Simplify a trigonometric expression

[anemone]

Simplify $\left(\sqrt{3}\sec \dfrac{\pi}{5}+\tan\dfrac{2\pi}{5}\right)\tan \dfrac{2\pi}{15}$.

 

[Amer]

Here is my work

Spoiler

$ \left( \sqrt{3} \sec \frac{\pi}{5} + \tan \frac{2\pi}{5}\right) \tan \frac{2\pi}{15}$

 

[anemone]

Here is my work

Spoiler

$ \left( \sqrt{3} \sec \frac{\pi}{5} + \tan \frac{2\pi}{5}\right) \tan \frac{2\pi}{15}$

Hi Amer,

I suspect you might not aware that you have not posted your solution due to some unforeseen technical problem, is that true?

 

[anemone]

I want to apologize for the careless typo in the original problem statement. I wish I would have noticed it much earlier...

The problem should read:

Simplify $\left(\sqrt{3}\sec \dfrac{\pi}{5}+\tan\dfrac{\pi}{30}\right)\tan \dfrac{2\pi}{15}$.

 

[anemone]

My solution:

Spoiler

The problem can also be interepreted as to simplify

$\begin{align*}(\sqrt{3}\sec 36^{\circ}+\tan6^{\circ})\tan 24^{\circ}&=(\dfrac{\sqrt{3}}{\cos 36^{\circ}}+\tan6^{\circ})\tan 24^{\circ}\\&=\dfrac{\sqrt{3}\tan 24^{\circ}}{\cos 36^{\circ}}+\tan6^{\circ}\tan 24^{\circ}\end{align*}$

It's really hard to "predict" what would be the next best step to simplify the given expression, but my plan of attack is to find an expression that relates between $\tan 24^{\circ}$ and $\cos 36^{\circ}$.

From the identity

$\tan (90^{\circ}-y)=\dfrac{1}{\tan y}$

and

$\tan x \tan (60^{\circ}-x) \tan (60^{\circ}-x)=\tan 3x$, we see that if we set $x=24^{\circ}$, we have

$\tan 24^{\circ} \tan 36^{\circ} \tan 84^{\circ}=\tan 72^{\circ}$

$\tan 24^{\circ} \tan 36^{\circ} \dfrac{1}{\tan 6^{\circ}}=\dfrac{1}{\tan 18^{\circ}}$

$\tan 24^{\circ} \tan 36^{\circ} \tan 18^{\circ}=\tan 6^{\circ}$

$\tan 24^{\circ} \dfrac{\sin 36^{\circ}}{\cos 36^{\circ}} \dfrac{\sin 18^{\circ}}{\cos 18^{\circ}}=\tan 6^{\circ}$

$\tan 24^{\circ} \dfrac{2\sin 18^{\circ}\cancel{\cos 18^{\circ}}}{\cos 36^{\circ}} \dfrac{\sin 18^{\circ}}{\cancel{\cos 18^{\circ}}}=\tan 6^{\circ}$

$\tan 24^{\circ}(2\sin^2 18^{\circ})=\tan 6^{\circ}\cos 36^{\circ}$

$\tan 24^{\circ}(1-\cos 36^{\circ})=\tan 6^{\circ}\cos 36^{\circ}$

$\dfrac{\sqrt{3}\tan 24^{\circ}}{\cos 36^{\circ}}=1-(\tan 6^{\circ}\tan 24^{\circ})$

$\begin{align*}\therefore (\sqrt{3}\sec 36^{\circ}+\tan6^{\circ})\tan 24^{\circ}&=\dfrac{\sqrt{3}\tan 24^{\circ}}{\cos 36^{\circ}}+\tan6^{\circ}\tan 24^{\circ}\\&=1-(\tan 6^{\circ}\tan 24^{\circ})+\tan6^{\circ}\tan 24^{\circ}\\&=1\end{align*}$

 

[Amer]

