Heptagonal triangle

A heptagonal triangle is an obtuse scalene triangle whose vertices coincide with the first, second, and fourth vertices of a regular heptagon (from an arbitrary starting vertex). Thus its sides coincide with one side and the adjacent shorter and longer diagonals of the regular heptagon. All heptagonal triangles are similar (have the same shape), and so they are collectively known as the heptagonal triangle. Its angles have measures $\pi /7,2\pi /7,$ and $4\pi /7,$ and it is the only triangle with angles in the ratios 1:2:4. The heptagonal triangle has various remarkable properties

A regular heptagon (with red sides), its longer diagonals (green), and its shorter diagonals (blue). Each of the fourteen congruent heptagonal triangles has one green side, one blue side, and one red side.

Key points

The heptagonal triangle's nine-point center is also its first Brocard point.[1]^{:Propos. 12}

The second Brocard point lies on the nine-point circle.[2]^{:p. 19}

The circumcenter and the Fermat points of a heptagonal triangle form an equilateral triangle.[1]^{:Thm. 22}

The distance between the circumcenter O and the orthocenter H is given by[2]^{:p. 19}

OH=R{\sqrt {2}},

where R is the circumradius. The squared distance from the incenter I to the orthocenter is[2]^{:p. 19}

IH^{2}={\frac {R^{2}+4r^{2}}{2}},

where r is the inradius.

The two tangents from the orthocenter to the circumcircle are mutually perpendicular.[2]^{:p. 19}

Relations of distances

Sides

The heptagonal triangle's sides a < b < c coincide respectively with the regular heptagon's side, shorter diagonal, and longer diagonal. They satisfy[3]^{:Lemma 1}

{\begin{aligned}a^{2}&=c(c-b),\\[5pt]b^{2}&=a(c+a),\\[5pt]c^{2}&=b(a+b),\\[5pt]{\frac {1}{a}}&={\frac {1}{b}}+{\frac {1}{c}}\end{aligned}}

(the latter[2]^{:p. 13} being the optic equation) and hence

ab+ac=bc,

and[3]^{:Coro. 2}

b^{3}+2b^{2}c-bc^{2}-c^{3}=0,

c^{3}-2c^{2}a-ca^{2}+a^{3}=0,

a^{3}-2a^{2}b-ab^{2}+b^{3}=0.

Thus –b/c, c/a, and a/b all satisfy the cubic equation

t^{3}-2t^{2}-t+1=0.

However, no algebraic expressions with purely real terms exist for the solutions of this equation, because it is an example of casus irreducibilis.

The approximate relation of the sides is

b\approx 1.80193\cdot a,\qquad c\approx 2.24698\cdot a.

We also have[4]

{\frac {a^{2}}{bc}},\quad -{\frac {b^{2}}{ca}},\quad -{\frac {c^{2}}{ab}}

satisfy the cubic equation

t^{3}+4t^{2}+3t-1=0.

We also have[4]

{\frac {a^{3}}{bc^{2}}},\quad -{\frac {b^{3}}{ca^{2}}},\quad {\frac {c^{3}}{ab^{2}}}

satisfy the cubic equation

t^{3}-t^{2}-9t+1=0.

We also have[4]

{\frac {a^{3}}{b^{2}c}},\quad {\frac {b^{3}}{c^{2}a}},\quad -{\frac {c^{3}}{a^{2}b}}

satisfy the cubic equation

t^{3}+5t^{2}-8t+1=0.

We also have[2]^{:p. 14}

b^{2}-a^{2}=ac,

c^{2}-b^{2}=ab,

a^{2}-c^{2}=-bc,

and[2]^{:p. 15}

{\frac {b^{2}}{a^{2}}}+{\frac {c^{2}}{b^{2}}}+{\frac {a^{2}}{c^{2}}}=5.

We also have[4]

ab-bc+ca=0,

a^{3}b-b^{3}c+c^{3}a=0,

a^{4}b+b^{4}c-c^{4}a=0,

a^{11}b^{3}-b^{11}c^{3}+c^{11}a^{3}=0.

There are no other (m, n), m, n > 0, m, n < 2000 such that

a^{m}b^{n}\pm b^{m}c^{n}\pm c^{m}a^{n}=0.

