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How can I simplify this trigonometric expression?

$$ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$$

I used

$$\sin \frac{\pi}{15}=2 \sin \frac{\pi}{30} \cos \frac{\pi}{30}$$ and $$\cos\frac{\pi}{15}=2\cos^2\frac{\pi}{30}-1$$

But these give me more complicated expressions.

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    $\begingroup$ What is the source of this uncommon problem? $\endgroup$ – lab bhattacharjee Sep 24 '19 at 10:39
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    $\begingroup$ I find the answer is $\dfrac{3\pi}{10}$ by wolfram alpha $\endgroup$ – Isaac YIU Math Studio Sep 24 '19 at 10:44
  • $\begingroup$ Reminds me of math.stackexchange.com/questions/805023/… $\endgroup$ – lab bhattacharjee Sep 24 '19 at 10:53
  • $\begingroup$ @labbhattacharjee A contest math problem.(in the book) I could not solve. (I tried for hours) WA can solve ,but unfortunately, doesnt show steps. So, I shared problem here. $\endgroup$ – Elementary Sep 24 '19 at 16:23
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    $\begingroup$ @Elementary, I need more time & concentration which I'm really sorry that I could not invest. I will be really happy to solve it. Thanks for the nice problem. Pray continue supplying such problems $\endgroup$ – lab bhattacharjee Sep 25 '19 at 11:55
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Let $\arctan\left(\dfrac{2\cos6^\circ\cos12^\circ}{1+2\cos6^\circ\sin12^\circ}\right)=y,-90^\circ<y<90^\circ$

$\implies\dfrac{2\cos6^\circ\cos12^\circ}{1+2\cos6^\circ\sin12^\circ}=\tan y=\dfrac{\sin y}{\cos y}$

Rearranging we get $$\cos(6^\circ+y)+\cos(18^\circ+y)=\sin y=\cos(90^\circ-y)$$

$$\iff\cos(90^\circ-y)=\cos(18^\circ+y)+\cos(6^\circ+y)=2\cos6^\circ\cos(12^\circ+y)\ \ \ \ (1)$$

Like Solve equation $\tan(x)=\sec(42^\circ)+\sqrt{3}$

using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$

$$\cos36^\circ=\cos72^\circ+\cos60^\circ=2\cos6^\circ\cos66^\circ\ \ \ \ (2)$$

Compare $(1),(2)$

$$\frac{\cos36^\circ}{\cos(90^\circ-y)}=\dfrac{2\cos6^\circ\cos66^\circ}{2\cos6^\circ\cos(12^\circ+y)}$$

$$\iff\dfrac{\cos(12^\circ+y)}{\cos(90^\circ-y)}=\dfrac{\cos66^\circ}{\cos36^\circ}$$

Apply Componendo et Dividendo and Prosthaphaeresis Formulas to find

$$\tan(y-39^\circ)=\tan15^\circ$$

The rest should be easy!

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$\begin{align} T &= \large \left(\frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}\right) \left({2\sin 6° \over 2\sin 6°}\right)\cr &= \large \frac{\sin 24°}{2\sin 6°+\;2\sin^2 12°}\cr\cr {1\over T}&= \large \frac{2\sin(30°-24°)+\;(1-\cos 24°)}{\sin 24°}\cr &=\large {(\cos 24° - \sqrt3\sin24°) + (1-\cos 24°) \over \sin 24°}\cr &=\large{1\over\sin(60°-36°)} - \sqrt3 \cr &=\large{4\over 2\sqrt3 \cos36° - 2\sin 36°}-\sqrt3 \cr \end{align}$

From wikipedia on Golden triangle, $\quad2\cos(36°)=\phi,\quad 2\sin(36°)=\sqrt{4-\phi^2}$

$\begin{align} \large{1\over T}&=\left({4\over \sqrt3\phi-\sqrt{4-\phi^2}}\right) \left({\sqrt3\phi+\sqrt{4-\phi^2} \over \sqrt3\phi+\sqrt{4-\phi^2}} \right) -\sqrt3 \cr &= {4(\sqrt3\phi + \sqrt{4-\phi^2}) \over 3\phi^2-(4-\phi^2)} -\sqrt3 \cr &= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi^2-1}-\sqrt3 \cr &= {\sqrt3\phi + \sqrt{4-\phi^2} \over \phi} -\sqrt3 \cr &= {\sqrt{4-\phi^2} \over \phi} = \tan(36°) \end{align}$

$$T = \tan(90°-36°) = \tan(54°)$$

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The identity can also be proved from: $\quad\cos36° = \cos 72° + {1\over2}$

$\cos 72° + {1\over2}= (2\cos^2 36° - 1)+ {1\over2}\;= \large{\phi^2\over2}-{1\over2} ={\phi \over 2}=\cos36°$


$T= \Large \frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}$

$\Large{1\over T} = {1\over 2\cos 6° \cos 12°} + \normalsize \tan 12°$

$\begin{align} \tan 36° - \tan12° &= {\sin36° \over \cos36°} - {\sin12° \over \cos12°} \cr &= {\sin36° \cos12° - \cos36° \sin12° \over \cos12° \cos36°} \cr &= {\sin24° \over \cos12° (\cos72° + 0.5)} \cr &= {\sin24° \over \cos12°(\sin18° + \sin30°)} \cr &= {\sin24° \over \cos12°(2 \sin24° \cos6°)} \cr \tan36° &= {1\over 2\cos 6° \cos 12°} + \tan12° \end{align}$

$$T = \cot 36° = \tan 54°$$

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