Inequality - how to prove that xcosx < sinx for all x

[martint]

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

Many thanks

 

[quasar987]

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

 

[martint]

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

sorry!

 

[quasar987]

Well, notice that the problem is equivalent to showing that xcotx<1 for all x in (0,pi). So use calculus.

 

[sutupidmath]

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

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

 

[martint]

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

 

[quasar987]

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

 

[martint]

yes thanks, have been using differential calculus to try and solve the problem I posted but have not been able to, else I wouldn't have posted it! just don't see the advantages of changing to xcotx<1

 

[Philcorp]

Consider the function f(x) = xcos(x)-sin(x). If you know calculus it should be fairly easy to show where this function is negative!

 

[martint]

Again, this is what I have been trying to do! are we for example taking f(x), noting that f(0) =0, and using the first derivative test to show that the function is decreasing ?

 

[sutupidmath]

ok, here it is how i would argue about showing the validity of that inequality using calculus.

Lets define the following function:

$$f(x)=xcosx-sinx,\forall x \in \left(0,\pi \right)$$

NOw,

$$f'(x)=cosx-xsinx-cosx=-xsinx$$

Now let's find the critical numbers

$$f'(x)=0=>xsinx=0=>x=0,and,x=k\pi$$

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 $$x=\frac{\pi}{2}$$

SO:

$$f'(x)=\frac{\pi}{2}sin\frac{\pi}{2}=-\frac{\pi}{2}<0$$

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

$$f(0)>f(0+\epsilon),\epsilon>0,\epsilon\to\ 0, \epsilon \in (0,\pi]$$


Hence,

$$f(0)=0*cos(0)-sin(0)=0$$ it means that

$$f(x)<0,\forall x \in (0,\pi)$$

so:

$$f(x)=xcosx-sinx<0=>xcosx<sinx$$

 

[martint]

ahh thank you very much stupidmath! Also, when solving inequalities like this in the future, should i always try to solve for critical nos, or is it enough to say for example in this case that since f(0)=0 and f'(x) = -xsinx, which is clearly negative over the whole the interval then f(x) is a decreasing function and hence less than 0?

 

[sutupidmath]

Take $$x=\frac{3 \pi}{2}$$ 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.