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Heptagonal triangle
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<h2 id="key-points">Key points</h2> <p>The heptagonal triangle's <a href="/facts/Nine-point_center/P8CPrFag">nine-point center</a> is also its first <a href="/facts/Brocard_point/zMXWMQY8">Brocard point</a>.<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a>: Propos. 12 </p><p>The second Brocard point lies on the nine-point circle.<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a>: p. 19 </p><p>The <a href="/facts/Circumcenter/RafYT4am">circumcenter</a> and the <a href="/facts/Fermat_points/QRvKxfE6">Fermat points</a> of a heptagonal triangle form an <a href="/facts/Equilateral_triangle/MJomsYDS">equilateral triangle</a>.<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a>: Thm. 22 </p><p>The distance between the circumcenter <i>O</i> and the <a href="/facts/Orthocenter/z9H8fFMz">orthocenter</a> <i>H</i> is given by<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a>: p. 19 </p> O H = R 2 , {\displaystyle OH=R{\sqrt {2}},} <p>where <i>R</i> is the <a href="/facts/Circumradius/RafYT4am">circumradius</a>. The squared distance from the <a href="/facts/Incenter/e933kPfL">incenter</a> <i>I</i> to the orthocenter is<a class="footnote-ref" id="fnref:5" href="#fn:5"><sup>5</sup></a>: p. 19 </p> I H 2 = R 2 + 4 r 2 2 , {\displaystyle IH^{2}={\frac {R^{2}+4r^{2}}{2}},} <p>where <i>r</i> is the <a href="/facts/Inradius/x9QCn9ML">inradius</a>. </p><p>The two tangents from the orthocenter to the circumcircle are mutually <a href="/facts/Perpendicular_lines/u0k8xqx4">perpendicular</a>.<a class="footnote-ref" id="fnref:6" href="#fn:6"><sup>6</sup></a>: p. 19 </p> <h2 id="relations-of-distances">Relations of distances</h2> <h3>Sides</h3> <p>The heptagonal triangle's sides <i>a</i> < <i>b</i> < <i>c</i> coincide respectively with the regular heptagon's side, shorter diagonal, and longer diagonal. They satisfy<a class="footnote-ref" id="fnref:7" href="#fn:7"><sup>7</sup></a>: Lemma 1 </p> 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}}} <p>(the latter<a class="footnote-ref" id="fnref:8" href="#fn:8"><sup>8</sup></a>: p. 13 being the <a href="/facts/Optic_equation/nSrH3aT5">optic equation</a>) and hence </p> a b + a c = b c , {\displaystyle ab+ac=bc,} <p>and<a class="footnote-ref" id="fnref:9" href="#fn:9"><sup>9</sup></a>: Coro. 2 </p> 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.} <p>Thus –<i>b</i>/<i>c</i>, <i>c</i>/<i>a</i>, and <i>a</i>/<i>b</i> all satisfy the <a href="/facts/Cubic_equation/Eqm3HCMx">cubic equation</a> </p> t 3 − 2 t 2 − t + 1 = 0. {\displaystyle t^{3}-2t^{2}-t+1=0.} <p>However, no <a href="/facts/Algebraic_expression/CK7D57Dm">algebraic expressions</a> with purely real terms exist for the solutions of this equation, because it is an example of <a href="/facts/Casus_irreducibilis/u2mpLr1L">casus irreducibilis</a>. </p><p>The approximate relation of the sides is </p> b ≈ 1.80193 ⋅ a , c ≈ 2.24698 ⋅ a . {\displaystyle b\approx 1.80193\cdot a,\qquad c\approx 2.24698\cdot a.