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Prove that for every $0<\alpha<\beta<(\pi/2)$:

I tried setting a function $f(x) = \tan(x)$ and using the Mean Value theorem to prove this, but that didn't work since $f(0)$ is undefined.

Does this involve some sort of trig identity?

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Ehr... $\tan(0)$ is perfectly defined: it's $0$. – Arturo Magidin Feb 23 '12 at 21:36
Isn't this just the fact that $\tan x$ rises more than linear (from $0$ till $\pi/2$)? – draks ... Feb 23 '12 at 21:40
You want to show that $\frac{\tan x}{x}$ is increasing. If you do it through the derivative, you want to show that $x-\cos x\sin x$ is positive. That can be done by comparing the area of a circular sector with angle $2x$ with the area of a certain triangle. – André Nicolas Feb 23 '12 at 21:46
up vote 5 down vote accepted


Consider $$f(x) = \frac{\tan x}{x}, x \in (0,\frac{\pi}{2})$$

and try to investigate if it is increasing or decreasing.

Also, try and prove that:

$$x \gt \sin x, \forall x \in (0, \pi)$$

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thanks, that makes things simpler. – nofe Feb 23 '12 at 21:46

The inequality is equivalent to

$$\frac{\tan\beta}{\beta} >\frac{\tan\alpha}{\alpha}.$$

This says that $(\tan x)/x$ is an increasing function on $(0,\pi/2)$. Differentiate:

$$\frac{d}{dx}\frac{\tan x}{x}=\frac{2x-\sin2x}{2}\left(\frac{\sec x}{x}\right)^2.$$

For $(\tan x)/x$ to be increasing we want to check that the above is be positive on the interval, which after dividing by obviously positive terms we deduce is equivalent to $u>\sin u$ on $u\in(0,\pi)$. We can deduce this by noting that both $u'=1=\sin'(u)$ at $u=0$, and $u'=1>\sin'u=\cos u$ on $[0,\pi)$. This implies $u>\sin u$ because, in general, $f\ge g$ on an interval entails $\int_a^u f(t)dt\ge \int_a^u g(t)dt$.

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