In Euclidean geometry , 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
π
/
7
,
2
π
/
7
,
{\displaystyle \pi /7,2\pi /7,}
4
π
/
7
,
{\displaystyle 4\pi /7,}
Shorter diagonals
Each of the fourteen congruent heptagonal triangles has one green side, one blue side, and one red side. Relations of distances
edit
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
a
2
=
c
(
c
−
b
)
,
b
2
=
a
(
c
+
a
)
,
c
2
=
b
(
a
+
b
)
,
1
a
=
1
b
+
1
c
{\displaystyle {\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 optic equation ) and hence
a
b
+
a
c
=
b
c
,
{\displaystyle ab+ac=bc,}
and[ 3] : Coro. 2
b
3
+
2
b
2
c
−
b
c
2
−
c
3
=
0
,
{\displaystyle b^{3}+2b^{2}c-bc^{2}-c^{3}=0,}
c
3
−
2
c
2
a
−
c
a
2
+
a
3
=
0
,
{\displaystyle c^{3}-2c^{2}a-ca^{2}+a^{3}=0,}
a
3
−
2
a
2
b
−
a
b
2
+
b
3
=
0.
{\displaystyle 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
−
2
t
2
−
t
+
1
=
0.
{\displaystyle 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
≈
1.80193
⋅
a
,
c
≈
2.24698
⋅
a
.
{\displaystyle b\approx 1.80193\cdot a,\qquad c\approx 2.24698\cdot a.}
We also have[ 4] [ 5]
a
2
b
c
,
−
b
2
c
a
,
−
c
2
a
b
{\displaystyle {\frac {a^{2}}{bc}},\quad -{\frac {b^{2}}{ca}},\quad -{\frac {c^{2}}{ab}}}
satisfy the cubic equation
t
3
+
4
t
2
+
3
t
−
1
=
0.
{\displaystyle t^{3}+4t^{2}+3t-1=0.}
We also have[ 4]
a
3
b
c
2
,
−
b
3
c
a
2
,
c
3
a
b
2
{\displaystyle {\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
−
9
t
+
1
=
0.
{\displaystyle t^{3}-t^{2}-9t+1=0.}
We also have[ 4]
a
3
b
2
c
,
b
3
c
2
a
,
−
c
3
a
2
b
{\displaystyle {\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
+
5
t
2
−
8
t
+
1
=
0.
{\displaystyle t^{3}+5t^{2}-8t+1=0.}
We also have[ 2] : p. 14
b
2
−
a
2
=
a
c
,
{\displaystyle b^{2}-a^{2}=ac,}
c
2
−
b
2
=
a
b
,
{\displaystyle c^{2}-b^{2}=ab,}
a
2
−
c
2
=
−
b
c
,
{\displaystyle a^{2}-c^{2}=-bc,}
and[ 2] : p. 15
b
2
a
2
+
c
2
b
2
+
a
2
c
2
=
5.
{\displaystyle {\frac {b^{2}}{a^{2}}}+{\frac {c^{2}}{b^{2}}}+{\frac {a^{2}}{c^{2}}}=5.}
We also have[ 4]
a
b
−
b
c
+
c
a
=
0
,
{\displaystyle ab-bc+ca=0,}
a
3
b
−
b
3
c
+
c
3
a
=
0
,
{\displaystyle a^{3}b-b^{3}c+c^{3}a=0,}
a
4
b
+
b
4
c
−
c
4
a
=
0
,
{\displaystyle a^{4}b+b^{4}c-c^{4}a=0,}
a
11
b
3
−
b
11
c
3
+
c
11
a
3
=
0.
{\displaystyle a^{11}b^{3}-b^{11}c^{3}+c^{11}a^{3}=0.}
The altitudes h a h b h c
h
a
=
h
b
+
h
c
{\displaystyle h_{a}=h_{b}+h_{c}}
[ 2] : p. 13 and
h
a
2
+
h
b
2
+
h
c
2
=
a
2
+
b
2
+
c
2
2
.
{\displaystyle 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
{\displaystyle w_{A}}
A :[ 2] : p. 19
2
h
b
=
w
A
.
{\displaystyle 2h_{b}=w_{A}.}
Here angle A is the smallest angle, and B is the second smallest.
Internal angle bisectors
edit
We have these properties of the internal angle bisectors
w
A
,
w
B
,
{\displaystyle w_{A},w_{B},}
w
C
{\displaystyle w_{C}}
A, B , and C respectively:[ 2] : p. 16
w
A
=
b
+
c
,
{\displaystyle w_{A}=b+c,}
w
B
=
c
−
a
,
{\displaystyle w_{B}=c-a,}
w
C
=
b
−
a
.
{\displaystyle w_{C}=b-a.}
edit
The triangle's area is[ 6]
A
=
7
4
R
2
,
{\displaystyle 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
=
7
R
2
.
{\displaystyle a^{2}+b^{2}+c^{2}=7R^{2}.}
We also have[ 7]
a
4
+
b
4
+
c
4
=
21
R
4
.
{\displaystyle a^{4}+b^{4}+c^{4}=21R^{4}.}
a
6
+
b
6
+
c
6
=
70
R
6
.
{\displaystyle 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[ 6]
8
x
3
+
28
x
2
+
14
x
−
7
=
0.
{\displaystyle 8x^{3}+28x^{2}+14x-7=0.}
In addition,[ 2] : p. 15
1
a
2
+
1
b
2
+
1
c
2
=
2
R
2
.
{\displaystyle {\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}+{\frac {1}{c^{2}}}={\frac {2}{R^{2}}}.}
We also have[ 7]
1
a
4
+
1
b
4
+
1
c
4
=
2
R
4
.
{\displaystyle {\frac {1}{a^{4}}}+{\frac {1}{b^{4}}}+{\frac {1}{c^{4}}}={\frac {2}{R^{4}}}.}
1
a
6
+
1
b
6
+
1
c
6
=
17
7
R
6
.
{\displaystyle {\frac {1}{a^{6}}}+{\frac {1}{b^{6}}}+{\frac {1}{c^{6}}}={\frac {17}{7R^{6}}}.}
In general for all integer n ,
a
2
n
+
b
2
n
+
c
2
n
=
g
(
n
)
(
2
R
)
2
n
{\displaystyle a^{2n}+b^{2n}+c^{2n}=g(n)(2R)^{2n}}
where
g
(
−
1
)
=
8
,
g
(
0
)
=
3
,
g
(
1
)
=
7
{\displaystyle g(-1)=8,\quad g(0)=3,\quad g(1)=7}
and
g
(
n
)
=
7
g
(
n
−
1
)
−
14
g
(
n
−
2
)
+
7
g
(
n
−
3
)
.
{\displaystyle g(n)=7g(n-1)-14g(n-2)+7g(n-3).}
We also have[ 7]
2
b
2
−
a
2
=
7
b
R
,
2
c
2
−
b
2
=
7
c
R
,
2
a
2
−
c
2
=
−
7
a
R
.