Altitudes

The altitudes h_a, h_b, and h_c satisfy

h_{a}=h_{b}+h_{c}

[2]^{:p. 13}

and

h_{a}^{2}+h_{b}^{2}+h_{c}^{2}={\frac {a^{2}+b^{2}+c^{2}}{2}}.

[2]^{:p. 14}

The altitude from side b (opposite angle B) is half the internal angle bisector $w_{A}$ of A:[2]^{:p. 19}

2h_{b}=w_{A}.

Here angle A is the smallest angle, and B is the second smallest.

Internal angle bisectors

We have these properties of the internal angle bisectors $w_{A},w_{B},$ and $w_{C}$ of angles A, B, and C respectively:[2]^{:p. 16}

w_{A}=b+c,

w_{B}=c-a,

w_{C}=b-a.

Circumradius, inradius, and exradius

The triangle's area is[5]

A={\frac {\sqrt {7}}{4}}R^{2},

where R is the triangle's circumradius.

We have[2]^{:p. 12}

a^{2}+b^{2}+c^{2}=7R^{2}.

We also have[6]

a^{4}+b^{4}+c^{4}=21R^{4}.

a^{6}+b^{6}+c^{6}=70R^{6}.

The ratio r /R of the inradius to the circumradius is the positive solution of the cubic equation[5]

8x^{3}+28x^{2}+14x-7=0.

In addition,[2]^{:p. 15}

{\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}+{\frac {1}{c^{2}}}={\frac {2}{R^{2}}}.

We also have[6]

{\frac {1}{a^{4}}}+{\frac {1}{b^{4}}}+{\frac {1}{c^{4}}}={\frac {2}{R^{4}}}.

{\frac {1}{a^{6}}}+{\frac {1}{b^{6}}}+{\frac {1}{c^{6}}}={\frac {17}{7R^{6}}}.

In general for all integer n,

a^{2n}+b^{2n}+c^{2n}=g(n)(2R)^{2n}

where

g(-1)=8,\quad g(0)=3,\quad g(1)=7

and

g(n)=7g(n-1)-14g(n-2)+7g(n-3).

We also have[6]

2b^{2}-a^{2}={\sqrt {7}}bR,\quad 2c^{2}-b^{2}={\sqrt {7}}cR,\quad 2a^{2}-c^{2}=-{\sqrt {7}}aR.

We also have[4]

a^{3}c+b^{3}a-c^{3}b=-7R^{4},

a^{4}c-b^{4}a+c^{4}b=7{\sqrt {7}}R^{5},

a^{11}c^{3}+b^{11}a{3}-c^{11}b^{3}=-7^{3}17R^{14}.

The exradius r_a corresponding to side a equals the radius of the nine-point circle of the heptagonal triangle.[2]^{:p. 15}

Orthic triangle

The heptagonal triangle's orthic triangle, with vertices at the feet of the altitudes, is similar to the heptagonal triangle, with similarity ratio 1:2. The heptagonal triangle is the only obtuse triangle that is similar to its orthic triangle (the equilateral triangle being the only acute one).[2]^{:pp. 12–13}

Trigonometric properties

The various trigonometric identities associated with the heptagonal triangle include these:[2]^{:pp. 13–14}[5]

A={\frac {\pi }{7}},\quad B={\frac {2\pi }{7}},\quad C={\frac {4\pi }{7}}.

\cos A=b/2a,\quad \cos B=c/2b,\quad \cos C=-a/2c,

[4]^{:Proposition 10}

\cos A\cos B\cos C=-{\frac {1}{8}},

\cos ^{2}A+\cos ^{2}B+\cos ^{2}C={\frac {5}{4}},

\cos ^{4}A+\cos ^{4}B+\cos ^{4}C={\frac {13}{16}},

\cot A+\cot B+\cot C={\sqrt {7}},

\cot ^{2}A+\cot ^{2}B+\cot ^{2}C=5,

\csc ^{2}A+\csc ^{2}B+\csc ^{2}C=8,

\csc ^{4}A+\csc ^{4}B+\csc ^{4}C=32,

\sec ^{2}A+\sec ^{2}B+\sec ^{2}C=24,

\sec ^{4}A+\sec ^{4}B+\sec ^{4}C=416,

\sin A\sin B\sin C={\frac {\sqrt {7}}{8}},

\sin ^{2}A\sin ^{2}B\sin ^{2}C={\frac {7}{64}},

\sin ^{2}A+\sin ^{2}B+\sin ^{2}C={\frac {7}{4}},

\sin ^{4}A+\sin ^{4}B+\sin ^{4}C={\frac {21}{16}},

\tan A\tan B\tan C=\tan A+\tan B+\tan C=-{\sqrt {7}},

\tan ^{2}A+\tan ^{2}B+\tan ^{2}C=21.

