<s>[INST] Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2\pi}{7}=\sqrt{7}$$
$$\tan\frac{\pi}{11}+4\sin\frac{3\pi}{11}=\tan\frac{3\pi}{11}+4\sin\frac{2\pi}{11}=\sqrt{11}$$
$$\tan\frac{6\pi}{13}-4\sin\frac{5\pi}{13}=\tan\frac{2\pi}{13}+4\sin\frac{6\pi}{13}=\sqrt{13+2\sqrt{13}}$$
Does anyone know of any analogous identities for larger primes? I have not been able to find anything similar for $p=17$ or $p=19$.
(I am not asking for proofs of the above equations.)
 [/INST] There is rule to satisfy either to find the identities or to find any particular value of a trigonometric function with that prime in the denominator and a multiple of $ \pi$ in the numerator.


*

*Considering the case of $5$, it is a Fermat prime and hence it can be constructed using straight edges and compass. Because, $F_n = 2^{2^n}+1$ and $F_1=5$. And since the divisor neglecting the $1$ is $ k=2$, it can be manipulated with simple bisectors or straight edges and compass. 

*Considering the number $7$, it is not a Fermat prime. But it is a Pierpont prime i.e. of the form $2^u 3^v+1$. for $7$, $u=v=1$. For those numbers which are Pierpont primes they are either constructible using angle trisectors (due to the factor $3$) or through neusis construction. 

*Considering the number $11$, it neither Fermat not Pierpont prime. Hence, it can neither be constructed using the bisectors (straight edges and a  compass) nor using an angle trisector. It is only possible through neusis construction. 

*For $13$-gon, $13$ is a Pierpont prime with $u=2, v=1$ and hence it can be constructed using angle trisectors or neusis but it can't be constructed using straight edges since it is not a Fermat prime. 

*Considering the number $15$, it is a product of distinct Fermat primes $3$ ($ F_0$) and $5$ ($F_1$) and hence it is constructible using straight edges and a compass. 

*Considering the number $17$, it is a Fermat prime ($F_2$) and hence it is constructible using straight edges and compass. 

*But $19$ is not a Fermat prime but a Piermont prime and hence it is constructible using angle trisectors or neusis. This holds for any prime number. Only the Fermat prime-sided polygons will have a defined value for all the trigonometric functions but Pierpont primes will not have a clearly defined value for any trigonometric function. Instead, they give the identity for a sum or product (or mixed) of different amplitudes of sine and cosine of angles of the format $\frac{n\pi}{p}$ where $p$ is the prime under consideration and the number $n$ need not be the same throughout the identity.
For example, consider the pentagon. It gives the value for $\cos{\frac{\pi}{10}}$ which may be used to give the value for $\cos{\frac{\pi}{5}}$ using trigonometric identity for $\cos{2x}$. The value of is $\cos{\frac{\pi}{10}}=\sqrt{\frac{5+\sqrt{5}}{8}}$. Similarly for the next Fermat prime 17, it is given as:
\begin{equation}
16\cos{\frac{2\pi}{17}}=-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}
\end{equation}
For the Pierpont number 13, it can't be given a defined value for a trigonometric function as mentioned above and hence the identity holds as:
\begin{equation}
\cos^2 {\frac{\pi}{13}}+\cos^2 {\frac{3\pi}{13}}+\cos^2 {\frac{4\pi}{13}}=\frac{11+\sqrt{13}}{8}
\end{equation}
\begin{equation}
\sin{\frac{\pi}{13}}+\sin{\frac{3\pi}{13}}+\sin{\frac{4\pi}{13}}=\sqrt{\frac{13+3\sqrt{13}}{8}}
\end{equation}
But Wolfram has revealed about the number $23$ as stated here. Also, its general solution is as follows (i.e. any trigonometric function can be written in terms of 1 and fractional powers of -1):
$-\frac{1}{2}(-1)^{\frac{22}{23}}\big[1+(-1)^{\frac{2}{23}}\big]$
It is the same case for $7$, a Pierpont prime as:
\begin{equation}
\prod_{k=1}^{3} \sin{\frac{k\pi}{7}} =\frac{\sqrt{7}}{8} 
\end{equation}
\begin{equation}
\prod_{k=1}^{3} \cos{\frac{k\pi}{7}} =\frac{1}{8} 
\end{equation}
\begin{equation}
\cos^2 {\frac{\pi}{7}}-\cos{\frac{\pi}{7}}\cos{\frac{2\pi}{7}}=\frac{1}{4}
\end{equation}
And another identity for the same is:
\begin{equation}
\cos^{\frac{1}{3}} {\frac{2\pi}{7}}-\Bigg[-\cos{\frac{4\pi}{7}}\Bigg]^{\frac{1}{3}}-\Bigg[-\cos{\frac{6\pi}{7}}\Bigg]^{\frac{1}{3}}=-\Bigg[\frac{1}{2} \bigg(3\times7^{\frac{1}{3}}-5\bigg)\Bigg]^{\frac{1}{3}}
\end{equation}
The reason for saying that a number of sides, $n$ which is not a Fermat prime but a Pierpont prime for a $n-$gon being constructible only by trisectors or neusis and not by bisectors is that Fermat primes allows to factor out 2 i.e. bisect since it is in that form of power of 2 and since $n$ is not a Fermat prime but a Pierpont prime, the only factor that distinguishes Pierpont from Fermat prime is 3. Hence, it is constructible using trisectors. That is, consider Fermat prime to be of the form $F_n = 2^{k(n)}+1; k(n)=2^n$ i.e. $k(n)=k$ in short. Pierpont prime adds another factor of 3 i.e. $2^u3^v+1$.
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