For all x∈]0;2pi[ : tan(x) > x

  • Thread starter Alexx1
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In summary, the conversation was about whether or not the statement "For all x∈]0;2pi[ : tan(x) > x" could be proven. It was concluded that this statement is not true on the given interval, as the tangent function is undefined at certain points and for all x in the specified intervals, tan(x) is less than x. The correct statement should be "For all x∈]0;pi/2[ : tan(x) > x". It was also discussed that this statement can be proven by reducing the interval to ]0, pi/2[ and looking at the function f(x)= tan(x)- x, whose derivative is positive for all x in this interval.
  • #1
Alexx1
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Can someone prove this?

For all x∈]0;2pi[ : tan(x) > x
 
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  • #2
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).
 
  • #3
Mark44 said:
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.
 
  • #4
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.
 
  • #5
HallsofIvy said:
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!
 
  • #6
Mark44 said:
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!
 
  • #7
See HallsOfIvy's post #4.
 

1. What does the inequality "tan(x) > x" mean in the given context?

The inequality "tan(x) > x" means that the tangent of x is greater than x for all values of x that are between 0 and 2pi (excluding 0 and 2pi). In other words, the tangent function is increasing on this interval, and the value of the tangent function at any given point is always greater than the value of x at that point.

2. How can we prove that "tan(x) > x" for all x∈]0;2pi[?

This inequality can be proven using the properties of the tangent function and the interval given. One approach is to use the derivative of the tangent function, which is sec^2(x), to show that the function is increasing on the given interval. Another approach is to use the geometric definition of the tangent function and the properties of triangles to show that the ratio of the opposite side to the adjacent side is always greater than the angle itself.

3. Is it possible for the inequality "tan(x) > x" to be true for values of x outside of the given interval?

No, this inequality is only true for values of x that are between 0 and 2pi, excluding 0 and 2pi. This is because the tangent function has a period of pi, and any values outside of this interval can be converted to values within the interval by adding or subtracting multiples of pi.

4. How does this inequality relate to the graph of the tangent function?

The inequality "tan(x) > x" means that the graph of the tangent function is always above the line y=x for values of x between 0 and 2pi (excluding 0 and 2pi). This can be seen on the graph of the tangent function, where the curve of the function is always above the line y=x within this interval.

5. Can we use this inequality to solve other trigonometric equations?

Yes, this inequality can be used to solve other trigonometric equations by applying the same techniques used to prove it. By understanding the properties of the tangent function and the given interval, we can use this inequality to make conclusions about other trigonometric functions and their relationships with angles in a similar interval.

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