The cubic equation

64y^{3}-112y^{2}+56y-7=0

has solutions[2]^{:p. 14} $\sin ^{2}{\frac {\pi }{7}},\sin ^{2}{\frac {2\pi }{7}},$ and $\sin ^{2}{\frac {4\pi }{7}},$ which are the squared sines of the angles of the triangle.

The positive solution of the cubic equation

x^{3}+x^{2}-2x-1=0

equals $2\cos {\frac {2\pi }{7}},$ which is twice the cosine of one of the triangle’s angles.[7]^{:p. 186–187}

Sin (2π / 7), sin (4π / 7), and sin (8π / 7) are the roots of[4]

x^{3}-{\frac {\sqrt {7}}{2}}x^{2}+{\frac {\sqrt {7}}{8}}=0.

We also have:[6]

\sin A-\sin B-\sin C=-{\frac {\sqrt {7}}{2}},

\sin A\sin B-\sin B\sin C+\sin C\sin A=0,

\sin A\sin B\sin C={\frac {\sqrt {7}}{8}}.

-\sin A,\sin B,\sin C{\text{ are the roots of }}x^{3}-{\frac {\sqrt {7}}{2}}x^{2}+{\frac {\sqrt {7}}{8}}=0.

For an integer n , let

S(n)=(-\sin {A})^{n}+\sin ^{n}{B}+\sin ^{n}{C}.

For n = 0,...,20,

S(n)=3,{\frac {\sqrt {7}}{2}},{\frac {7}{2^{2}}},{\frac {\sqrt {7}}{2}},{\frac {7\cdot 3}{2^{4}}},{\frac {7{\sqrt {7}}}{2^{4}}},{\frac {7\cdot 5}{2^{5}}},{\frac {7^{2}{\sqrt {7}}}{2^{7}}},{\frac {7^{2}\cdot 5}{2^{8}}},{\frac {7\cdot 25{\sqrt {7}}}{2^{9}}},{\frac {7^{2}\cdot 9}{2^{9}}},{\frac {7^{2}\cdot 13{\sqrt {7}}}{2^{11}}},

{\frac {7^{2}\cdot 33}{2^{11}}},{\frac {7^{2}\cdot 3{\sqrt {7}}}{2^{9}}},{\frac {7^{4}\cdot 5}{2^{14}}},{\frac {7^{2}\cdot 179{\sqrt {7}}}{2^{15}}},{\frac {7^{3}\cdot 131}{2^{16}}},{\frac {7^{3}\cdot 3{\sqrt {7}}}{2^{12}}},{\frac {7^{3}\cdot 493}{2^{18}}},{\frac {7^{3}\cdot 181{\sqrt {7}}}{2^{18}}},{\frac {7^{5}\cdot 19}{2^{19}}}.

For n= 0, -1, ,..-20,

S(n)=3,0,2^{3},-{\frac {2^{3}\cdot 3{\sqrt {7}}}{7}},2^{5},-{\frac {2^{5}\cdot 5{\sqrt {7}}}{7}},{\frac {2^{6}\cdot 17}{7}},-2^{7}{\sqrt {7}},{\frac {2^{9}\cdot 11}{7}},-{\frac {2^{10}\cdot 33{\sqrt {7}}}{7^{2}}},{\frac {2^{10}\cdot 29}{7}},-{\frac {2^{14}\cdot 11{\sqrt {7}}}{7^{2}}},{\frac {2^{12}\cdot 269}{7^{2}}},

-{\frac {2^{13}\cdot 117{\sqrt {7}}}{7^{2}}},{\frac {2^{14}\cdot 51}{7}},-{\frac {2^{21}\cdot 17{\sqrt {7}}}{7^{3}}},{\frac {2^{17}\cdot 237}{7^{2}}},-{\frac {2^{17}\cdot 1445{\sqrt {7}}}{7^{3}}},{\frac {2^{19}\cdot 2203}{7^{3}}},-{\frac {2^{19}\cdot 1919{\sqrt {7}}}{7^{3}}},{\frac {2^{20}\cdot 5851}{7^{3}}}.

