached file)

Answer 16

It is the consequence or the first inequality.

Answer 17

Ok. I'll say thank you so much. Feel like i got it. Ones again thank you

Answer 18

No problem. Ask me again you you'd like!