MHB Can You Prove That $\tan 50^{\circ}>1.18$ Without a Calculator?

[anemone]

Without the help of calculator, show that $\tan 50^{\circ}>1.18$

 

[chisigma]

Without the help of calculator, show that $\tan 50^{\circ}>1.18$

[sp]Considering the first eight of the McLaurin series of the function tan x...

$\displaystyle \tan x \sim x + \frac{1}{3}\ x^{3} + \frac{2}{15}\ x^{5} + \frac {17}{315}\ x^{7}\ (1)$... setting in (1) $x = \frac{5}{18}\ \pi$ You obtain... $\displaystyle \tan x \sim 1.182468$[/sp]

Kind regards

$\chi$ $\sigma$

 

[anemone]

[sp]Considering the first eight of the McLaurin series of the function tan x...

$\displaystyle \tan x \sim x + \frac{1}{3}\ x^{3} + \frac{2}{15}\ x^{5} + \frac {17}{315}\ x^{7}\ (1)$... setting in (1) $x = \frac{5}{18}\ \pi$ You obtain... $\displaystyle \tan x \sim 1.182468$[/sp]

Kind regards

$\chi$ $\sigma$

Thanks for participating, chisigma and I think your solution is a good one!

I still welcome others to give a stab at it, as I can tell there is other way(perhaps many other ways) to prove the inequality to be true.

 

[magneto1]

Spoiler

I will omit the degree sign ($^\circ$) in the proof. Note
\[
\tan 150 = \tan (180 - 30) = \frac{\tan 180 - \tan 30}{1 + \tan 180 \tan 30} = -\frac{1}{\sqrt{3}}.
\]
We will use the formula:
\[
\tan 3a = \frac{3 \tan a - \tan^3 a}{1 - 3 \tan^2 a}.
\]
So we have:
\[
-\frac{1}{\sqrt{3}} = \tan 150 = \tan (50 \cdot 3) = \frac{3 \tan 50 - \tan^3 50}{1 - 3 \tan^2 50}
\]
Let $x = \tan 50$, that equation becomes:
\[
-\frac{1}{\sqrt{3}} = \frac{3 x- x^3}{1 - 3x^2} \Longrightarrow 3x^3 + 3\sqrt{3} x^2 - 9x - \sqrt{3} = 0.
\]

Define $f(x) := 3x^3 + 3\sqrt{3} x^2 - 9x - \sqrt{3}$, note that $\sqrt{3} > 1.7$.
$f(\sqrt{3}) = 9\sqrt{3} + 9\sqrt{3} - 9\sqrt{3} - \sqrt{3} > 9(1.7) - \sqrt{3} > 0$.

We can rewrite $f$ as $x(3x(x+\sqrt{3})-9)-\sqrt{3}$, so
$f(1.18) = 1.18(1.18\cdot 3(1.18+\sqrt{3})-9)-\sqrt{3} < 0$. By IVT, there is some $x \in (1.18,\sqrt{3})$
where $f(x) = 0$. So, $\tan 50 > 1.18$. Note that over the range $[0,\frac{\pi}{2})$, $\tan(\cdot)$ is monotonically
increasing, so if there is a root in that interval, it is unique.

 

[lfdahl]

Spoiler

\[tan(50) = tan(\frac{\pi}{4}+\frac{\pi}{36})=\frac{1+tan(
\frac{\pi}{36})}{1-tan(\frac{\pi}{36})} \\\\ tan(x)\approx x + \frac{1}{3}x^3 +...\;\; (MacLaurin \;series) \; gives: tan(x) > x \;\;for\;\; small\;\;positive \;\;x \\\\ Thus, I get: \\\\ tan(50) > \frac{1+\frac{\pi}{36}}{1-\frac{\pi}{36}}=\frac{36+\pi}{36-\pi}> \frac{36+3}{36-3}=39/33=13/11\approx 1.18\]

 

[anemone]

Hi magneto and lfdahl,

Thanks for the solution and participation to my challenge problem!:)

A solution I saw that is provided by other:

Spoiler

$\tan 50^{\circ}=\tan (45^{\circ}+5^{\circ})=\dfrac{\tan45^{\circ}+\tan5^{\circ}}{1-\tan45^{\circ}\tan5^{\circ}}=\dfrac{1+\tan5^{\circ}}{1-\tan5^{\circ}}$

But note that $\tan 5^{\circ}=\tan \left( \dfrac{\pi}{36} \right) >\dfrac{\pi}{36}>\dfrac{3}{36}=\dfrac{1}{12}$ which gives

$1+\tan 5^{\circ}>\dfrac{13}{12}$ and $\dfrac{1}{1-\tan 5^{\circ}}>\dfrac{12}{11}$

This implies $\dfrac{1+\tan5^{\circ}}{1-\tan5^{\circ}}>\dfrac{13}{11}$ so $\tan 50^{\circ}>\dfrac{13}{11}>1.18$