} <p>We also have<a class="footnote-ref" id="fnref:10" href="#fn:10"><sup>10</sup></a><a class="footnote-ref" id="fnref:11" href="#fn:11"><sup>11</sup></a> </p> 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}}} <p>satisfy the <a href="/facts/Cubic_equation/Eqm3HCMx">cubic equation</a> </p> t 3 + 4 t 2 + 3 t − 1 = 0. {\displaystyle t^{3}+4t^{2}+3t-1=0.} <p>We also have<a class="footnote-ref" id="fnref:12" href="#fn:12"><sup>12</sup></a> </p> 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}}}} <p>satisfy the <a href="/facts/Cubic_equation/Eqm3HCMx">cubic equation</a> </p> t 3 − t 2 − 9 t + 1 = 0. {\displaystyle t^{3}-t^{2}-9t+1=0.} <p>We also have<a class="footnote-ref" id="fnref:13" href="#fn:13"><sup>13</sup></a> </p> 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}}} <p>satisfy the <a href="/facts/Cubic_equation/Eqm3HCMx">cubic equation</a> </p> t 3 + 5 t 2 − 8 t + 1 = 0. {\displaystyle t^{3}+5t^{2}-8t+1=0.} <p>We also have<a class="footnote-ref" id="fnref:14" href="#fn:14"><sup>14</sup></a>: p. 14 </p> 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,} <p>and<a class="footnote-ref" id="fnref:15" href="#fn:15"><sup>15</sup></a>: p. 15 </p> 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.} <p>We also have<a class="footnote-ref" id="fnref:16" href="#fn:16"><sup>16</sup></a> </p> 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.} <h3>Altitudes</h3> <p>The altitudes <i>h</i><i>a</i>, <i>h</i><i>b</i>, and <i>h</i><i>c</i> satisfy </p> h a = h b + h c {\displaystyle h_{a}=h_{b}+h_{c}} <a class="footnote-ref" id="fnref:17" href="#fn:17"><sup>17</sup></a>: p. 13 <p>and </p> 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}}.} <a class="footnote-ref" id="fnref:18" href="#fn:18"><sup>18</sup></a>: p. 14 <p>The altitude from side <i>b</i> (opposite angle <i>B</i>) is half the internal angle bisector w A {\displaystyle w_{A}} of <i>A</i>:<a class="footnote-ref" id="fnref:19" href="#fn:19"><sup>19</sup></a>: p. 19 </p> 2 h b = w A . {\displaystyle 2h_{b}=w_{A}.} <p>Here angle <i>A</i> is the smallest angle, and <i>B</i> is the second smallest. </p> <h3>Internal angle bisectors</h3> <p>We have these properties of the <a href="/facts/Bisection/JTsya7Ca">internal angle bisectors</a> w A , w B , {\displaystyle w_{A},w_{B},} and w C {\displaystyle w_{C}} of angles <i>A, B</i>, and <i>C</i> respectively:<a class="footnote-ref" id="fnref:20" href="#fn:20"><sup>20</sup></a>: p. 16 </p> 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.