{\displaystyle 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
=
−
7
R
4
,
{\displaystyle a^{3}c+b^{3}a-c^{3}b=-7R^{4},}
a
4
c
−
b
4
a
+
c
4
b
=
7
7
R
5
,
{\displaystyle 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
17
R
14
.
{\displaystyle a^{11}c^{3}+b^{11}a^{3}-c^{11}b^{3}=-7^{3}17R^{14}.}
The exradius r a a equals the radius of the nine-point circle of the heptagonal triangle.[ 2] : p. 15
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
The rectangular hyperbola through
A
,
B
,
C
,
G
=
X
(
2
)
,
H
=
X
(
4
)
{\displaystyle A,B,C,G=X(2),H=X(4)}
first focus
F
1
=
X
(
5
)
{\displaystyle F_{1}=X(5)}
center
U
{\displaystyle U}
(
O
,
F
1
)
{\displaystyle (O,F_{1})}
second focus
F
2
{\displaystyle F_{2}}
Trigonometric properties
edit
Trigonometric identities
edit
The various trigonometric identities associated with the heptagonal triangle include these:[ 2] : pp. 13–14 [ 6] [ 7]
A
=
π
7
cos
A
=
b
2
a
B
=
2
π
7
cos
B
=
c
2
b
C
=
4
π
7
cos
C
=
−
a
2
c
{\displaystyle {\begin{aligned}A&={\frac {\pi }{7}}\\[6pt]\cos A&={\frac {b}{2a}}\end{aligned}}\quad {\begin{aligned}B&={\frac {2\pi }{7}}\\[6pt]\cos B&={\frac {c}{2b}}\end{aligned}}\quad {\begin{aligned}C&={\frac {4\pi }{7}}\\[6pt]\cos C&=-{\frac {a}{2c}}\end{aligned}}}
[ 4] : Proposition 10
sin
A
×
sin
B
×
sin
C
=
7
8
sin
A
−
sin
B
−
sin
C
=
−
7
2
cos
A
×
cos
B
×
cos
C
=
−
1
8
tan
A
×
tan
B
×
tan
C
=
−
7
tan
A
+
tan
B
+
tan
C
=
−
7
cot
A
+
cot
B
+
cot
C
=
7
sin
2
A
×
sin
2
B
×
sin
2
C
=
7
64
sin
2
A
+
sin
2
B
+
sin
2
C
=
7
4
cos
2
A
+
cos
2
B
+
cos
2
C
=
5
4
tan
2
A
+
tan
2
B
+
tan
2
C
=
21
sec
2
A
+
sec
2
B
+
sec
2
C
=
24
csc
2
A
+
csc
2
B
+
csc
2
C
=
8
cot
2
A
+
cot
2
B
+
cot
2
C
=
5
sin
4
A
+
sin
4
B
+
sin
4
C
=
21
16
cos
4
A
+
cos
4
B
+
cos
4
C
=
13
16
sec
4
A
+
sec
4
B
+
sec
4
C
=
416
csc
4
A
+
csc
4
B
+
csc
4
C
=
32
{\displaystyle {\begin{array}{rcccccl}\sin A\!&\!\times \!&\!\sin B\!&\!\times \!&\!\sin C\!&\!=\!&\!{\frac {\sqrt {7}}{8}}\\[2pt]\sin A\!&\!-\!&\!\sin B\!&\!-\!&\!\sin C\!&\!=\!&\!-{\frac {\sqrt {7}}{2}}\\[2pt]\cos A\!&\!\times \!&\!\cos B\!&\!\times \!&\!\cos C\!&\!=\!&\!-{\frac {1}{8}}\\[2pt]\tan A\!&\!\times \!&\!\tan B\!&\!\times \!&\!\tan C\!&\!=\!&\!-{\sqrt {7}}\\[2pt]\tan A\!&\!+\!&\!\tan B\!&\!+\!&\!\tan C\!&\!=\!&\!-{\sqrt {7}}\\[2pt]\cot A\!&\!+\!&\!\cot B\!&\!+\!&\!\cot C\!&\!=\!&\!{\sqrt {7}}\\[8pt]\sin ^{2}\!A\!&\!\times \!&\!\sin ^{2}\!B\!&\!\times \!&\!\sin ^{2}\!C\!&\!=\!&\!{\frac {7}{64}}\\[2pt]\sin ^{2}\!A\!&\!+\!&\!\sin ^{2}\!B\!&\!+\!&\!\sin ^{2}\!C\!&\!=\!&\!{\frac {7}{4}}\\[2pt]\cos ^{2}\!A\!&\!+\!&\!\cos ^{2}\!B\!&\!+\!&\!\cos ^{2}\!C\!&\!=\!&\!{\frac {5}{4}}\\[2pt]\tan ^{2}\!A\!&\!+\!&\!\tan ^{2}\!B\!&\!+\!&\!\tan ^{2}\!C\!&\!=\!&\!21\\[2pt]\sec ^{2}\!A\!&\!+\!&\!\sec ^{2}\!B\!&\!+\!&\!\sec ^{2}\!C\!&\!=\!&\!24\\[2pt]\csc ^{2}\!A\!&\!+\!&\!\csc ^{2}\!B\!&\!+\!&\!\csc ^{2}\!C\!&\!=\!&\!8\\[2pt]\cot ^{2}\!A\!&\!+\!&\!\cot ^{2}\!B\!&\!+\!&\!\cot ^{2}\!C\!&\!=\!&\!5\\[8pt]\sin ^{4}\!A\!&\!+\!&\!\sin ^{4}\!B\!&\!+\!&\!\sin ^{4}\!C\!&\!=\!&\!{\frac {21}{16}}\\[2pt]\cos ^{4}\!A\!&\!+\!&\!\cos ^{4}\!B\!&\!+\!&\!\cos ^{4}\!C\!&\!=\!&\!{\frac {13}{16}}\\[2pt]\sec ^{4}\!A\!&\!+\!&\!\sec ^{4}\!B\!&\!+\!&\!\sec ^{4}\!C\!&\!=\!&\!416\\[2pt]\csc ^{4}\!A\!&\!+\!&\!\csc ^{4}\!B\!&\!+\!&\!\csc ^{4}\!C\!&\!=\!&\!32\\[8pt]\end{array}}}
tan
A
−
4
sin
B
=
−
7
tan
B
−
4
sin
C
=
−
7
tan
C
+
4
sin
A
=
−
7
{\displaystyle {\begin{array}{ccccl}\tan A\!&\!-\!&\!4\sin B\!&\!=\!&\!-{\sqrt {7}}\\[2pt]\tan B\!&\!-\!&\!4\sin C\!&\!=\!&\!-{\sqrt {7}}\\[2pt]\tan C\!&\!+\!&\!4\sin A\!&\!=\!&\!-{\sqrt {7}}\end{array}}}
[ 7] [ 8]
cot
2
A
=
1
−
2
tan
C
7
cot
2
B
=
1
−
2
tan
A
7
cot
2
C
=
1
−
2
tan
B
7
