A Swedish Olympiad Problem [tan v = 2v]

 The problem is

Suppose $v \in (0, \frac{\pi}{2})$ is the unique number such that $\tan v = 2v$.

Determine whether $\sin v < \frac{20}{21}$.

Scroll down for a solution.

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Another strange and interesting problem.

Note that the function $f(x) = \tan x - 2x$ decreases till $\frac{\pi}{4}$ and then increases.

Thus for $x \in [0, v]$, $f(x) \leq 0$ and for $x \in (v, \frac{\pi}{2}), f(x) > 0$.

if $a \in (0, \frac{\pi}{2})$ is such that $\sin a = \frac{20}{21}$, then $\tan a  = \frac{20}{\sqrt{41}} > 3$.

We will now show that $v < 1.5$ which implies $\tan v < 3$ and thus $\tan v < \tan a$, implying $\sin v < \frac{20}{21}$

We have that $v > \frac{\pi}{3}$, because $\tan \frac{\pi}{3} = \sqrt{3} <  \frac{2\pi}{3}$ and so $f(\frac{\pi}{3}) < 0 $.

Now for $x \in (\frac{\pi}{3}, \frac{\pi}{2})$, by the mean value theorem, we have that for some $\eta \in (\frac{\pi}{3}, x)$

$$\frac{\tan x - \sqrt{3}}{x - \frac{\pi}{3}} = \sec^2 \eta  > \sec^2 \frac{\pi}{3} = 4$$

Thus for $x \in  (\frac{\pi}{3}, \frac{\pi}{2})$ we have that

$$\tan x > 4(x - \frac{\pi}{3}) + \sqrt{3}$$

Putiing $x = v$. and $\tan v = 2v$ gives us

$$2v > 4(v - \frac{\pi}{3}) + \sqrt{3}$$

Giving 

$$v < \frac{4\pi - 3 \sqrt{3}}{6} \approx 1.23$$