} <h3>Circumradius, inradius, and exradius</h3> <p>The triangle's area is<a class="footnote-ref" id="fnref:21" href="#fn:21"><sup>21</sup></a> </p> A = 7 4 R 2 , {\displaystyle A={\frac {\sqrt {7}}{4}}R^{2},} <p>where <i>R</i> is the triangle's <a href="/facts/Circumradius/RafYT4am">circumradius</a>. </p><p>We have<a class="footnote-ref" id="fnref:22" href="#fn:22"><sup>22</sup></a>: p. 12 </p> a 2 + b 2 + c 2 = 7 R 2 . {\displaystyle a^{2}+b^{2}+c^{2}=7R^{2}.} <p>We also have<a class="footnote-ref" id="fnref:23" href="#fn:23"><sup>23</sup></a> </p> 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}.} <p>The ratio <i>r</i> /<i>R</i> of the <a href="/facts/Inradius/x9QCn9ML">inradius</a> to the circumradius is the positive solution of the cubic equation<a class="footnote-ref" id="fnref:24" href="#fn:24"><sup>24</sup></a> </p> 8 x 3 + 28 x 2 + 14 x − 7 = 0. {\displaystyle 8x^{3}+28x^{2}+14x-7=0.} <p>In addition,<a class="footnote-ref" id="fnref:25" href="#fn:25"><sup>25</sup></a>: p. 15 </p> 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}}}.} <p>We also have<a class="footnote-ref" id="fnref:26" href="#fn:26"><sup>26</sup></a> </p> 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}}}.} <p>In general for all integer <i>n</i>, </p> 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}} <p>where </p> g ( − 1 ) = 8 , g ( 0 ) = 3 , g ( 1 ) = 7 {\displaystyle g(-1)=8,\quad g(0)=3,\quad g(1)=7} <p>and </p> 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).} <p>We also have<a class="footnote-ref" id="fnref:27" href="#fn:27"><sup>27</sup></a> </p> 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.} <p>We also have<a class="footnote-ref" id="fnref:28" href="#fn:28"><sup>28</sup></a> </p> 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}.} <p>The <a href="/facts/Excircle/x9QCn9ML">exradius</a> <i>r</i><i>a</i> corresponding to side <i>a</i> equals the radius of the <a href="/facts/Nine-point_circle/0UfSxFX9">nine-point circle</a> of the heptagonal triangle.<a class="footnote-ref" id="fnref:29" href="#fn:29"><sup>29</sup></a>: p. 15 </p> <h2 id="orthic-triangle">Orthic triangle</h2> <p>The heptagonal triangle's <a href="/facts/Orthic_triangle/z9H8fFMz">orthic triangle</a>, with vertices at the feet of the <a href="/facts/Altitude_(geometry)/z9H8fFMz">altitudes</a>, is <a href="/facts/Similarity_(geometry)/Sbd6hEAN">similar</a> 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 <a href="/facts/Equilateral_triangle/MJomsYDS">equilateral triangle</a> being the only acute one).<a class="footnote-ref" id="fnref:30" href="#fn:30"><sup>30</sup></a>: pp. 12–13 </p> <h2 id="hyperbola">Hyperbola</h2> <p>The rectangular hyperbola through A , B , C , G = X ( 2 ) , H = X ( 4 ) {\displaystyle A,B,C,G=X(2),H=X(4)} has the following properties: </p> <ul><li>first focus F 1 = X ( 5 ) {\displaystyle F_{1}=X(5)} </li> <li>center U {\displaystyle U} is on Euler circle (general property) and on circle ( O , F 1 ) {\displaystyle (O,F_{1})} </li> <li>second focus F 2 {\displaystyle F_{2}} is on the circumcircle</li></ul> <h2 id="trigonometric-properties">Trigonometric