Can someone prove this?
For all x∈]0;2pi[ : tan(x) > x
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.
But [itex]\pi/2[/itex] and [itex]3\pi/2[/itex] do belong to that interval! That's what Mark44 was saying. And, in fact, for [itex]\pi/2< x< \pi[/itex], tan(x) is negative and can't possibly be larger than x!
You need to reduce to [itex]]0, \pi/2[[/itex] 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 [itex]]0, \pi/2[[/itex], 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!
I'm very sorry, I made a mistake. You were right
It had to be: For all x∈]0;pi/2[ : tan(x) > x
See HallsOfIvy's post #4.
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