I need to know if $$\cos(\pi/5) \in \mathbb{Q} (\sin(\pi/5))?$$
I can compute explicitly such $\cos$ and $\sin$, but I have some difficulties how to deduce from this an answer.
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$\begingroup$ Somehow we should exploit: $\cos \frac{\pi}{5} = \sqrt{1 - \sin^2 \frac{\pi}{5}}$ or something. I been looking at these proofwiki.org/wiki/Quintuple_Angle_Formulas $\endgroup$– cactus314Aug 13, 2015 at 10:57
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$\begingroup$ I use this thing: $(\cos(\frac{\pi}{5}) + i\sin{\frac{\pi}{5}})^5 = -1$. $\endgroup$– Alex-omskAug 13, 2015 at 11:01
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3 Answers
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As it turns out ${\displaystyle \cos\bigg({\pi \over 5}\bigg) = {1 + \sqrt{5} \over 4}}$, which can be seen for example by using the equation $\cos(3x) = \cos(2x)$ for ${\displaystyle x = {\pi \over 5}}$, writing everything in terms of $\cos(x)$ using double and triple angle formulas, and then solving for $\cos(x)$. So we have $$\sin^2\bigg({\pi \over 5}\bigg) = 1 - \bigg({1 + \sqrt{5} \over 4}\bigg)^2$$ This is of the form $a + b \sqrt{5}$ for rationals $a$ and $b$ with $b \neq 0$. Thus ${\displaystyle \sqrt{5} \in Q\big(\sin{\pi \over 5}\big)}$, as is ${\displaystyle \cos({\pi \over 5})}$ since it is just ${\displaystyle {1 + \sqrt{5} \over 4}}$.
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$\begingroup$ Easiest to write twice the cosine as $\zeta+\zeta^{-1}$, where $\zeta$ is a primitive tenth root of unity, satisfying a fourth-degree polynomial. $\endgroup$– LubinAug 13, 2015 at 19:19
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What you’re missing is the observation that $c=\cos(\pi/5)$ is a quadratic irrationality over $\Bbb Q$. Once you know that $c^2+Ac+B=0$ for well-chosen rational numbers $A$ and $B$ with $A$ nonzero, the equation $c=(B-c^2)/A$ becomes $c=(s^2+B-1)/A$, where $s=\sin(\pi/5)$, and you have it.
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Hint : show that $c=\frac{3}{2}-2s^2$, where $c=\cos(\frac{\pi}{5})$ and $s=\sin(\frac{\pi}{5})$. You can check it using Euler's identities for example.
answered Aug 13, 2015 at 11:00
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$\begingroup$ I have two equations: $c^5 - 10c^3s^2+5cs^4+1=0$ and $s^4 - 10c^2s^2+5c^4 = 0$. It is easy to solve the latter and find that $s = \sqrt{5 - 2\sqrt{5}}c$, but I don't know how to solve the first one. $\endgroup$ Aug 13, 2015 at 11:20
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$\begingroup$ @Alex-omsk I think you're making it more complicated than it is. You don't like my suggestion ? You don't like Euler's identities ? $\endgroup$ Aug 13, 2015 at 11:28
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$\begingroup$ Hmm, I use Euler's identity to get this equations.. $\endgroup$ Aug 13, 2015 at 11:32
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$\begingroup$ Ok, I see that it is true. But how can I get it? $\endgroup$ Aug 13, 2015 at 11:47