properties</h2> <h3>Trigonometric identities</h3> <p>The various <a href="/facts/Trigonometric_identities/Xw9Jx8Vz">trigonometric identities</a> associated with the heptagonal triangle include these:<a class="footnote-ref" id="fnref:31" href="#fn:31"><sup>31</sup></a>: pp. 13–14 <a class="footnote-ref" id="fnref:32" href="#fn:32"><sup>32</sup></a><a class="footnote-ref" id="fnref:33" href="#fn:33"><sup>33</sup></a> </p><p> 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}}} <a class="footnote-ref" id="fnref:34" href="#fn:34"><sup>34</sup></a>: Proposition 10 </p><p> 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}}} </p><p> 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}}} <a class="footnote-ref" id="fnref:35" href="#fn:35"><sup>35</sup></a><a class="footnote-ref" id="fnref:36" href="#fn:36"><sup>36</sup></a> </p><p> 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}}} <a class="footnote-ref" id="fnref:37" href="#fn:37"><sup>37</sup></a> </p><p> 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}}} <a class="footnote-ref" id="fnref:38" href="#fn:38"><sup>38</sup></a> </p><p> 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}}} <a class="footnote-ref" id="fnref:39" href="#fn:39"><sup>39</sup></a> </p> <h3>Cubic polynomials</h3> <p>The <a href="/facts/Cubic_equation/Eqm3HCMx">cubic equation</a> 64 y 3 − 112 y 2 + 56 y − 7 = 0 {\displaystyle 64y^{3}-112y^{2}+56y-7=0} has solutions<a class="footnote-ref" id="fnref:40" href="#fn:40"><sup>40</sup></a>: p. 14 sin 2 A , sin 2 B , sin 2 C . {\displaystyle \sin ^{2}\!A,\ \sin ^{2}\!B,\ \sin ^{2}\!C.} </p><p>The positive solution of the cubic equation x 3 + x 2 − 2 x − 1 = 0 {\displaystyle x^{3}+x^{2}-2x-1=0} equals 2 cos B . {\displaystyle 2\cos B.} <a class="footnote-ref" id="fnref:41" href="#fn:41"><sup>41</sup></a>: p. 186–187 </p><p>The <a href="/facts/Root_of_a_polynomial/mlK3dhoT">roots</a> 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} are<a class="footnote-ref" id="fnref:42" href="#fn:42"><sup>42</sup></a> sin 2 A , sin 2 B , sin 2 C . {\displaystyle \sin 2A,\ \sin 2B,\ \sin 2C.} </p><p>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} are − sin A , sin B , sin C . {\displaystyle -\sin A,\ \sin B,\ \sin C.} </p><p>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} are − cos A , cos B , cos C . {\displaystyle -\cos A,\ \cos B,\ \cos C.} </p><p>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} are tan A , tan B , tan C . {\displaystyle \tan A,\ \tan B,\ \tan C.} </p><p>The roots of the cubic equation x 3 − 21 x 2 + 35 x − 7 = 0 {\displaystyle x^{3}-21x^{2}+35x-7=0} are tan 2 A , tan 2 B , tan 2 C . {\displaystyle \tan ^{2}\!A,\ \tan ^{2}\!B,\ \tan ^{2}\!C.