{\displaystyle {\begin{aligned}\cot ^{2}\!A&=1-{\frac {2\tan C}{\sqrt {7}}}\\[2pt]\cot ^{2}\!B&=1-{\frac {2\tan A}{\sqrt {7}}}\\[2pt]\cot ^{2}\!C&=1-{\frac {2\tan B}{\sqrt {7}}}\end{aligned}}}
[ 4]
cos
A
=
−
1
2
+
4
7
×
sin
3
C
sec
A
=
2
+
4
×
cos
C
sec
A
=
6
−
8
×
sin
2
B
sec
A
=
4
−
16
7
×
sin
3
B
cot
A
=
7
+
8
7
×
sin
2
B
cot
A
=
3
7
+
4
7
×
cos
B
sin
2
A
=
1
2
+
1
2
×
cos
B
cos
2
A
=
3
4
+
2
7
×
sin
3
A
cot
2
A
=
3
+
8
7
×
sin
A
sin
3
A
=
−
7
8
+
7
4
×
cos
B
csc
3
A
=
−
6
7
+
2
7
×
tan
2
C
{\displaystyle {\begin{array}{rcccccl}\cos A\!&\!=\!&\!{\frac {-1}{2}}\!&\!+\!&\!{\frac {4}{\sqrt {7}}}\!&\!\times \!&\!\sin ^{3}\!C\\[2pt]\sec A\!&\!=\!&\!2\!&\!+\!&\!4\!&\!\times \!&\!\cos C\\[4pt]\sec A\!&\!=\!&\!6\!&\!-\!&\!8\!&\!\times \!&\!\sin ^{2}\!B\\[4pt]\sec A\!&\!=\!&\!4\!&\!-\!&\!{\frac {16}{\sqrt {7}}}\!&\!\times \!&\!\sin ^{3}\!B\\[2pt]\cot A\!&\!=\!&\!{\sqrt {7}}\!&\!+\!&\!{\frac {8}{\sqrt {7}}}\!&\!\times \!&\!\sin ^{2}\!B\\[2pt]\cot A\!&\!=\!&\!{\frac {3}{\sqrt {7}}}\!&\!+\!&\!{\frac {4}{\sqrt {7}}}\!&\!\times \!&\!\cos B\\[2pt]\sin ^{2}\!A\!&\!=\!&\!{\frac {1}{2}}\!&\!+\!&\!{\frac {1}{2}}\!&\!\times \!&\!\cos B\\[2pt]\cos ^{2}\!A\!&\!=\!&\!{\frac {3}{4}}\!&\!+\!&\!{\frac {2}{\sqrt {7}}}\!&\!\times \!&\!\sin ^{3}\!A\\[2pt]\cot ^{2}\!A\!&\!=\!&\!3\!&\!+\!&\!{\frac {8}{\sqrt {7}}}\!&\!\times \!&\!\sin A\\[2pt]\sin ^{3}\!A\!&\!=\!&\!{\frac {-{\sqrt {7}}}{8}}\!&\!+\!&\!{\frac {\sqrt {7}}{4}}\!&\!\times \!&\!\cos B\\[2pt]\csc ^{3}\!A\!&\!=\!&\!{\frac {-6}{\sqrt {7}}}\!&\!+\!&\!{\frac {2}{\sqrt {7}}}\!&\!\times \!&\!\tan ^{2}\!C\end{array}}}
[ 4]
sin
A
sin
B
−
sin
B
sin
C
+
sin
C
sin
A
=
0
{\displaystyle \sin A\sin B-\sin B\sin C+\sin C\sin A=0}
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
=
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
=
7
7
2
5
{\displaystyle {\begin{aligned}\sin ^{3}\!B\sin C-\sin ^{3}\!C\sin A-\sin ^{3}\!A\sin B&=0\\[3pt]\sin B\sin ^{3}\!C-\sin C\sin ^{3}\!A-\sin A\sin ^{3}\!B&={\frac {7}{2^{4}\!}}\\[2pt]\sin ^{4}\!B\sin C-\sin ^{4}\!C\sin A+\sin ^{4}\!A\sin B&=0\\[2pt]\sin B\sin ^{4}\!C+\sin C\sin ^{4}\!A-\sin A\sin ^{4}\!B&={\frac {7{\sqrt {7}}}{2^{5}}}\end{aligned}}}
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
=
7
3
⋅
17
2
14
{\displaystyle {\begin{aligned}\sin ^{11}\!B\sin ^{3}\!C-\sin ^{11}\!C\sin ^{3}\!A-\sin ^{11}\!A\sin ^{3}\!B&=0\\[2pt]\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}}}\end{aligned}}}
[ 9]
The cubic equation
64
y
3
−
112
y
2
+
56
y
−
7
=
0
{\displaystyle 64y^{3}-112y^{2}+56y-7=0}
[ 2] : p. 14
sin
2
A
,
sin
2
B
,
sin
2
C
.
{\displaystyle \sin ^{2}\!A,\ \sin ^{2}\!B,\ \sin ^{2}\!C.}
The positive solution of the cubic equation
x
3
+
x
2
−
2
x
−
1
=
0
{\displaystyle x^{3}+x^{2}-2x-1=0}
2
cos
B
.
{\displaystyle 2\cos B.}
[ 10] : p. 186–187
The roots of the cubic equation
x
3
−
7
2
x
2
+
7
8
=
0
{\displaystyle x^{3}-{\tfrac {\sqrt {7}}{2}}x^{2}+{\tfrac {\sqrt {7}}{8}}=0}
[ 4]
sin
2
A
,
sin
2
B
,
sin
2
C
.
{\displaystyle \sin 2A,\ \sin 2B,\ \sin 2C.}
The roots of the cubic equation
x
3
−
7
2
x
2
+
7
8
=
0
{\displaystyle x^{3}-{\tfrac {\sqrt {7}}{2}}x^{2}+{\tfrac {\sqrt {7}}{8}}=0}
−
sin
A
,
sin
B
,
sin
C
.
{\displaystyle -\sin A,\ \sin B,\ \sin C.}
The roots of the cubic equation
x
3
+
1
2
x
2
−
1
2
x
−
1
8
=
0
{\displaystyle x^{3}+{\tfrac {1}{2}}x^{2}-{\tfrac {1}{2}}x-{\tfrac {1}{8}}=0}
−
cos
A
,
cos
B
,
cos
C
.
{\displaystyle -\cos A,\ \cos B,\ \cos C.}
The roots of the cubic equation
x
3
+
7
x
2
−
7
x
+
7
=
0
{\displaystyle x^{3}+{\sqrt {7}}x^{2}-7x+{\sqrt {7}}=0}
tan
A
,
tan
B
,
tan
C
.
{\displaystyle \tan A,\ \tan B,\ \tan C.}
The roots of the cubic equation
x
3
−
21
x
2
+
35
x
−
7
=
0
{\displaystyle x^{3}-21x^{2}+35x-7=0}
tan
2
A
,
tan
2
B
,
tan
2
C
.