-\cos A,\cos B,\cos C{\text{ are the roots of }}x^{3}+{\frac {1}{2}}x^{2}-{\frac {1}{2}}x-{\frac {1}{8}}=0.

For an integer n , let

C(n)=(-\cos {A})^{n}+\cos ^{n}{B}+\cos ^{n}{C}.

For n= 0, 1, ,..10,

C(n)=3,-{\frac {1}{2}},{\frac {5}{4}},-{\frac {1}{2}},{\frac {13}{16}},-{\frac {1}{2}},{\frac {19}{32}},-{\frac {57}{128}},{\frac {117}{256}},-{\frac {193}{512}},{\frac {185}{512}},...

C(-n)=3,-4,24,-88,416,-1824,8256,-36992,166400,-747520,3359744,...

\tan A,\tan B,\tan C{\text{ are the roots of }}x^{3}+{\sqrt {7}}x^{2}-7x+{\sqrt {7}}=0.

\tan ^{2}A,\tan ^{2}B,\tan ^{2}C{\text{ are the roots of }}x^{3}-21x^{2}+35x-7=0.

For an integer n , let

T(n)=\tan ^{n}{A}+\tan ^{n}{B}+\tan ^{n}{C}.

For n= 0, 1, ,..10,

T(n)=3,-{\sqrt {7}},7\cdot 3,-31{\sqrt {7}},7\cdot 53,-7\cdot 87{\sqrt {7}},7\cdot 1011,-7^{2}\cdot 239{\sqrt {7}},7^{2}\cdot 2771,-7\cdot 32119{\sqrt {7}},7^{2}\cdot 53189,

T(-n)=3,{\sqrt {7}},5,{\frac {25{\sqrt {7}}}{7}},19,{\frac {103{\sqrt {7}}}{7}},{\frac {563}{7}},7\cdot 9{\sqrt {7}},{\frac {2421}{7}},{\frac {13297{\sqrt {7}}}{7^{2}}},{\frac {10435}{7}},...

We also have[6][8]

\tan A-4\sin B=-{\sqrt {7}},

\tan B-4\sin C=-{\sqrt {7}},

\tan C+4\sin A=-{\sqrt {7}}.

We also have[4]

\cot ^{2}A=1-{\frac {2\tan C}{\sqrt {7}}},

\cot ^{2}B=1-{\frac {2\tan A}{\sqrt {7}}},

\cot ^{2}C=1-{\frac {2\tan B}{\sqrt {7}}}.

We also have[4]

\cos A=-{\frac {1}{2}}+{\frac {4}{\sqrt {7}}}\sin ^{3}C,

\cos ^{2}A={\frac {3}{4}}+{\frac {2}{\sqrt {7}}}\sin ^{3}A,

\cot A={\frac {3}{\sqrt {7}}}+{\frac {4}{\sqrt {7}}}\cos B,

\cot ^{2}A=3+{\frac {8}{\sqrt {7}}}\sin A,

\cot A={\sqrt {7}}+{\frac {8}{\sqrt {7}}}\sin ^{2}B,

\csc ^{3}A=-{\frac {6}{\sqrt {7}}}+{\frac {2}{\sqrt {7}}}\tan ^{2}C,

\sec A=2+4\cos C,

\sec A=6-8\sin ^{2}B,

\sec A=4-{\frac {16}{\sqrt {7}}}\sin ^{3}B,

\sin ^{2}A={\frac {1}{2}}+{\frac {1}{2}}\cos B,

\sin ^{3}A=-{\frac {\sqrt {7}}{8}}+{\frac {\sqrt {7}}{4}}\cos B,

We also have[9]

