When $x$ is an acute angle ,prove $\tan x > n\tan(\frac {x}{n})$

Question

Suppose $x$ is acute angle and $n \in \mathbb{N}$ . I want to prove $\tan x >n.\tan (\frac{x}{n})$ ,But I don't have clue . Can anybody help me o,or give an idea to find show that ?

Thanks in advance for any help,clue or solution .

Answer 1

For fun, here's a simple trigonograph:

$$0 <\theta < \frac{\pi}{2}\quad\text{and}\quad 1 < n \in \mathbb{N} \qquad\implies\qquad \tan\theta > n \tan\frac{\theta}{n}$$

Answer 2

Let $f(x)=\tan x.$ Since $f'(x)=\frac1{\cos^2 x}$ is monotone increasing in $[0,\pi/2)$, $f(x)$ is convex, so $f(\lambda\,x_1+(1-\lambda)\,x_2)\le\lambda\,f(x_1)+(1-\lambda)\,f(x_2)$ for $\lambda\in[0,1]$ and $x_1,x_2\in[0,\pi/2)$. Now put $\lambda=1/n$, $x_1=x$ and $x_2=0$: $\tan\frac{x}{n}\le\frac1{n}\,\tan x$. Obviously, we can't have a strict inequality for $n=1$.

Answer 3

Using $\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y}\ge \tan x + \tan y$ with equality iff $\tan x \tan y = 0$ establishes the result by induction on $n$.

Answer 4

Let $$ f(x)=\tan(\frac{x}{n})-\frac1n\tan x. $$ Then $$ f'(x)=\frac1n\sec^2(\frac{x}{n})-\frac1n\sec^2 x=\frac1n\left(\frac1{\cos^2(\frac xn)}-\frac{1}{\cos^2x}\right). $$ Noting that $\frac{1}{\cos^2x}$ is increasing and $\frac{x}{n}<x$, one has $f'(x)<0$ or $f(x)$ is decreasing and hence $f(x)<f(0)=0$ for $x>0$, namely $$ \tan(\frac{x}{n})<\frac1n\tan x. $$

Answer 5

I'll mention another strategy using the function $\text{tanc}x:=\frac{\tan x}{x}$. The required result is $\text{tanc}x>\text{tanc}\frac{x}{n}$. We need only show $\text{tanc}x$ increases on acute $x$. Its derivative is $\frac{\sec^2 x }{x}-\frac{\tan x}{x^2}=\frac{x-\sin x\cos x}{x^2\cos^2 x}=\frac{2x-\sin 2x}{2x^2\cos^2 x}$, which is positive because $\sin y<y$ for $y>0$.