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I recently saw on this site, the identity $$\sec\frac\pi{30}=\sqrt{2-\sqrt{5}+\sqrt{15-6\sqrt{5}}}$$ which I instantly wanted to prove.

I know that I can "reduce" the problem to the evaluation of $\cos\frac\pi{15},$ as the rest is easy with the use of the half-angle formula.

I know that $\cos$ obeys the 'nice' relation $$\cos nx=T_n(\cos x)$$ where $$T_n(x)=\frac{n}2\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{n-k}{n-k\choose k}(2x)^{n-2k}.$$ Thus, setting $t=\cos\frac\pi{15}$, $$T_{15}(t)=-1.$$ The only thing left to do is solve for $t$. We can narrow down our search to the values $0<t<1.$

I have never dealt with degree-$15$ polynomials before, so I was hoping one of you could help me out. Thanks!

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  • $\begingroup$ how about $n=5$ instead of $15$? $\endgroup$ – J. W. Tanner Sep 15 '19 at 3:35
  • $\begingroup$ @J.W.Tanner is there a $1/3$-angle formula? $\endgroup$ – clathratus Sep 15 '19 at 3:35
  • $\begingroup$ I was just saying $T_5(t)=\cos\frac{\pi}3=\frac12$ is a polynomial equation with degree significantly less than $15$ $\endgroup$ – J. W. Tanner Sep 15 '19 at 3:47
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    $\begingroup$ See a good methodology description here where you will find the reference to this astonishing table $\endgroup$ – Jean Marie Sep 15 '19 at 5:22
  • $\begingroup$ sec(pi/30) identity does not hold ! LHS ≈ 1.00551, but RHS ≈ 1.01111 $\endgroup$ – albert chan Sep 15 '19 at 22:17
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Another way:

$$\cos6^\circ=+\sqrt{\dfrac{1+\cos12^\circ}2}$$

$$\cos(30^\circ-18^\circ)=?$$

Use this to find $\cos18^\circ,\sin18^\circ$

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Staying with your approach (which could be made simpler as said in comments and answers), you have $$T_{15}(x)=-15 x+560 x^3-6048 x^5+28800 x^7-70400 x^9+92160 x^{11}-61440 x^{13}+16384 x^{15}$$ but $T_{15}(x)+1$ can be factorized as $$T_{15}(x)+1=(x+1) (2 x-1)^2 \left(4 x^2-2 x-1\right)^2 \left(16 x^4+8 x^3-16 x^2-8 x+1\right)^2$$ which means that you are left with the quartic $$16 x^4+8 x^3-16 x^2-8 x+1=0$$ which can be solved with radicals. Its solutions are $$\left\{x= \frac{1}{8} \left(-1-\sqrt{5}-\sqrt{6 \left(5-\sqrt{5}\right)}\right)\right\},\left\{x= \frac{1}{8} \left(-1-\sqrt{5}+\sqrt{6 \left(5-\sqrt{5}\right)}\right)\right\},\left\{x= \frac{1}{8} \left(-1+\sqrt{5}-\sqrt{6 \left(5+\sqrt{5}\right)}\right)\right\},\left\{x= \frac{1}{8} \left(-1+\sqrt{5}+\sqrt{6 \left(5+\sqrt{5}\right)}\right)\right\}$$

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  • $\begingroup$ Thanks, I thought it would be able to factor somewhat nicely. $\endgroup$ – clathratus Sep 15 '19 at 4:24
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    $\begingroup$ You are welcome ! It was clear to me that there was a quartic somewhere. Cheers. $\endgroup$ – Claude Leibovici Sep 15 '19 at 5:06
  • $\begingroup$ FYI, the last x solution (sign - + + +) = cos(12°) $\endgroup$ – albert chan Sep 15 '19 at 22:34
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It is impossible to prove your identity. The identity does not hold.

I did derive a compact expression for $\cos(6°)$
see Find the value of $4\cos 6\cos 42 \cos 60 \cos 66 \cos 78$ All angles in degrees

$\begin{align} \cos 6° &= \cos(36°-30°) \cr &= \cos36°\cos30° + \sin36°\sin30° \cr &= {\phi \over 2}\times{\sqrt3 \over 2} + \sqrt{1-({\phi \over 2})^2}\times{1 \over 2} \cr 4 \cos 6° &= \sqrt3\;\phi + \sqrt{4-\phi^2} = \sqrt3\;\phi + \sqrt{4-(\phi+1)} \cr\cr \cos 6° &= {\sqrt3\;\phi + \sqrt{3-\phi} \over 4} ≈ 0.99452\cr \end{align}$

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