} </p> <h3>Sequences</h3> <p>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}}} </p> <table><tbody><tr><th>Value of n:</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>9</th><th>10</th><th>11</th><th>12</th><th>13</th><th>14</th><th>15</th><th>16</th><th>17</th><th>18</th><th>19</th><th>20</th></tr><tr><th> S ( n ) {\displaystyle S(n)} </th><td> 3 {\displaystyle \ 3\ } </td><td> 7 2 {\displaystyle {\tfrac {\sqrt {7}}{2}}} </td><td> 7 2 2 {\displaystyle {\tfrac {7}{2^{2}}}} </td><td> 7 2 {\displaystyle {\tfrac {\sqrt {7}}{2}}} </td><td> 7 ⋅ 3 2 4 {\displaystyle {\tfrac {7\cdot 3}{2^{4}}}} </td><td> 7 7 2 4 {\displaystyle {\tfrac {7{\sqrt {7}}}{2^{4}}}} </td><td> 7 ⋅ 5 2 5 {\displaystyle {\tfrac {7\cdot 5}{2^{5}}}} </td><td> 7 2 7 2 7 {\displaystyle {\tfrac {7^{2}{\sqrt {7}}}{2^{7}}}} </td><td> 7 2 ⋅ 5 2 8 {\displaystyle {\tfrac {7^{2}\cdot 5}{2^{8}}}} </td><td> 7 ⋅ 25 7 2 9 {\displaystyle {\tfrac {7\cdot 25{\sqrt {7}}}{2^{9}}}} </td><td> 7 2 ⋅ 9 2 9 {\displaystyle {\tfrac {7^{2}\cdot 9}{2^{9}}}} </td><td> 7 2 ⋅ 13 7 2 11 {\displaystyle {\tfrac {7^{2}\cdot 13{\sqrt {7}}}{2^{11}}}} </td><td> 7 2 ⋅ 33 2 11 {\displaystyle {\tfrac {7^{2}\cdot 33}{2^{11}}}} </td><td> 7 2 ⋅ 3 7 2 9 {\displaystyle {\tfrac {7^{2}\cdot 3{\sqrt {7}}}{2^{9}}}} </td><td> 7 4 ⋅ 5 2 14 {\displaystyle {\tfrac {7^{4}\cdot 5}{2^{14}}}} </td><td> 7 2 ⋅ 179 7 2 15 {\displaystyle {\tfrac {7^{2}\cdot 179{\sqrt {7}}}{2^{15}}}} </td><td> 7 3 ⋅ 131 2 16 {\displaystyle {\tfrac {7^{3}\cdot 131}{2^{16}}}} </td><td> 7 3 ⋅ 3 7 2 12 {\displaystyle {\tfrac {7^{3}\cdot 3{\sqrt {7}}}{2^{12}}}} </td><td> 7 3 ⋅ 493 2 18 {\displaystyle {\tfrac {7^{3}\cdot 493}{2^{18}}}} </td><td> 7 3 ⋅ 181 7 2 18 {\displaystyle {\tfrac {7^{3}\cdot 181{\sqrt {7}}}{2^{18}}}} </td><td> 7 5 ⋅ 19 2 19 {\displaystyle {\tfrac {7^{5}\cdot 19}{2^{19}}}} </td></tr><tr><th> S ( − n ) {\displaystyle S(-n)} </th><td> 3 {\displaystyle 3} </td><td> 0 {\displaystyle 0} </td><td> 2 3 {\displaystyle 2^{3}} </td><td> − 2 3 ⋅ 3 7 7 {\displaystyle -{\tfrac {2^{3}\cdot 3{\sqrt {7}}}{7}}} </td><td> 2 5 {\displaystyle 2^{5}} </td><td> − 2 5 ⋅ 5 7 7 {\displaystyle -{\tfrac {2^{5}\cdot 5{\sqrt {7}}}{7}}} </td><td> 2 6 ⋅ 17 7 {\displaystyle {\tfrac {2^{6}\cdot 17}{7}}} </td><td> − 2 7 7 {\displaystyle -2^{7}{\sqrt {7}}} </td><td> 2 9 ⋅ 11 7 {\displaystyle {\tfrac {2^{9}\cdot 11}{7}}} </td><td> − 2 10 ⋅ 33 7 7 2 {\displaystyle -{\tfrac {2^{10}\cdot 33{\sqrt {7}}}{7^{2}}}} </td><td> 2 10 ⋅ 29 7 {\displaystyle {\tfrac {2^{10}\cdot 29}{7}}} </td><td> − 2 14 ⋅ 11 7 7 2 {\displaystyle -{\tfrac {2^{14}\cdot 11{\sqrt {7}}}{7^{2}}}} </td><td> 2 12 ⋅ 269 7 2 {\displaystyle {\tfrac {2^{12}\cdot 269}{7^{2}}}} </td><td> − 2 13 ⋅ 117 7 7 2 {\displaystyle -{\tfrac {2^{13}\cdot 117{\sqrt {7}}}{7^{2}}}} </td><td> 2 14 ⋅ 51 7 {\displaystyle {\tfrac {2^{14}\cdot 51}{7}}} </td><td> − 2 21 ⋅ 17 7 7 3 {\displaystyle -{\tfrac {2^{21}\cdot 17{\sqrt {7}}}{7^{3}}}} </td><td> 2 17 ⋅ 237 7 2 {\displaystyle {\tfrac {2^{17}\cdot 237}{7^{2}}}} </td><td> − 2 17 ⋅ 1445 7 7 3 {\displaystyle -{\tfrac {2^{17}\cdot 1445{\sqrt {7}}}{7^{3}}}} </td><td> 2 19 ⋅ 2203 7 3 {\displaystyle {\tfrac {2^{19}\cdot 