Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$

Question

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.)

Answer 1

The case of multiples of $\pi/11$ actually involves five "symmetrically equivalent" forms:

$4\sin(5\pi/11)-\tan(2\pi/11)=\sqrt{11}$

$4\sin(\pi/11)+\tan(4\pi/11)=\sqrt{11}$

$4\sin(4\pi/11)-\tan(5\pi/11)=-\sqrt{11}$

$4\sin(2\pi/11)+\tan(3\pi/11)=\sqrt{11}$

$4\sin(3\pi/11)+\tan(\pi/11)=\sqrt{11}$

All are derivable from the quadratic Gauss sum corresponding to the prime number $11$.

These are manifestations of the more compact, "symmetric" relation

$\color{blue}{4\sin(3\theta)-\tan(\theta)=(k|11)\sqrt{11}}$

where $\theta=(2k\pi/11)$ and $(k|11)$ is the Legendre symbol of residue $k$ modulo $11$, with the specific equations quoted above representing $k=1,2,3,4,5$ respectively.

There is a hidden feature in the equation rendered in blue above. In addition to multiples of $\pi/11$ we get one more value of $\theta$ between $0$ and $\pi$ where the function on the left evaluates to $+\sqrt{11}$. Correspondingly there is an additional value of $\theta$ between $\pi$ and $2\pi$, having the same cosine, for which the function value is $-\sqrt{11}$. Now, suppose we plug in $x=2\cos\theta$. Squaring the blue equation, expressing the quantities in terms of $x$ and clearing fractions yields an eighth-degree polynomial equation, which factors as follows:

$(x^5+x^4-4x^3-3x^2+3x+1)(x^3-x^2-x-1)=0$

The quintic factor is just the minimal polynomial for $2\cos(2k\pi/11)$ for $k\in\{1,2,3,4,5\}$. The cubic factor, containing the "extra" roots for $\theta$, has been coupled into the quintic one through the Gauss sum and the "sine-tangent relations" derived from it.

In 2014 the regular hendecagon was discovered to be neusis constructible. The authors "miraculously" found that the neusis construction, which requires finding the roots of the quintic factor given above, can be rendered in terms of cubic roots for which a neusis construction is guaranteed. It turns out that in the construction found by the authors, the distance from the pole of the neusis (a fixed point through which the marked ruler passes) to the straight line that includes one of the marks satisfies the equation

$a^3+a^2+a-1=0.$

This corresponds exactly to the reciprocal of the root of $x^3-x^2-x-1=0$, the latter equation being the coupled cubic factor emerging above from the Gauss sum. The construction is still a bit of a miracle, but we see that its cubic roots do not just appear out of nowhere. They are derived from the Gauss sum!

Answer 2

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. $2^u3^v+1$.

Answer 3

$$\tan\frac{2\pi}{29}+4\left(-\sin\frac{2\pi}{29}+\sin\frac{6\pi}{29}+\sin\frac{8\pi}{29}-\sin\frac{20\pi}{29}+\sin\frac{22\pi}{29}\right)=\sqrt{29-2\sqrt{29}} $$

Answer 4

$$\tan\frac{2\pi}{37}+4\left(-\sin\frac{4\pi}{37}+\sin\frac{14\pi}{37}+\sin\frac{16\pi}{37}+\sin\frac{24\pi}{37}-\sin\frac{28\pi}{37}+\sin\frac{32\pi}{37}-\sin\frac{36\pi}{37}\right)=\sqrt{37+6\sqrt{37}} $$

Answer 5

There is also the following: $$\tan\frac{\pi}{19}+4\sin\frac{7\pi}{19}+4\sin\frac{8\pi}{19}-4\sin\frac{6\pi}{19}=\sqrt{19},$$ $$-\tan\frac{3\pi}{19}+4\sin\frac{\pi}{19}+4\sin\frac{2\pi}{19}+4\sin\frac{5\pi}{19}=\sqrt{19},$$ $$\tan\frac{5\pi}{19}+4\sin\frac{2\pi}{19}-4\sin\frac{3\pi}{19}-4\sin\frac{8\pi}{19}=\sqrt{19},$$ $$\tan\frac{7\pi}{19}+4\sin\frac{\pi}{19}-4\sin\frac{4\pi}{19}+4\sin\frac{8\pi}{19}=\sqrt{19}.$$