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Prove that $\frac{\tan{x}}{\tan{y}}>\frac{x}{y} : \forall (0<y<x<\frac{\pi}{2})$.

My try, considering $f(t)=\frac{\tan{x}}{\tan{y}}-\frac{x}{y}$ and derivating it to see whether the function is increasing in the given interval.

I should be sure that $\lim_{x,y\rightarrow0}\frac{\tan{x}}{\tan{y}}-\frac{x}{y}\geq0$ for the previous derivative check to be useful, which I'm not yet, but I'm assuming it's $0$ since I'd say that since both $x,y$ approach $0$ equally then the quotient of both their tangents and themselves is $1$, hence the substraction being $0$.

However, the trouble arrives at the time of derivating it because of the 2 variables, I'm not sure if I have to fix one and derivate in terms of the other one, or what to do. I have to say I'm currently coursing a module on real single-variable analysis, so it can't have anything to do with multivariable analysis.

Martin Sleziak
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F.Webber
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5 Answers5

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In fact, we can show a slightly stronger result.

If $f(0)=0$ and $f(x)$ is strictly convex then for $x<y$ we have $$\frac{f(x)}x<\frac{f(y)}y.$$

This result is shown here for differentiable functions. But it can be shown for any convex function.

Proof. Since $x=\frac xy \cdot y + \frac{y-x}y \cdot 0$ we get $$f(x) < \frac xy f(y) + \frac{y-x}y f(0)$$ from convexity. Hence $$\frac{f(x)}x < \frac{f(y)}y.$$

So for $f(x)=\tan x$ we have $$\frac{\tan x}x < \frac{\tan y}y$$ whenever $0<x<y<\frac\pi2$. This is equivalent to the inequality in the original question.

Community
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Martin Sleziak
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3

Consider the function

$$f(x) = \frac{\tan x}{x}.$$

You need to show that function is increasing on $(0,\pi/2)$.

If you differentiate that, you find

$$f'(x) = \frac{x(1+\tan^2 x) - \tan x}{x^2},$$

and you need to show $x(1+\tan^2 x) > \tan x$ for $x\in (0,\pi/2)$. A little trick helps showing that.

Daniel Fischer
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  • Thanks Daniel, I took a while to answer because I was trying out all methods by myself. Seems like I was making the function selection really bad, but now I think I've got a better idea on how to approach this kind of problems. As for the end, I used that $x\geq\sin(x):\forall x\in(0,\frac{\pi}{2})$. That worked pretty nicely though I'm not sure if that's what you meant. – F.Webber Jan 08 '14 at 23:14
  • If you multiply with $\cos^2 x$, you get $x > \sin x \cos x$, that was what I meant. Close enough. – Daniel Fischer Jan 08 '14 at 23:30
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The inequality is equivalent to $$ \frac{\cot y}{\cot x}>\frac{x}{y} $$ or $$ y\cot y>x\cot x $$ so one could try proving that $f(x)=x\cot x$ is decreasing in $(0,\pi/2)$. Now

$$ f'(x)=\cot x-\frac{x}{\sin^2x}=\frac{\sin x\cos x-x}{\sin^2 x}=\cdots $$ (a simple transformation of the numerator will do).

egreg
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  • Thanks for your answer as well, its interesting how all 3 solutions are based on a similar idea, but because of the nature of trigonometric functions each one gives you a different point of view. I again used at the end that $x\geq\sin(x):\forall x\in(0,\frac{\pi}{2})$ for the numerator. – F.Webber Jan 08 '14 at 23:20
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If we can show that $\dfrac{\tan(x)}{x}$ is monotonically increasing, then if $0\lt y\lt x\lt\dfrac\pi2$, we have $$ \frac{\tan(x)}{x}\gt\frac{\tan(y)}{y}\quad\text{and therefore}\quad\frac{\tan(x)}{\tan(y)}\gt\frac xy $$ The derivative of $\dfrac{\tan(x)}{x}$ is $$ \frac{x\sec^2(x)-\tan(x)}{x^2}=\frac{\sec^2(x)}{2x^2}\left(2x-\sin(2x)\right) $$ Now all you need to show is that $2x-\sin(2x)\gt0$ on $(0,\frac\pi2)$.

robjohn
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  • Is this any different from Daniel's? – egreg Jan 08 '14 at 23:11
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    @egreg: certainly the beginnings are the same, but I think that writing the derivative as $$\frac{\sec^2(x)}{2x^2}\left(2x-\sin(2x)\right)$$ makes things a bit easier. If Daniel Fischer thinks it is too close, I will delete mine. – robjohn Jan 08 '14 at 23:14
  • Thanks for your answer as well robjohn, it sure does, by using that $x\geq\sin(x):\forall x\in(0,\frac{\pi}{2})$ the result comes pretty straightfoward, thanks! – F.Webber Jan 08 '14 at 23:15
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The function $x \cot x$ is strictly decreasing on $[0,\pi)$, as in other answers.

It turns out that the Taylor expansion at $0$ is $$x \cot x = 1 - \frac{x^2}{3}- \frac{x^4}{45} - \frac{2 x^6}{945} - \cdots $$ valid on $(-\pi, \pi)$, check the general formula here. Notice that all of the coefficients after the free term are $\le 0$.

orangeskid
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