{\displaystyle \tan ^{2}\!A,\ \tan ^{2}\!B,\ \tan ^{2}\!C.}
For an integer n , let
S
(
n
)
=
(
−
sin
A
)
n
+
sin
n
B
+
sin
n
C
C
(
n
)
=
(
−
cos
A
)
n
+
cos
n
B
+
cos
n
C
T
(
n
)
=
tan
n
A
+
tan
n
B
+
tan
n
C
{\displaystyle {\begin{aligned}S(n)&=(-\sin A)^{n}+\sin ^{n}\!B+\sin ^{n}\!C\\[4pt]C(n)&=(-\cos A)^{n}+\cos ^{n}\!B+\cos ^{n}\!C\\[4pt]T(n)&=\tan ^{n}\!A+\tan ^{n}\!B+\tan ^{n}\!C\end{aligned}}}
Value of n :
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
S
(
n
)
{\displaystyle S(n)}
3
{\displaystyle \ 3\ }
7
2
{\displaystyle {\tfrac {\sqrt {7}}{2}}}
7
2
2
{\displaystyle {\tfrac {7}{2^{2}}}}
7
2
{\displaystyle {\tfrac {\sqrt {7}}{2}}}
7
⋅
3
2
4
{\displaystyle {\tfrac {7\cdot 3}{2^{4}}}}
7
7
2
4
{\displaystyle {\tfrac {7{\sqrt {7}}}{2^{4}}}}
7
⋅
5
2
5
{\displaystyle {\tfrac {7\cdot 5}{2^{5}}}}
7
2
7
2
7
{\displaystyle {\tfrac {7^{2}{\sqrt {7}}}{2^{7}}}}
7
2
⋅
5
2
8
{\displaystyle {\tfrac {7^{2}\cdot 5}{2^{8}}}}
7
⋅
25
7
2
9
{\displaystyle {\tfrac {7\cdot 25{\sqrt {7}}}{2^{9}}}}
7
2
⋅
9
2
9
{\displaystyle {\tfrac {7^{2}\cdot 9}{2^{9}}}}
7
2
⋅
13
7
2
11
{\displaystyle {\tfrac {7^{2}\cdot 13{\sqrt {7}}}{2^{11}}}}
7
2
⋅
33
2
11
{\displaystyle {\tfrac {7^{2}\cdot 33}{2^{11}}}}
7
2
⋅
3
7
2
9
{\displaystyle {\tfrac {7^{2}\cdot 3{\sqrt {7}}}{2^{9}}}}
7
4
⋅
5
2
14
{\displaystyle {\tfrac {7^{4}\cdot 5}{2^{14}}}}
7
2
⋅
179
7
2
15
{\displaystyle {\tfrac {7^{2}\cdot 179{\sqrt {7}}}{2^{15}}}}
7
3
⋅
131
2
16
{\displaystyle {\tfrac {7^{3}\cdot 131}{2^{16}}}}
7
3
⋅
3
7
2
12
{\displaystyle {\tfrac {7^{3}\cdot 3{\sqrt {7}}}{2^{12}}}}
7
3
⋅
493
2
18
{\displaystyle {\tfrac {7^{3}\cdot 493}{2^{18}}}}
7
3
⋅
181
7
2
18
{\displaystyle {\tfrac {7^{3}\cdot 181{\sqrt {7}}}{2^{18}}}}
7
5
⋅
19
2
19
{\displaystyle {\tfrac {7^{5}\cdot 19}{2^{19}}}}
S
(
−
n
)
{\displaystyle S(-n)}
3
{\displaystyle 3}
0
{\displaystyle 0}
2
3
{\displaystyle 2^{3}}
−
2
3
⋅
3
7
7
{\displaystyle -{\tfrac {2^{3}\cdot 3{\sqrt {7}}}{7}}}
2
5
{\displaystyle 2^{5}}
−
2
5
⋅
5
7
7
{\displaystyle -{\tfrac {2^{5}\cdot 5{\sqrt {7}}}{7}}}
2
6
⋅
17
7
{\displaystyle {\tfrac {2^{6}\cdot 17}{7}}}
−
2
7
7
{\displaystyle -2^{7}{\sqrt {7}}}
2
9
⋅
11
7
{\displaystyle {\tfrac {2^{9}\cdot 11}{7}}}
−
2
10
⋅
33
7
7
2
{\displaystyle -{\tfrac {2^{10}\cdot 33{\sqrt {7}}}{7^{2}}}}
2
10
⋅
29
7
{\displaystyle {\tfrac {2^{10}\cdot 29}{7}}}
−
2
14
⋅
11
7
7
2
{\displaystyle -{\tfrac {2^{14}\cdot 11{\sqrt {7}}}{7^{2}}}}
2
12
⋅
269
7
2
{\displaystyle {\tfrac {2^{12}\cdot 269}{7^{2}}}}
−
2
13
⋅
117
7
7
2
{\displaystyle -{\tfrac {2^{13}\cdot 117{\sqrt {7}}}{7^{2}}}}
2
14
⋅
51
7
{\displaystyle {\tfrac {2^{14}\cdot 51}{7}}}
−
2
21
⋅
17
7
7
3
{\displaystyle -{\tfrac {2^{21}\cdot 17{\sqrt {7}}}{7^{3}}}}
2
17
⋅
237
7
2
{\displaystyle {\tfrac {2^{17}\cdot 237}{7^{2}}}}
−
2
17
⋅
1445
7
7
3
{\displaystyle -{\tfrac {2^{17}\cdot 1445{\sqrt {7}}}{7^{3}}}}
2
19
⋅
2203
7
3
{\displaystyle {\tfrac {2^{19}\cdot 2203}{7^{3}}}}
−
2
19
⋅
1919
7
7
3
{\displaystyle -{\tfrac {2^{19}\cdot 1919{\sqrt {7}}}{7^{3}}}}
2
20
⋅
5851
7
3
{\displaystyle {\tfrac {2^{20}\cdot 5851}{7^{3}}}}
C
(
n
)
{\displaystyle C(n)}
3
{\displaystyle 3}
−
1
2
{\displaystyle -{\tfrac {1}{2}}}
5
4
{\displaystyle {\tfrac {5}{4}}}
−
1
2
{\displaystyle -{\tfrac {1}{2}}}
13
16
{\displaystyle {\tfrac {13}{16}}}
−
1
2
{\displaystyle -{\tfrac {1}{2}}}
19
32
{\displaystyle {\tfrac {19}{32}}}
−
57
128
{\displaystyle -{\tfrac {57}{128}}}
117
256
{\displaystyle {\tfrac {117}{256}}}
−
193
512
{\displaystyle -{\tfrac {193}{512}}}
185
512
{\displaystyle {\tfrac {185}{512}}}
C
(
−
n
)
{\displaystyle C(-n)}
3
{\displaystyle 3}
−
4
{\displaystyle -4}
24
{\displaystyle 24}
−
88
{\displaystyle -88}
416
{\displaystyle 416}
−
1824
{\displaystyle -1824}
8256
{\displaystyle 8256}
−
36992
{\displaystyle -36992}
166400
{\displaystyle 166400}
−
747520
{\displaystyle -747520}
3359744
{\displaystyle 3359744}
T
(
n
)
{\displaystyle T(n)}
3
{\displaystyle 3}
−
7
{\displaystyle -{\sqrt {7}}}
7
⋅
3
{\displaystyle 7\cdot 3}
−
31
7
{\displaystyle -31{\sqrt {7}}}
7
⋅
53
{\displaystyle 7\cdot 53}
−
7
⋅
87
7
{\displaystyle -7\cdot 87{\sqrt {7}}}
7
⋅
1011
{\displaystyle 7\cdot 1011}
−
7
2
⋅
239
7
{\displaystyle -7^{2}\cdot 239{\sqrt {7}}}
7
2
⋅
2771
{\displaystyle 7^{2}\cdot 2771}
−
7
