For all x∈]0;2pi[ : tan(x) > x

[Alexx1]

Can someone prove this?

For all x∈]0;2pi[ : tan(x) > x

 

[Mark44]

No, because it isn't true on the interval you gave. The tangent function is undefined at x = pi/2 and x = 3pi/2. For all x in (pi/2, pi), tan(x) < x, and the same is true for the interval (3pi/2, 2pi).

 

[Alexx1]

No, because it isn't true on the interval you gave. The tangent function is undefined at x = pi/2 and x = 3pi/2. For all x in (pi/2, pi), tan(x) < x, and the same is true for the interval (3pi/2, 2pi).

No, 0 and 2pi don't belong to the interval (see the ] [ ), so it is true.
We've seen a similar example with sin x: for all x > 0: sin x < x
But with tan x I can't prove it.

 

[HallsofIvy]

But $\pi/2$ and $3\pi/2$ do belong to that interval! That's what Mark44 was saying. And, in fact, for $\pi/2< x< \pi$, tan(x) is negative and can't possibly be larger than x!

You need to reduce to $]0, \pi/2[$ in order to have tan(x)< x.
Look at the function f(x)= tan(x)- x. It's derivative is sec²(x)- 1 and since sec(x)> 1 for all x in $]0, \pi/2[$, that derivative is positive.

 

[Alexx1]

But $\pi/2$ and $3\pi/2$ do belong to that interval! That's what Mark44 was saying. And, in fact, for $\pi/2< x< \pi$, tan(x) is negative and can't possibly be larger than x!

You need to reduce to $]0, \pi/2[$ in order to have tan(x)< x.
Look at the function f(x)= tan(x)- x. It's derivative is sec²(x)- 1 and since sec(x)> 1 for all x in $]0, \pi/2[$, that derivative is positive.

Oh sorry I made a mistake.. it had to be: For all x∈]0;pi/2[ : tan(x) > x

I'm sorry, you were both right
Thanks for the answer!

 

[Alexx1]

No, because it isn't true on the interval you gave. The tangent function is undefined at x = pi/2 and x = 3pi/2. For all x in (pi/2, pi), tan(x) < x, and the same is true for the interval (3pi/2, 2pi).

I'm very sorry, I made a mistake. You were right

It had to be: For all x∈]0;pi/2[ : tan(x) > x

Sorry!

 

[Mark44]

See HallsOfIvy's post #4.