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From this geometry problem, I can not find geometry solution. However the answer is $X=\frac{2\pi}{15}$ by geometry method. enter image description here

Then I get the identity $$\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1.$$

How to prove it by trigonometric method ?

Thank in advances.

Martin Sleziak
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kong
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  • In general, squaring and applying known trigonometric identities can sometimes be of help. Might not be of help here though. – pshmath0 Jan 15 '15 at 08:17
  • @kong, Related : http://math.stackexchange.com/questions/805023/how-to-prove-dfrac1-sin6-circ-cos12-circ-cos6-circ-sin12-circ?rq=1 – lab bhattacharjee Jan 15 '15 at 08:29

3 Answers3

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I'd try using that $$\frac{1}{5} - \frac{1}{2} + \frac{1}{3} = \frac{1}{30},$$ $$\frac{4}{3} - \frac{6}{5} = \frac{2}{15}.$$ And the formula: $$ \tan(a\pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a\tan b}. $$

hjhjhj57
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This is not a fully sketched answer.I enjoyed while going through it and hence thought others might like it.

Since $\displaystyle\sin\frac{\pi}{10}=\frac{1}{4}(\sqrt{5}-1)$ we obtain $\displaystyle\sin\frac{\pi}{30}=\frac{1}{8}(\sqrt{30-6\sqrt{5}}-1-\sqrt{5})$ and consequently $$\tan\frac{\pi}{30}=\sqrt{7-2\sqrt{5}-2\sqrt{15-6\sqrt{5}}}$$ Now since $\displaystyle\cos\frac{\pi}{5}=\frac{1}{4}(\sqrt{5}+1)$ we obtain $\displaystyle\cos\frac{\pi}{15}=\frac{1}{8}(\sqrt{30+6\sqrt{5}}-1+\sqrt{5})$ and consequently $$\tan\frac{\pi}{15}=\frac{1}{\sqrt{7+2\sqrt{5}+2\sqrt{15+6\sqrt{5}}}}$$ which in turn implies $$\tan\frac{2\pi}{15}=\frac{\sqrt{7+2\sqrt{5}+2\sqrt{15+6\sqrt{5}}}}{3+\sqrt{5}+\sqrt{15+6\sqrt{5}}}$$ Finally note that $\displaystyle\cos\frac{\pi}{5}=\frac{1}{4}(\sqrt{5}+1)$. Putting all these together and after some good bit of time spent on simplification we obtain the result.

Math-fun
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We need $\displaystyle\sqrt3\sec\frac\pi5+\tan\frac\pi{30}=\cot\frac{2\pi}{15}=\tan\dfrac{11\pi}{30}$ as $\dfrac{2\pi}{15}+\dfrac{11\pi}{30}=\dfrac\pi2$

$\iff\displaystyle\sqrt3\sec\dfrac\pi5=\tan\dfrac{11\pi}{30}-\tan\dfrac\pi{30}$

$\iff\displaystyle\dfrac{\sqrt3}{\cos\dfrac\pi5}=\frac{\sin\left(\dfrac{11\pi}{30}-\dfrac\pi{30}\right)}{\cos\dfrac\pi{30}\cos\dfrac{11\pi}{30}}$

As $\sin\left(\dfrac{11\pi}{30}-\dfrac\pi{30}\right)=\sin\dfrac\pi3=\dfrac{\sqrt3}2,$

$\iff\displaystyle\cos\frac\pi5=2\cos\frac\pi{30}\cos\dfrac{11\pi}{30}=\cos\dfrac\pi3+\cos\frac{2\pi}5$

$\iff\displaystyle\cos\frac\pi5-\cos\frac{2\pi}5=\cos\dfrac\pi3=\dfrac12$

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

Community
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lab bhattacharjee
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