⋅
32119
7
{\displaystyle -7\cdot 32119{\sqrt {7}}}
7
2
⋅
53189
{\displaystyle 7^{2}\cdot 53189}
T
(
−
n
)
{\displaystyle T(-n)}
3
{\displaystyle 3}
7
{\displaystyle {\sqrt {7}}}
5
{\displaystyle 5}
25
7
7
{\displaystyle {\tfrac {25{\sqrt {7}}}{7}}}
19
{\displaystyle 19}
103
7
7
{\displaystyle {\tfrac {103{\sqrt {7}}}{7}}}
563
7
{\displaystyle {\tfrac {563}{7}}}
7
⋅
9
7
{\displaystyle 7\cdot 9{\sqrt {7}}}
2421
7
{\displaystyle {\tfrac {2421}{7}}}
13297
7
7
2
{\displaystyle {\tfrac {13297{\sqrt {7}}}{7^{2}}}}
10435
7
{\displaystyle {\tfrac {10435}{7}}}
Ramanujan identities
edit
We also have Ramanujan type identities,[ 7] [ 11]
2
sin
2
A
3
+
2
sin
2
B
3
+
2
sin
2
C
3
=
−
7
18
×
−
7
3
+
6
+
3
(
5
−
3
7
3
3
+
4
−
3
7
3
3
)
3
2
sin
2
A
3
+
2
sin
2
B
3
+
2
sin
2
C
3
=
−
7
18
×
−
7
3
+
6
+
3
(
5
−
3
7
3
3
+
4
−
3
7
3
3
)
3
4
sin
2
2
A
3
+
4
sin
2
2
B
3
+
4
sin
2
2
C
3
=
49
18
×
49
3
+
6
+
3
(
12
+
3
(
49
3
+
2
7
3
)
3
+
11
+
3
(
49
3
+
2
7
3
)
3
)
3
2
cos
2
A
3
+
2
cos
2
B
3
+
2
cos
2
C
3
=
5
−
3
7
3
3
4
cos
2
2
A
3
+
4
cos
2
2
B
3
+
4
cos
2
2
C
3
=
11
+
3
(
2
7
3
+
49
3
)
3
tan
2
A
3
+
tan
2
B
3
+
tan
2
C
3
=
−
7
18
×
7
3
+
6
+
3
(
5
+
3
(
7
3
−
49
3
)
3
+
−
3
+
3
(
7
3
−
49
3
)
3
)
3
tan
2
2
A
3
+
tan
2
2
B
3
+
tan
2
2
C
3
=
49
18
×
3
49
3
+
6
+
3
(
89
+
3
(
3
49
3
+
5
7
3
)
3
+
25
+
3
(
3
49
3
+
5
7
3
)
3
)
3
{\displaystyle {\begin{array}{ccccccl}{\sqrt[{3}]{2\sin 2A}}\!&\!+\!&\!{\sqrt[{3}]{2\sin 2B}}\!&\!+\!&\!{\sqrt[{3}]{2\sin 2C}}\!&\!=\!&\!-{\sqrt[{18}]{7}}\times {\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}\\[2pt]{\sqrt[{3}]{2\sin 2A}}\!&\!+\!&\!{\sqrt[{3}]{2\sin 2B}}\!&\!+\!&\!{\sqrt[{3}]{2\sin 2C}}\!&\!=\!&\!-{\sqrt[{18}]{7}}\times {\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}\\[2pt]{\sqrt[{3}]{4\sin ^{2}2A}}\!&\!+\!&\!{\sqrt[{3}]{4\sin ^{2}2B}}\!&\!+\!&\!{\sqrt[{3}]{4\sin ^{2}2C}}\!&\!=\!&\!{\sqrt[{18}]{49}}\times {\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)}}\\[6pt]{\sqrt[{3}]{2\cos 2A}}\!&\!+\!&\!{\sqrt[{3}]{2\cos 2B}}\!&\!+\!&\!{\sqrt[{3}]{2\cos 2C}}\!&\!=\!&\!{\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}\\[8pt]{\sqrt[{3}]{4\cos ^{2}2A}}\!&\!+\!&\!{\sqrt[{3}]{4\cos ^{2}2B}}\!&\!+\!&\!{\sqrt[{3}]{4\cos ^{2}2C}}\!&\!=\!&\!{\sqrt[{3}]{11+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}\\[6pt]{\sqrt[{3}]{\tan 2A}}\!&\!+\!&\!{\sqrt[{3}]{\tan 2B}}\!&\!+\!&\!{\sqrt[{3}]{\tan 2C}}\!&\!=\!&\!-{\sqrt[{18}]{7}}\times {\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)}}\\[2pt]{\sqrt[{3}]{\tan ^{2}2A}}\!&\!+\!&\!{\sqrt[{3}]{\tan ^{2}2B}}\!&\!+\!&\!{\sqrt[{3}]{\tan ^{2}2C}}\!&\!=\!&\!{\sqrt[{18}]{49}}\times {\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)}}\end{array}}}
1
2
sin
2
A
3
+
1
2
sin
2
B
3
+
1
2
sin
2
C
3
=
−
1
7
18
×
6
+
3
(
5
−
3
7
3
3
+
4
−
3
7
3
3
)
3
1
4
sin
2
2
A
3
+
1
4
sin
2
2
B
3
+
1
4
sin
2
2
C
3
=
1
49
18
×
2
7
3
+
6
+
3
(
12
+
3
(
49
3
+
2
7
3
)
3
+
11
+
3
(
49
3
+
2
7
3
)
3
)
3
1
2
cos
2
A
3
+
1
2
cos
2
B
3
+
1
2
cos
2
C
3
=
4
−
3
7
3
3
1
4
cos
2
2
A
3
+
1
4
cos
2
2
B
3
+
1
4
cos
2
2
C
3
=
12
+
3
(
2
7
3
+
49
3
)
3
1
tan
2
A
3
+
1
tan
2
B
3
+
1
tan
2
C
3
=
−
1
7
18
×
−
49
3
+
6
+
3
(
5
+
3
(
7
3
−
49
3
)
3
+
−
3
+
3
(
7
3
−
49
3
)
3
)
3
1
tan
2
2
A
3
+
1
tan
2
2
B
3
+
1
tan
2
2
C
3
=
1
49
18
×
5
7
3
+
6
+
3
(
89
+
3
(
3
49
3
+
5
7
3
)
3
+
25
+
3
(
3
49
3
+
5
7
3
)
3
)
3
{\displaystyle {\begin{array}{ccccccl}{\frac {1}{\sqrt[{3}]{2\sin 2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{2\sin 2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{2\sin 2C}}}\!&\!=\!&\!-{\frac {1}{\sqrt[{18}]{7}}}\times {\sqrt[{3}]{6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}\\[2pt]{\frac {1}{\sqrt[{3}]{4\sin ^{2}2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{4\sin ^{2}2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{4\sin ^{2}2C}}}\!&\!=\!&\!{\frac {1}{\sqrt[{18}]{49}}}\times {\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)}}\\[2pt]{\frac {1}{\sqrt[{3}]{2\cos 2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{2\cos 2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{2\cos 2C}}}\!&\!=\!&\!{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\\[6pt]{\frac {1}{\sqrt[{3}]{4\cos ^{2}2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{4\cos ^{2}2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{4\cos ^{2}2C}}}\!