\sin ^{3}B\sin C-\sin ^{3}C\sin A-\sin ^{3}A\sin B=0,

\sin B\sin ^{3}C-\sin C\sin ^{3}A-\sin A\sin ^{3}B={\frac {7}{2^{4}}},

\sin ^{4}B\sin C-\sin ^{4}C\sin A+\sin ^{4}A\sin B=0,

\sin B\sin ^{4}C+\sin C\sin ^{4}A-\sin A\sin ^{4}B={\frac {7{\sqrt {7}}}{2^{5}}},

\sin ^{11}B\sin ^{3}C-\sin ^{11}C\sin ^{3}A-\sin ^{11}A\sin ^{3}B=0,

\sin ^{3}B\sin ^{11}C-\sin ^{3}C\sin ^{11}A-\sin ^{3}A\sin ^{11}B={\frac {7^{3}\cdot 17}{2^{14}}}.

We also have Ramanujan type identities,[10][11]

{\sqrt[{3}]{2\sin({\frac {2\pi }{7}})}}+{\sqrt[{3}]{2\sin({\frac {4\pi }{7}})}}+{\sqrt[{3}]{2\sin({\frac {8\pi }{7}})}}=

{\text{.......}}\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}

{\frac {1}{\sqrt[{3}]{2\sin({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\sin({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\sin({\frac {8\pi }{7}})}}}=

{\text{.......}}\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}

{\sqrt[{3}]{4\sin ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{4\sin ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{4\sin ^{2}({\frac {8\pi }{7}})}}=

{\text{.......}}\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}

{\frac {1}{\sqrt[{3}]{4\sin ^{2}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\sin ^{2}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\sin ^{2}({\frac {8\pi }{7}})}}}=

{\text{.......}}\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{2{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}

{\sqrt[{3}]{2\cos({\frac {2\pi }{7}})}}+{\sqrt[{3}]{2\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{2\cos({\frac {8\pi }{7}})}}={\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}

{\frac {1}{\sqrt[{3}]{2\cos({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\cos({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\cos({\frac {8\pi }{7}})}}}={\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}

{\sqrt[{3}]{4\cos ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{4\cos ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{4\cos ^{2}({\frac {8\pi }{7}})}}={\sqrt[{3}]{11+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}

{\frac {1}{\sqrt[{3}]{4\cos ^{2}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}({\frac {8\pi }{7}})}}}={\sqrt[{3}]{12+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}

{\sqrt[{3}]{\tan({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\tan({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\tan({\frac {8\pi }{7}})}}=

{\text{.......}}\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}

{\frac {1}{\sqrt[{3}]{\tan({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan({\frac {8\pi }{7}})}}}=

{\text{.......}}\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{-{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}

{\sqrt[{3}]{\tan ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\tan ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\tan ^{2}({\frac {8\pi }{7}})}}=

{\text{.......}}\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{3{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}

{\frac {1}{\sqrt[{3}]{\tan ^{2}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}({\frac {8\pi }{7}})}}}=

{\text{.......}}\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{5{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}

We also have[9]

{\sqrt[{3}]{\cos({\frac {2\pi }{7}})/\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {4\pi }{7}})/\cos({\frac {8\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {8\pi }{7}})/\cos({\frac {2\pi }{7}})}}=-{\sqrt[{3}]{7}}.

{\sqrt[{3}]{\cos({\frac {4\pi }{7}})/\cos({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {8\pi }{7}})/\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {2\pi }{7}})/\cos({\frac {8\pi }{7}})}}=0.

{\sqrt[{3}]{2\sin({2\pi }{7}}}+{\sqrt[{3}]{2\sin({4\pi }{7}}}+{\sqrt[{3}]{2\sin({8\pi }{7}}}=\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}

{\sqrt[{3}]{\cos ^{4}({\frac {4\pi }{7}})/\cos({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos ^{4}({\frac {8\pi }{7}})/\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{4}({\frac {2\pi }{7}})/\cos({\frac {8\pi }{7}})}}=-{\sqrt[{3}]{49}}/2.

{\sqrt[{3}]{\cos ^{5}({\frac {2\pi }{7}})/\cos ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {4\pi }{7}})/\cos ^{2}({\frac {8\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {8\pi }{7}})/\cos ^{2}({\frac {2\pi }{7}})}}=0.

