MHB Simplify a trigonometric expression

AI Thread Summary
The discussion revolves around simplifying the trigonometric expression $\left(\sqrt{3}\sec \dfrac{\pi}{5}+\tan\dfrac{\pi}{30}\right)\tan \dfrac{2\pi}{15}$. A participant points out a typo in the original problem, clarifying that it should involve $\tan \dfrac{\pi}{30}$ instead of $\tan \dfrac{2\pi}{5}$. Multiple solutions are shared, with one participant expressing appreciation for the different approaches. The conversation highlights the importance of accuracy in problem statements for effective collaboration. Overall, the thread emphasizes problem-solving in trigonometry while addressing minor errors in the original query.
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Simplify $\left(\sqrt{3}\sec \dfrac{\pi}{5}+\tan\dfrac{2\pi}{5}\right)\tan \dfrac{2\pi}{15}$.
 
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Here is my work

$ \left( \sqrt{3} \sec \frac{\pi}{5} + \tan \frac{2\pi}{5}\right) \tan \frac{2\pi}{15}$
 
Last edited by a moderator:
Amer said:
Here is my work

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

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 [TABLE="class: grid, width: 800"]
[TR]
[TD]$\tan (90^{\circ}-y)=\dfrac{1}{\tan y}$[/TD]
[TD]and[/TD]
[TD]$\tan x \tan (60^{\circ}-x) \tan (60^{\circ}-x)=\tan 3x$, we see that if we set $x=24^{\circ}$, we have[/TD]
[/TR]
[/TABLE]


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

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

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

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

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

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

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

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

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

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