&\!=\!&\!{\sqrt[{3}]{12+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}\\[2pt]{\frac {1}{\sqrt[{3}]{\tan 2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{\tan 2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{\tan 2C}}}\!&\!=\!&\!-{\frac {1}{\sqrt[{18}]{7}}}\times {\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)}}\\[2pt]{\frac {1}{\sqrt[{3}]{\tan ^{2}2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{\tan ^{2}2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{\tan ^{2}2C}}}\!&\!=\!&\!{\frac {1}{\sqrt[{18}]{49}}}\times {\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)}}\end{array}}}
cos
2
A
cos
2
B
3
+
cos
2
B
cos
2
C
3
+
cos
2
C
cos
2
A
3
=
−
7
3
cos
2
B
cos
2
A
3
+
cos
2
C
cos
2
B
3
+
cos
2
A
cos
2
C
3
=
0
cos
4
2
B
cos
2
A
3
+
cos
4
2
C
cos
2
B
3
+
cos
4
2
A
cos
2
C
3
=
−
49
3
2
cos
5
2
A
cos
2
2
B
3
+
cos
5
2
B
cos
2
2
C
3
+
cos
5
2
C
cos
2
2
A
3
=
0
cos
5
2
B
cos
2
2
A
3
+
cos
5
2
C
cos
2
2
B
3
+
cos
5
2
A
cos
2
2
C
3
=
−
3
×
7
3
2
cos
14
2
A
cos
5
2
B
3
+
cos
14
2
B
cos
5
2
C
3
+
cos
14
2
C
cos
5
2
A
3
=
0
cos
14
2
B
cos
5
2
A
3
+
cos
14
2
C
cos
5
2
B
3
+
cos
14
2
A
cos
5
2
C
3
=
−
61
×
7
3
8
.
{\displaystyle {\begin{array}{ccccccl}{\sqrt[{3}]{\frac {\cos 2A}{\cos 2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos 2B}{\cos 2C}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos 2C}{\cos 2A}}}\!&\!=\!&\!-{\sqrt[{3}]{7}}\\[2pt]{\sqrt[{3}]{\frac {\cos 2B}{\cos 2A}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos 2C}{\cos 2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos 2A}{\cos 2C}}}\!&\!=\!&\!0\\[2pt]{\sqrt[{3}]{\frac {\cos ^{4}2B}{\cos 2A}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{4}2C}{\cos 2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{4}2A}{\cos 2C}}}\!&\!=\!&\!-{\frac {\sqrt[{3}]{49}}{2}}\\[2pt]{\sqrt[{3}]{\frac {\cos ^{5}2A}{\cos ^{2}2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{5}2B}{\cos ^{2}2C}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{5}2C}{\cos ^{2}2A}}}\!&\!=\!&\!0\\[2pt]{\sqrt[{3}]{\frac {\cos ^{5}2B}{\cos ^{2}2A}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{5}2C}{\cos ^{2}2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{5}2A}{\cos ^{2}2C}}}\!&\!=\!&\!-3\times {\frac {\sqrt[{3}]{7}}{2}}\\[2pt]{\sqrt[{3}]{\frac {\cos ^{14}2A}{\cos ^{5}2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{14}2B}{\cos ^{5}2C}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{14}2C}{\cos ^{5}2A}}}\!&\!=\!&\!0\\[2pt]{\sqrt[{3}]{\frac {\cos ^{14}2B}{\cos ^{5}2A}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{14}2C}{\cos ^{5}2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{14}2A}{\cos ^{5}2C}}}\!&\!=\!&\!-61\times {\frac {\sqrt[{3}]{7}}{8}}.\end{array}}}
[ 9]
{\displaystyle }
^ a b Yiu, Paul (2009). "Heptagonal Triangles and Their Companions" (PDF) . Forum Geometricorum 9 : 125–148. ^ a b c d e f g h i j k l m n o p q Bankoff, Leon; Garfunkel, Jack (1973). "The Heptagonal Triangle". Mathematics Magazine 46 (1): 7–19. doi :10.2307/2688574 . JSTOR 2688574 . ^ a b Altintas, Abdilkadir (2016). "Some Collinearities in the Heptagonal Triangle" (PDF) . Forum Geometricorum 16 : 249–256. ^ a b c d e f g h i Wang, Kai (2019). "Heptagonal Triangle and Trigonometric Identities". Forum Geometricorum 19 : 29–38. ^ Wang, Kai (August 2019). "On cubic equations with zero sums of cubic roots of roots" – via ResearchGate. ^ a b c Weisstein, Eric W. "Heptagonal Triangle" . mathworld.wolfram.com . Retrieved 2024-08-02 . ^ a b c d e f Wang, Kai (September 2018). "Trigonometric Properties For Heptagonal Triangle" – via ResearchGate. ^ Moll, Victor H. (2007-09-24). "An elementary trigonometric equation". arXiv :0709.3755 math.NT ]. ^ a b Wang, Kai (October 2019). "On Ramanujan Type Identities For PI/7" – via ResearchGate. ^ 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 . JSTOR 2323624 . Archived from the original (PDF) on 2015-12-19. ^ Witula, Roman; Slota, Damian (2007). "New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7" (PDF) . Journal of Integer Sequences 10 (5) 07.5.6. Bibcode :2007JIntS..10...56W .