{\sqrt[{3}]{\cos ^{5}({\frac {4\pi }{7}})/\cos ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {8\pi }{7}})/\cos ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {2\pi }{7}})/\cos ^{2}({\frac {9\pi }{7}})}}=-3*{\sqrt[{3}]{7}}/2.

{\sqrt[{3}]{\cos ^{14}({\frac {2\pi }{7}})/\cos ^{5}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {4\pi }{7}})/\cos ^{5}({\frac {8\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {8\pi }{7}})/\cos ^{5}({\frac {2\pi }{7}}}}=0.

{\sqrt[{3}]{\cos ^{14}({\frac {4\pi }{7}})/\cos ^{5}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {8\pi }{7}})/\cos ^{5}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {2\pi }{7}})/\cos ^{5}({\frac {8\pi }{7}})}}=-61*{\sqrt[{3}]{7}}/8.

Paul Yiu, "Heptagonal Triangles and Their Companions", Forum Geometricorum 9, 2009, 125–148. http://forumgeom.fau.edu/FG2009volume9/FG200912.pdf
Leon Bankoff and Jack Garfunkel, "The heptagonal triangle", Mathematics Magazine 46 (1), January 1973, 7–19.
Abdilkadir Altintas, "Some Collinearities in the Heptagonal Triangle", Forum Geometricorum 16, 2016, 249–256.http://forumgeom.fau.edu/FG2016volume16/FG201630.pdf
Wang, Kai. “Heptagonal Triangle and Trigonometric Identities”, Forum Geometricorum 19, 2019, 29–38.
Weisstein, Eric W. "Heptagonal Triangle." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/HeptagonalTriangle.html
Wang, Kai. https://www.researchgate.net/publication/327825153_Trigonometric_Properties_For_Heptagonal_Triangle
Gleason, Andrew Mattei (March 1988). "Angle trisection, the heptagon, and the triskaidecagon" (PDF). The American Mathematical Monthly. 95 (3): 185–194. doi:10.2307/2323624. Archived from the original (PDF) on 2015-12-19.

Kai Wang, "Heptagonal Triangle and Trigonometric Identities", Forum Geometricorum 19, 2019, 29-38. http://forumgeom.fau.edu/FG2019volume19/FG201904.pdf
Kai Wang, https://www.researchgate.net/publication/335392159_On_cubic_equations_with_zero_sums_of_cubic_roots_of_roots
Kai Wang, https://www.researchgate.net/publication/336813631_Topics_of_Ramanujan_type_identities_for_PI7
Victor H. Moll, An elementary trigonometric equation, https://arxiv.org/abs/0709.3755, 2007
Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007).

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[Yiu-1] Paul Yiu, "Heptagonal Triangles and Their Companions", Forum Geometricorum 9, 2009, 125–148. http://forumgeom.fau.edu/FG2009volume9/FG200912.pdf

[BG-2] Leon Bankoff and Jack Garfunkel, "The heptagonal triangle", Mathematics Magazine 46 (1), January 1973, 7–19.

[Altintas-3] Abdilkadir Altintas, "Some Collinearities in the Heptagonal Triangle", Forum Geometricorum 16, 2016, 249–256.http://forumgeom.fau.edu/FG2016volume16/FG201630.pdf

[Wang2-4] Wang, Kai. “Heptagonal Triangle and Trigonometric Identities”, Forum Geometricorum 19, 2019, 29–38.

[Weisstein-5] Weisstein, Eric W. "Heptagonal Triangle." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/HeptagonalTriangle.html

[Wang1-6] Wang, Kai. https://www.researchgate.net/publication/327825153_Trigonometric_Properties_For_Heptagonal_Triangle

[Gleason-7] Gleason, Andrew Mattei (March 1988). "Angle trisection, the heptagon, and the triskaidecagon" (PDF). The American Mathematical Monthly. 95 (3): 185–194. doi:10.2307/2323624. Archived from the original (PDF) on 2015-12-19.

[Wang1-12] Kai Wang, "Heptagonal Triangle and Trigonometric Identities", Forum Geometricorum 19, 2019, 29-38. http://forumgeom.fau.edu/FG2019volume19/FG201904.pdf

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