- #1

- 9

- 0

Hello!! does anyone know how to prove that xcosx < sinx for all x greater than 0?

Many thanks

Many thanks

- Thread starter martint
- Start date

- #1

- 9

- 0

Hello!! does anyone know how to prove that xcosx < sinx for all x greater than 0?

Many thanks

Many thanks

- #2

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,780

- 12

Well it's obviously false, for take x=2pi.

- #3

- 9

- 0

sorry my error, the interval for x should be 0 < x < (pi)

sorry!

sorry!

- #4

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,780

- 12

- #5

- 1,631

- 4

[tex] tan(x)>x=>\frac{sin(x)}{cos(x)}>x=>sin(x)>xcos(x)[/tex] THis is valid however on the interval [tex]\left(0,\frac{\pi}{2}\right)[/tex]

Edit: Or better use quasar987's suggestion, it is better and valid on the whole interval.

Edit: Or better use quasar987's suggestion, it is better and valid on the whole interval.

Last edited:

- #6

- 9

- 0

thanks guys, but im failing to see how its easier to solve the inequality when written as xcotx<1?

- #7

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,780

- 12

Do you know calculus? (differentiation and things like that)

- #8

- 9

- 0

- #9

- 23

- 0

- #10

- 9

- 0

- #11

- 1,631

- 4

Lets define the following function:

[tex] f(x)=xcosx-sinx,\forall x \in \left(0,\pi \right)[/tex]

NOw,

[tex]f'(x)=cosx-xsinx-cosx=-xsinx[/tex]

Now let's find the critical numbers

[tex] f'(x)=0=>xsinx=0=>x=0,and,x=k\pi[/tex]

But, notice that none of these numbers are in the domain of our function, so it means that f, has no cr numbers at all.

THis actually means also that f'(x) does not change sign at all on the given interval, so it is sufficient to plug in any value from the interval (0,pi) to determine the sign of f' on the whole interval, that is let [tex]x=\frac{\pi}{2}[/tex]

SO:

[tex]f'(x)=\frac{\pi}{2}sin\frac{\pi}{2}=-\frac{\pi}{2}<0[/tex]

THis means that our function is decreasing on the given interval. Now from this information we know that:

[tex] f(0)>f(0+\epsilon),\epsilon>0,\epsilon\to\ 0, \epsilon \in (0,\pi][/tex]

Hence,

[tex]f(0)=0*cos(0)-sin(0)=0[/tex] it means that

[tex] f(x)<0,\forall x \in (0,\pi)[/tex]

so:

[tex]f(x)=xcosx-sinx<0=>xcosx<sinx[/tex]

- #12

- 9

- 0

- #13

- 1,631

- 4

Take [tex] x=\frac{3 \pi}{2}[/tex] f' is clearly >0 for this value. Or take any x in the interval (pi,2pi) what do you get for f'=-xsinx ??????.SO it means that our funcitno is decreasing only on (0,pi) but this does not guarantee us that it is decreasing everywhere in its domain, like in this case which is not ture, since f is not decreasing everywhere but rather only on (0,pi) for our purporse.

So actually finding cr. points and seeing where function is increasing and decreasing guarantees us where f' changes sign and where not.

So actually finding cr. points and seeing where function is increasing and decreasing guarantees us where f' changes sign and where not.

Last edited:

- Replies
- 5

- Views
- 15K

- Replies
- 8

- Views
- 4K

- Replies
- 2

- Views
- 2K

- Replies
- 6

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 6K

- Replies
- 2

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 7

- Views
- 22K

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 5

- Views
- 2K