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63107bd555aab31d4dbd45b3bea63fcb044aeb43)
![{\displaystyle {\begin{aligned}A&={\frac {\pi }{7}}\\[6pt]\cos A&={\frac {b}{2a}}\end{aligned}}\quad {\begin{aligned}B&={\frac {2\pi }{7}}\\[6pt]\cos B&={\frac {c}{2b}}\end{aligned}}\quad {\begin{aligned}C&={\frac {4\pi }{7}}\\[6pt]\cos C&=-{\frac {a}{2c}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95dd39d97a0beffeae6cda91521c6eca64ade86a)
![{\displaystyle {\begin{array}{rcccccl}\sin A\!&\!\times \!&\!\sin B\!&\!\times \!&\!\sin C\!&\!=\!&\!{\frac {\sqrt {7}}{8}}\\[2pt]\sin A\!&\!-\!&\!\sin B\!&\!-\!&\!\sin C\!&\!=\!&\!-{\frac {\sqrt {7}}{2}}\\[2pt]\cos A\!&\!\times \!&\!\cos B\!&\!\times \!&\!\cos C\!&\!=\!&\!-{\frac {1}{8}}\\[2pt]\tan A\!&\!\times \!&\!\tan B\!&\!\times \!&\!\tan C\!&\!=\!&\!-{\sqrt {7}}\\[2pt]\tan A\!&\!+\!&\!\tan B\!&\!+\!&\!\tan C\!&\!=\!&\!-{\sqrt {7}}\\[2pt]\cot A\!&\!+\!&\!\cot B\!&\!+\!&\!\cot C\!&\!=\!&\!{\sqrt {7}}\\[8pt]\sin ^{2}\!A\!&\!\times \!&\!\sin ^{2}\!B\!&\!\times \!&\!\sin ^{2}\!C\!&\!=\!&\!{\frac {7}{64}}\\[2pt]\sin ^{2}\!A\!&\!+\!&\!\sin ^{2}\!B\!&\!+\!&\!\sin ^{2}\!C\!&\!=\!&\!{\frac {7}{4}}\\[2pt]\cos ^{2}\!A\!&\!+\!&\!\cos ^{2}\!B\!&\!+\!&\!\cos ^{2}\!C\!&\!=\!&\!{\frac {5}{4}}\\[2pt]\tan ^{2}\!A\!&\!+\!&\!\tan ^{2}\!B\!&\!+\!&\!\tan ^{2}\!C\!&\!=\!&\!21\\[2pt]\sec ^{2}\!A\!&\!+\!&\!\sec ^{2}\!B\!&\!+\!&\!\sec ^{2}\!C\!&\!=\!&\!24\\[2pt]\csc ^{2}\!A\!&\!+\!&\!\csc ^{2}\!B\!&\!+\!&\!\csc ^{2}\!C\!&\!=\!&\!8\\[2pt]\cot ^{2}\!A\!&\!+\!&\!\cot ^{2}\!B\!&\!+\!&\!\cot ^{2}\!C\!&\!=\!&\!5\\[8pt]\sin ^{4}\!A\!&\!+\!&\!\sin ^{4}\!B\!&\!+\!&\!\sin ^{4}\!C\!&\!=\!&\!{\frac {21}{16}}\\[2pt]\cos ^{4}\!A\!&\!+\!&\!\cos ^{4}\!B\!&\!+\!&\!\cos ^{4}\!C\!&\!=\!&\!{\frac {13}{16}}\\[2pt]\sec ^{4}\!A\!&\!+\!&\!\sec ^{4}\!B\!&\!+\!&\!\sec ^{4}\!C\!&\!=\!&\!416\\[2pt]\csc ^{4}\!A\!&\!+\!&\!\csc ^{4}\!B\!&\!+\!&\!\csc ^{4}\!C\!&\!=\!&\!32\\[8pt]\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11372d01f064a3ac9ffba8d9f82ccf67392e62ed)
![{\displaystyle {\begin{array}{ccccl}\tan A\!&\!-\!&\!4\sin B\!&\!=\!&\!-{\sqrt {7}}\\[2pt]\tan B\!&\!-\!&\!4\sin C\!&\!=\!&\!-{\sqrt {7}}\\[2pt]\tan C\!&\!+\!&\!4\sin A\!&\!=\!&\!-{\sqrt {7}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e81c6d96fd69af254e9ce675fad313b333fd7904)
![{\displaystyle {\begin{aligned}\cot ^{2}\!A&=1-{\frac {2\tan C}{\sqrt {7}}}\\[2pt]\cot ^{2}\!B&=1-{\frac {2\tan A}{\sqrt {7}}}\\[2pt]\cot ^{2}\!C&=1-{\frac {2\tan B}{\sqrt {7}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50839aa6272436a1b2888172a7c95235431e7688)
![{\displaystyle {\begin{array}{rcccccl}\cos A\!&\!=\!&\!{\frac {-1}{2}}\!&\!+\!&\!{\frac {4}{\sqrt {7}}}\!&\!\times \!&\!\sin ^{3}\!C\\[2pt]\sec A\!&\!=\!&\!2\!&\!+\!&\!4\!&\!\times \!&\!\cos C\\[4pt]\sec A\!&\!=\!&\!6\!&\!-\!&\!8\!&\!\times \!&\!\sin ^{2}\!B\\[4pt]\sec A\!&\!=\!&\!4\!&\!-\!&\!{\frac {16}{\sqrt {7}}}\!&\!\times \!&\!\sin ^{3}\!B\\[2pt]\cot A\!&\!=\!&\!{\sqrt {7}}\!&\!+\!&\!{\frac {8}{\sqrt {7}}}\!&\!\times \!&\!\sin ^{2}\!B\\[2pt]\cot A\!&\!=\!&\!{\frac {3}{\sqrt {7}}}\!&\!+\!&\!{\frac {4}{\sqrt {7}}}\!&\!\times \!&\!\cos B\\[2pt]\sin ^{2}\!A\!&\!=\!&\!{\frac {1}{2}}\!&\!+\!&\!{\frac {1}{2}}\!&\!\times \!&\!\cos B\\[2pt]\cos ^{2}\!A\!&\!=\!&\!{\frac {3}{4}}\!&\!+\!&\!{\frac {2}{\sqrt {7}}}\!&\!\times \!&\!\sin ^{3}\!A\\[2pt]\cot ^{2}\!A\!&\!=\!&\!3\!&\!+\!&\!{\frac {8}{\sqrt {7}}}\!&\!\times \!&\!\sin A\\[2pt]\sin ^{3}\!A\!&\!=\!&\!{\frac {-{\sqrt {7}}}{8}}\!&\!+\!&\!{\frac {\sqrt {7}}{4}}\!&\!\times \!&\!\cos B\\[2pt]\csc ^{3}\!A\!&\!=\!&\!{\frac {-6}{\sqrt {7}}}\!&\!+\!&\!{\frac {2}{\sqrt {7}}}\!&\!\times \!&\!\tan ^{2}\!C\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b36fe1e7d4ad29454a0aa79868851380c420711c)
![{\displaystyle {\begin{aligned}\sin ^{3}\!B\sin C-\sin ^{3}\!C\sin A-\sin ^{3}\!A\sin B&=0\\[3pt]\sin B\sin ^{3}\!C-\sin C\sin ^{3}\!A-\sin A\sin ^{3}\!B&={\frac {7}{2^{4}\!}}\\[2pt]\sin ^{4}\!B\sin C-\sin ^{4}\!C\sin A+\sin ^{4}\!A\sin B&=0\\[2pt]\sin B\sin ^{4}\!C+\sin C\sin ^{4}\!A-\sin A\sin ^{4}\!B&={\frac {7{\sqrt {7}}}{2^{5}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/313eecf9c55707e7d80590c5350d7c1c461b8b1d)
![{\displaystyle {\begin{aligned}\sin ^{11}\!B\sin ^{3}\!C-\sin ^{11}\!C\sin ^{3}\!A-\sin ^{11}\!A\sin ^{3}\!B&=0\\[2pt]\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}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04ff05b82052d8332beefe762209d7d4e1f67d72)
![{\displaystyle {\begin{aligned}S(n)&=(-\sin A)^{n}+\sin ^{n}\!B+\sin ^{n}\!C\\[4pt]C(n)&=(-\cos A)^{n}+\cos ^{n}\!B+\cos ^{n}\!C\\[4pt]T(n)&=\tan ^{n}\!A+\tan ^{n}\!B+\tan ^{n}\!C\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30746680912c548893246cbf950126c4bed81f68)
