How to prove that $\tan 55^\circ<\pi/2$ How to prove that $\tan 55^\circ<\pi/2$? I checked it on a calculator, but how to prove it though? Is it some trigonometric substitution?
 A: If you know that the tangent function is convex on $[0,90^\circ)$, you can do this:
\begin{align*}
\tan(55^\circ)
= \tan(\tfrac13\cdot 45^\circ + \tfrac23\cdot 60^\circ)
\le \tfrac13\tan(45^\circ) + \tfrac23\tan(60^\circ)
= \frac{1+2\sqrt3}{3}
< \frac32
< \frac\pi2
\end{align*}
To prove the convexity, the simplest thing would be to calculate the second derivative; if you don't have calculus, you can argue that
$$ \tfrac12\tan a+\tfrac12\tan b
= \tan(\tfrac{a+b}{2}) \cdot \frac{1+\tan^2(\tfrac{a-b}{2})}{1-\tan^2(\tfrac{a+b}{2})\tan^2(\tfrac{a-b}{2})}
\ge \tan(\tfrac{a+b}{2}) $$
when $\tan(\frac{a+b}{2})\ge 0$.  (The trigonometric identity comes from the usual identities for $\tan(u\pm v)$ with $u,v=\frac{a+b}{2},\frac{a-b}{2}$ and a bit of algebra; the inequality comes from the fact that $\frac{1+x}{1-c^2x}$ is an increasing function of $x$ and $\tan^2(\frac{a-b}{2})\ge 0$.)  Then the convexity for weights other than $\frac12,\frac12$ follows by continuity.
A: This is a variant on AsdrubalBeltran's answer, starting from the equation
$$-2+\sqrt{3}=\tan(165^\circ)=\tan(3\cdot55^\circ)=\frac{3\tan(55^\circ)-\tan^3(55^\circ)}{1-3\tan^2(55^\circ)}$$
This implies that $\tan(55^\circ)$ is a root of the cubic polynomial
$$P(t)=t^3+(6-3\sqrt3)t^2-3t+(\sqrt3-2)$$
Note that 
$$P(0)=\sqrt3-2\lt0$$ 
and 
$$P(-1)=-1+(6-3\sqrt3)+3+(\sqrt3-2)=6-2\sqrt3\gt0$$  
These inequalities imply $P$ has two negative roots and one positive root.  It suffices therefore to note that
$$P(3/2)={1\over8}\left(27+18(6-3\sqrt3)-36+8(\sqrt3-2) \right)={1\over8}(83-46\sqrt3)\gt0$$
to conclude that
$$\tan(55^\circ)\lt{3\over2}\lt{\pi\over2}$$
In fact, if you're willing to do a little arithmetic, it's possible to show $P(\sqrt2)\lt0$ and $P(10/7)\gt0$, so that
$$1.41421\approx\sqrt2\lt\tan(55^\circ)\lt{10\over7}\approx1.42857$$
A: I found a solution:
Note that : $\tan(3\cdot x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$
Then $$-2+\sqrt{3}=\tan(165^\circ)=\tan(3\cdot55^\circ)=\frac{3\tan(55^\circ)-\tan^3(55^\circ)}{1-3\tan^2(55^\circ)}$$
Now  how:
$$-\frac{18}{73}=\tan(3\cdot\tan^{-1}(1.5))=\frac{3\cdot1.5-(1.5)^3}{1-3(1.5)^2}$$
but it is known that $\tan{3x}$ is  increasign, if $$-2+\sqrt{3}<-\frac{18}{73}\Longrightarrow 55^\circ<\tan^{-1}(1.5)\Longrightarrow \tan{55^\circ}<1.5\Longrightarrow \tan{55^\circ}<\frac{\pi}{2} $$
