$\tan \frac{x}{2}>\frac{x}{2}$ for all $x \in (0, \pi)$

Question

I was trying to prove that $\tan \frac{x}{2}>\frac{x}{2}$ for all $x \in (0, \pi)$

I am trying to use some trigonometric identity that gives me lights to be able to prove this, could you give me some way?

Answer 1

Geometric proof is the best way as you can see there

• How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

as an alternative by $y=\frac x 2$

$$\tan (y)>y \iff f(y)=\tan (y)-y>0$$

which is true indeed $f(0)=0$ and

$$f'(y)=\tan^2 y>0$$

Answer 2

Define $f(x)=\tan\frac{x}2$. Then $f'(x)=\frac12\sec^2\frac{x}2$ and $f''(x)=\frac12\sec^2\frac{x}2\tan\frac{x}2$.

We know that $f(0)=0$ and $f'(0)=\frac12$.

For $x\in(0,\pi)$, we have $f'(x)>\frac12$ and $f''(x)>0$. Due to the nature of convex functions, this confirms our inequality.

Answer 3

Equivalently, $\tan y>y$ for acute $y$. Let $OA,\,OB$ be two radii of a circle, and extend $OB$ to meet the tangent at $A$, say at $C$. Then $y\cot y$ is the proportion of $\triangle OAC$'s area in the circle.

Answer 4

Hint: Use substitution and prove that $\tan u>u\enspace \forall u \in\bigl(0,\frac\pi 2\bigr)$. Using this corollary of the Mean value theorem, you can shorten the proof:

Let $f,g$ two differentiable functions defined on an interval $I$, $x_0\in I$, and suppose

• $f(x_0)\ge g(x_0)$,
• $f'(x)>g'(x)$ for all $x\in I$, $x>x_0$.

Then $f(x)>g(x)$ for all $x\in I$, $x>x_0$.