![{\displaystyle {\begin{array}{ccccccl}{\sqrt[{3}]{2\sin 2A}}\!&\!+\!&\!{\sqrt[{3}]{2\sin 2B}}\!&\!+\!&\!{\sqrt[{3}]{2\sin 2C}}\!&\!=\!&\!-{\sqrt[{18}]{7}}\times {\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}\\[2pt]{\sqrt[{3}]{2\sin 2A}}\!&\!+\!&\!{\sqrt[{3}]{2\sin 2B}}\!&\!+\!&\!{\sqrt[{3}]{2\sin 2C}}\!&\!=\!&\!-{\sqrt[{18}]{7}}\times {\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}\\[2pt]{\sqrt[{3}]{4\sin ^{2}2A}}\!&\!+\!&\!{\sqrt[{3}]{4\sin ^{2}2B}}\!&\!+\!&\!{\sqrt[{3}]{4\sin ^{2}2C}}\!&\!=\!&\!{\sqrt[{18}]{49}}\times {\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)}}\\[6pt]{\sqrt[{3}]{2\cos 2A}}\!&\!+\!&\!{\sqrt[{3}]{2\cos 2B}}\!&\!+\!&\!{\sqrt[{3}]{2\cos 2C}}\!&\!=\!&\!{\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}\\[8pt]{\sqrt[{3}]{4\cos ^{2}2A}}\!&\!+\!&\!{\sqrt[{3}]{4\cos ^{2}2B}}\!&\!+\!&\!{\sqrt[{3}]{4\cos ^{2}2C}}\!&\!=\!&\!{\sqrt[{3}]{11+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}\\[6pt]{\sqrt[{3}]{\tan 2A}}\!&\!+\!&\!{\sqrt[{3}]{\tan 2B}}\!&\!+\!&\!{\sqrt[{3}]{\tan 2C}}\!&\!=\!&\!-{\sqrt[{18}]{7}}\times {\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)}}\\[2pt]{\sqrt[{3}]{\tan ^{2}2A}}\!&\!+\!&\!{\sqrt[{3}]{\tan ^{2}2B}}\!&\!+\!&\!{\sqrt[{3}]{\tan ^{2}2C}}\!&\!=\!&\!{\sqrt[{18}]{49}}\times {\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)}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/810c85d0d2d0261c86d4ad110e55c03d4310c4f0)
![{\displaystyle {\begin{array}{ccccccl}{\frac {1}{\sqrt[{3}]{2\sin 2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{2\sin 2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{2\sin 2C}}}\!&\!=\!&\!-{\frac {1}{\sqrt[{18}]{7}}}\times {\sqrt[{3}]{6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}\\[2pt]{\frac {1}{\sqrt[{3}]{4\sin ^{2}2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{4\sin ^{2}2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{4\sin ^{2}2C}}}\!&\!=\!&\!{\frac {1}{\sqrt[{18}]{49}}}\times {\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)}}\\[2pt]{\frac {1}{\sqrt[{3}]{2\cos 2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{2\cos 2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{2\cos 2C}}}\!&\!=\!&\!{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\\[6pt]{\frac {1}{\sqrt[{3}]{4\cos ^{2}2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{4\cos ^{2}2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{4\cos ^{2}2C}}}\!&\!=\!&\!{\sqrt[{3}]{12+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}\\[2pt]{\frac {1}{\sqrt[{3}]{\tan 2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{\tan 2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{\tan 2C}}}\!&\!=\!&\!-{\frac {1}{\sqrt[{18}]{7}}}\times {\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)}}\\[2pt]{\frac {1}{\sqrt[{3}]{\tan ^{2}2A}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{\tan ^{2}2B}}}\!&\!+\!&\!{\frac {1}{\sqrt[{3}]{\tan ^{2}2C}}}\!&\!=\!&\!{\frac {1}{\sqrt[{18}]{49}}}\times {\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)}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/092ce3486cbd674aa014182b557e6dbbd0a058b9)
![{\displaystyle {\begin{array}{ccccccl}{\sqrt[{3}]{\frac {\cos 2A}{\cos 2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos 2B}{\cos 2C}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos 2C}{\cos 2A}}}\!&\!=\!&\!-{\sqrt[{3}]{7}}\\[2pt]{\sqrt[{3}]{\frac {\cos 2B}{\cos 2A}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos 2C}{\cos 2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos 2A}{\cos 2C}}}\!&\!=\!&\!0\\[2pt]{\sqrt[{3}]{\frac {\cos ^{4}2B}{\cos 2A}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{4}2C}{\cos 2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{4}2A}{\cos 2C}}}\!&\!=\!&\!-{\frac {\sqrt[{3}]{49}}{2}}\\[2pt]{\sqrt[{3}]{\frac {\cos ^{5}2A}{\cos ^{2}2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{5}2B}{\cos ^{2}2C}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{5}2C}{\cos ^{2}2A}}}\!&\!=\!&\!0\\[2pt]{\sqrt[{3}]{\frac {\cos ^{5}2B}{\cos ^{2}2A}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{5}2C}{\cos ^{2}2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{5}2A}{\cos ^{2}2C}}}\!&\!=\!&\!-3\times {\frac {\sqrt[{3}]{7}}{2}}\\[2pt]{\sqrt[{3}]{\frac {\cos ^{14}2A}{\cos ^{5}2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{14}2B}{\cos ^{5}2C}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{14}2C}{\cos ^{5}2A}}}\!&\!=\!&\!0\\[2pt]{\sqrt[{3}]{\frac {\cos ^{14}2B}{\cos ^{5}2A}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{14}2C}{\cos ^{5}2B}}}\!&\!+\!&\!{\sqrt[{3}]{\frac {\cos ^{14}2A}{\cos ^{5}2C}}}\!&\!=\!&\!-61\times {\frac {\sqrt[{3}]{7}}{8}}.\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9537ae68fa934954aeb5597d1f2df4209374a8a8)