My solution

Spoiler

Let $\displaystyle x = \frac{\pi}{30}$

The expression with the assumption above

$ \left( \sqrt{3} \sec 6x + \tan x \right) \tan 4x $

$\displaystyle \frac{\sqrt{3} \cos(x) \sin(4x) + \sin(x) \sin(4x) \cos(6x)}{\cos (6x) \cos(4x) \cos(x)}$

$\displaystyle \frac{ \sqrt{3}/2 ( \sin (5x) + \sin (3x) ) + 0.5 \cos(6x) ( \cos(3x) - \cos(5x) ) }{\cos (6x) \cos(4x) \cos(x)}$

Since $5x = \frac{\pi}{6}$

$\displaystyle \frac{\sqrt{3}/2 ( 1/2 + \sin (3x) ) +0.5 \cos(6x) ( \cos(3x) - \sqrt{3}/2) }{\cos (6x) \cos(4x) \cos(x)} $

$\displaystyle \frac{\sqrt{3} + 2\sqrt{3} \sin(3x) + \cos(6x) (2\cos(3x) - \sqrt{3} )}{\cos(6x) (\sqrt{3} + 2\cos(3x))}$

$\displaystyle \frac{\sqrt{3} + 2\sqrt{3} \sin(3x)- 2\sqrt{3}\cos(6x) + \cos(6x) (2\cos(3x) + \sqrt{3} )}{\cos(6x) (\sqrt{3} + 2\cos(3x))} = \frac{\sqrt{3} + 2\sqrt{3} \sin(3x)- 2\sqrt{3}\cos(6x)}{\cos(6x) (\sqrt{3} + 2\cos(3x))} +1 = 1 $

Since
$\displaystyle 1 + 2 \sin(3x) - 2 \cos (6x) = 1 + \frac{2}{4} ( \sqrt{5} -1) - \frac{2}{4} ( 1 + \sqrt{5}) = 0 $

 

[anemone]

My solution

Spoiler

Let $\displaystyle x = \frac{\pi}{30}$

The expression with the assumption above

$ \left( \sqrt{3} \sec 6x + \tan x \right) \tan 4x $

$\displaystyle \frac{\sqrt{3} \cos(x) \sin(4x) + \sin(x) \sin(4x) \cos(6x)}{\cos (6x) \cos(4x) \cos(x)}$

$\displaystyle \frac{ \sqrt{3}/2 ( \sin (5x) + \sin (3x) ) + 0.5 \cos(6x) ( \cos(3x) - \cos(5x) ) }{\cos (6x) \cos(4x) \cos(x)}$

Since $5x = \frac{\pi}{6}$

$\displaystyle \frac{\sqrt{3}/2 ( 1/2 + \sin (3x) ) +0.5 \cos(6x) ( \cos(3x) - \sqrt{3}/2) }{\cos (6x) \cos(4x) \cos(x)} $

$\displaystyle \frac{\sqrt{3} + 2\sqrt{3} \sin(3x) + \cos(6x) (2\cos(3x) - \sqrt{3} )}{\cos(6x) (\sqrt{3} + 2\cos(3x))}$

$\displaystyle \frac{\sqrt{3} + 2\sqrt{3} \sin(3x)- 2\sqrt{3}\cos(6x) + \cos(6x) (2\cos(3x) + \sqrt{3} )}{\cos(6x) (\sqrt{3} + 2\cos(3x))} = \frac{\sqrt{3} + 2\sqrt{3} \sin(3x)- 2\sqrt{3}\cos(6x)}{\cos(6x) (\sqrt{3} + 2\cos(3x))} +1 = 1 $

Since
$\displaystyle 1 + 2 \sin(3x) - 2 \cos (6x) = 1 + \frac{2}{4} ( \sqrt{5} -1) - \frac{2}{4} ( 1 + \sqrt{5}) = 0 $

Thanks Amer for participating and I am glad to receive another solution that works differently than mine.:)