MHB Sum of two trigonometric terms

[anemone]

Prove that $\tan \left( \dfrac{3 \pi}{11} \right)+ 4\sin \left( \dfrac{2 \pi}{11} \right)=\sqrt{11}$.

I know this problem may be stale as it has been posted countless times at other math forums, but I've seen one brilliant method to attack this problem recently, and I'm so eager to share it with the folks here at our site.

Despite my saying so, I would still welcome anyone who would want to take a stab at this challenge.

 

[Albert1]

Prove that $\tan \left( \dfrac{3 \pi}{11} \right)+ 4\sin \left( \dfrac{2 \pi}{11} \right)=\sqrt{11}$.

I know this problem may be stale as it has been posted countless times at other math forums, but I've seen one brilliant method to attack this problem recently, and I'm so eager to share it with the folks here at our site.

Despite my saying so, I would still welcome anyone who would want to take a stab at this challenge.

Spoiler

View attachment 1862

 

[anemone]

Spoiler

View attachment 1862

Hi Albert:), thanks for participating.

According to your diagram, I have found $\tan 3\theta=\dfrac{1}{t}$. And I am curious and wish to know if you have found the value of $t$ yet? I tried with no luck so far.

 

[mathbalarka]

Spoiler

Since I see arguments with period division by 11, I am pretty sure they are roots of an (reducible, yes, I'd think) eleventh degree equation. Then I would have just factorized that thing out and end up with the correct quadratic

Just a heuristic. I am pretty sure it works, though. Perhaps someone can finish through this line?

 

[Albert1]

Hi Albert:), thanks for participating.

According to your diagram, I have found $\tan 3\theta=\dfrac{1}{t}$. And I am curious and wish to know if you have found the value of $t$ yet? I tried with no luck so far.

$in \triangle BCD ,\,\, tan 3\theta =\dfrac {BD}{CD}=\dfrac {2t}{1-t^2+t^2} =2t$

 

[mathbalarka]

Eh, I derived this one by some kind of twisted Gauss sum, but perhaps this ain't what you want..., is it?

 

[Albert1]

Prove that $\tan \left( \dfrac{3 \pi}{11} \right)+ 4\sin \left( \dfrac{2 \pi}{11} \right)=\sqrt{11}$.

I know this problem may be stale as it has been posted countless times at other math forums, but I've seen one brilliant method to attack this problem recently, and I'm so eager to share it with the folks here at our site.

Despite my saying so, I would still welcome anyone who would want to take a stab at this challenge.

I use another method (this is easier ,but the calculation still very tedious)

Spoiler

let :$t=tan \,\,x ,\,\,(x=\dfrac {\pi}{11})$
then :$tan \,\, 3x=\dfrac{3t-t^3}{1-3t^2}----(a)$
$4sin \,\, 2x=4\times \dfrac{2t}{1+t^2}=\dfrac {8t}{1+t^2}----(b)$
$tan \,\, 2x=\dfrac{2t}{1-t^2}----(c)$
for :
$tan \,\, 5x +tan \,\, 6x=0$
$\dfrac {tan \,\, 3x+tan\,\, 2x}{1-tan\,\,3x\,tan\,2x}+\dfrac {2\times\, tan \,\, 3x}{1-tan^2 3x}=0---(d)$
put (a) and (c) to (d) we may solve for t (one of its solution will be t$\approx 0.2936)$
put t to (a) and (b) we may get the result:$(a)+(b)$and compare with $\sqrt {11}$

 

[anemone]

Thanks to you, Albert for providing me two different methods to tackle this particular problem.

@Balarka, you're free to use any method that you like to attack the problem and I am looking forward to read your solution post, as I am sure many will get benefited by looking at your solution, if there is any.

Prove that $\tan \left( \dfrac{3 \pi}{11} \right)+ 4\sin \left( \dfrac{2 \pi}{11} \right)=\sqrt{11}$.

I know this problem may be stale as it has been posted countless times at other math forums, but I've seen one brilliant method to attack this problem recently, and I'm so eager to share it with the folks here at our site.

Despite my saying so, I would still welcome anyone who would want to take a stab at this challenge.

Spoiler

According to the tangents of sums formulae, we have

$\tan 11x=\dfrac{11 \tan x-165\tan^3x+462\tan^5x-330\tan^7x+55\tan^2x-tan^{11}x}{1-55\tan^2x+330\tan^4x-462\tan^6x+165\tan^8x-11\tan^{10}x}$

For $x=\dfrac{\pi}{11}$, we have $11 \tan x-165\tan^3x+462\tan^5x-330\tan^7x+55\tan^2x-tan^{11}x=0$

Hence,

$\tan^{10}x=55\tan^8x-330\tan^6x+462\tan^4x-165\tan^2x+11$

Notice that

$\begin{align*}\sin 2x&=2\sin x\cos x\\&=\dfrac{2\sin x \cos x \cos x}{\cos x}\\&=2\tan x \cos^2 x\\&=2\tan x\dfrac{\cos^2 x}{1}\\&=2\tan x\dfrac{\cos^2 x}{\sin^2 x+\cos^2 x}\\&=2\tan x\dfrac{1}{\tan^2 x+1}\end{align*}$

hence we get

$\begin{align*}(\tan 3x+4\sin 2x)^2&=\left( \dfrac{3\tan x-\tan^3 x}{1-3\tan^2 x}+\dfrac{8\tan x}{1+\tan^2 x} \right)^2\\&=\dfrac{(11\tan x-22\tan^3 x-\tan^5 x)^2}{((1-3\tan^2 x)(\tan^2 x+1))^2}\\&=\dfrac{\tan^{10} x+44\tan^8 x+462\tan^6 x-484\tan^4 x+121\tan^2 x}{9\tan^8 x+12\tan^6 x-2\tan^4 x-4\tan^2 x+1}\\&=\dfrac{11(9\tan^8 x+12\tan^6 x-2\tan^4 x-4\tan^2 x+1)}{9\tan^8 x+12\tan^6 x-2\tan^4 x-4\tan^2 x+1}\\&=11\end{align*}$

Note that we replaced $\tan^{10}x$ by $55\tan^8x-330\tan^6x+462\tan^4x-165\tan^2x+11$ in the second last step above since we're dealing with $x=\dfrac{\pi}{11}$.

therefore we can conclude that
$\tan \left( \dfrac{3 \pi}{11} \right)+ 4\sin \left( \dfrac{2 \pi}{11} \right)=\sqrt{11}$.