2203}{7^{3}}}} </td><td> − 2 19 ⋅ 1919 7 7 3 {\displaystyle -{\tfrac {2^{19}\cdot 1919{\sqrt {7}}}{7^{3}}}} </td><td> 2 20 ⋅ 5851 7 3 {\displaystyle {\tfrac {2^{20}\cdot 5851}{7^{3}}}} </td></tr><tr><th> C ( n ) {\displaystyle C(n)} </th><td> 3 {\displaystyle 3} </td><td> − 1 2 {\displaystyle -{\tfrac {1}{2}}} </td><td> 5 4 {\displaystyle {\tfrac {5}{4}}} </td><td> − 1 2 {\displaystyle -{\tfrac {1}{2}}} </td><td> 13 16 {\displaystyle {\tfrac {13}{16}}} </td><td> − 1 2 {\displaystyle -{\tfrac {1}{2}}} </td><td> 19 32 {\displaystyle {\tfrac {19}{32}}} </td><td> − 57 128 {\displaystyle -{\tfrac {57}{128}}} </td><td> 117 256 {\displaystyle {\tfrac {117}{256}}} </td><td> − 193 512 {\displaystyle -{\tfrac {193}{512}}} </td><td> 185 512 {\displaystyle {\tfrac {185}{512}}} </td></tr><tr><th> C ( − n ) {\displaystyle C(-n)} </th><td> 3 {\displaystyle 3} </td><td> − 4 {\displaystyle -4} </td><td> 24 {\displaystyle 24} </td><td> − 88 {\displaystyle -88} </td><td> 416 {\displaystyle 416} </td><td> − 1824 {\displaystyle -1824} </td><td> 8256 {\displaystyle 8256} </td><td> − 36992 {\displaystyle -36992} </td><td> 166400 {\displaystyle 166400} </td><td> − 747520 {\displaystyle -747520} </td><td> 3359744 {\displaystyle 3359744} </td></tr><tr><th> T ( n ) {\displaystyle T(n)} </th><td> 3 {\displaystyle 3} </td><td> − 7 {\displaystyle -{\sqrt {7}}} </td><td> 7 ⋅ 3 {\displaystyle 7\cdot 3} </td><td> − 31 7 {\displaystyle -31{\sqrt {7}}} </td><td> 7 ⋅ 53 {\displaystyle 7\cdot 53} </td><td> − 7 ⋅ 87 7 {\displaystyle -7\cdot 87{\sqrt {7}}} </td><td> 7 ⋅ 1011 {\displaystyle 7\cdot 1011} </td><td> − 7 2 ⋅ 239 7 {\displaystyle -7^{2}\cdot 239{\sqrt {7}}} </td><td> 7 2 ⋅ 2771 {\displaystyle 7^{2}\cdot 2771} </td><td> − 7 ⋅ 32119 7 {\displaystyle -7\cdot 32119{\sqrt {7}}} </td><td> 7 2 ⋅ 53189 {\displaystyle 7^{2}\cdot 53189} </td></tr><tr><th> T ( − n ) {\displaystyle T(-n)} </th><td> 3 {\displaystyle 3} </td><td> 7 {\displaystyle {\sqrt {7}}} </td><td> 5 {\displaystyle 5} </td><td> 25 7 7 {\displaystyle {\tfrac {25{\sqrt {7}}}{7}}} </td><td> 19 {\displaystyle 19} </td><td> 103 7 7 {\displaystyle {\tfrac {103{\sqrt {7}}}{7}}} </td><td> 563 7 {\displaystyle {\tfrac {563}{7}}} </td><td> 7 ⋅ 9 7 {\displaystyle 7\cdot 9{\sqrt {7}}} </td><td> 2421 7 {\displaystyle {\tfrac {2421}{7}}} </td><td> 13297 7 7 2 {\displaystyle {\tfrac {13297{\sqrt {7}}}{7^{2}}}} </td><td> 10435 7 {\displaystyle {\tfrac {10435}{7}}} </td></tr></tbody></table> <h3>Ramanujan identities</h3> <p>We also have Ramanujan type identities,<a class="footnote-ref" id="fnref:43" href="#fn:43"><sup>43</sup></a><a class="footnote-ref" id="fnref:44" href="#fn:44"><sup>44</sup></a> </p><p> 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}}} </p><p> 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}}} </p><p> 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}}} <a class="footnote-ref" id="fnref:45" href="#fn:45"><sup>45</sup></a> </p><p> {\displaystyle } </p> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Yiu, Paul (2009). "Heptagonal Triangles and Their Companions" (PDF). Forum Geometricorum. 9: 125–148. <a href="http://forumgeom.fau.edu/FG2009volume9/FG200912.pdf" target="_blank">http://forumgeom.fau.edu/FG2009volume9/FG200912.pdf</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Bankoff, Leon; Garfunkel, Jack (1973). "The Heptagonal Triangle". Mathematics Magazine. 46 (1): 7–19. doi:10.2307/2688574. JSTOR 2688574. <a href="/wiki/Mathematics_Magazine" target="_blank">/wiki/Mathematics_Magazine</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>Yiu, Paul (2009). "Heptagonal Triangles and Their Companions" (PDF). Forum Geometricorum. 9: 125–148. <a href="http://forumgeom.fau.edu/FG2009volume9/FG200912.pdf" target="_blank">http://forumgeom.fau.edu/FG2009volume9/FG200912.pdf</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>Bankoff, Leon; Garfunkel, Jack (1973). "The Heptagonal Triangle". Mathematics Magazine. 46 (1): 7–19. doi:10.2307/2688574. JSTOR 2688574. <a href="/wiki/Mathematics_Magazine" target="_blank">/wiki/Mathematics_Magazine</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> <li id="fn:5"><p>Bankoff, Leon; Garfunkel, Jack (1973). "The Heptagonal Triangle". Mathematics Magazine. 46 (1): 7–19. doi:10.2307/2688574. JSTOR 2688574. <a href="/wiki/Mathematics_Magazine" target="_blank">/wiki/Mathematics_Magazine</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li> <li id="fn:6"><p>Bankoff, Leon; Garfunkel, Jack (1973). "The Heptagonal Triangle". Mathematics Magazine. 46 (1): 7–19. doi:10.2307/2688574. JSTOR 2688574. <a href="/wiki/Mathematics_Magazine" target="_blank">/wiki/Mathematics_Magazine</a> <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li> <li id="fn:7"><p>Altintas, Abdilkadir (2016). "Some Collinearities in the Heptagonal Triangle" (PDF). Forum Geometricorum. 16: 249–256. <a href="http://forumgeom.fau.edu/FG2016volume16/FG201630.pdf" target="_blank">http://forumgeom.fau.edu/FG2016volume16/FG201630.pdf</a> <a href="#fnref:7" class="footnote-back-ref">↩</a></p></li> <li id="fn:8"><p>Bankoff, Leon; Garfunkel, Jack (1973). "The Heptagonal Triangle". Mathematics Magazine. 46 (1): 7–19. doi:10.2307/2688574. JSTOR 2688574. <a href="/wiki/Mathematics_Magazine" target="_blank">/wiki/Mathematics_Magazine</a> <a href="#fnref:8" class="footnote-back-ref">↩</a></p></li> <li id="fn:9"><p>Altintas, Abdilkadir (2016). "Some Collinearities in the Heptagonal Triangle" (PDF). Forum Geometricorum. 16: 249–256. <a href="http://forumgeom.fau.edu/FG2016volume16/FG201630.pdf" target="_blank">http://forumgeom.fau.edu/FG2016volume16/FG201630.pdf</a> <a href="#fnref:9" class="footnote-back-ref">↩</a></p></li> <li id="fn:10"><p>Wang, Kai (2019). "Heptagonal Triangle and Trigonometric Identities". 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