Как связан тангенс и синусов


sin ⁡ π 60 = как связан тангенс и синусов cos ⁡ 29 π 60 = sin ⁡ 3 ∘ = cos ⁡ 87 ∘ = 2 ( 3 + 1 ) ( 5 − 1 ) − 2 ( 3 − 1 ) 5 + 5 16, {\displaystyle \sin {\frac {\pi }{60}}=\cos {\frac {29\,\pi }{60}}=\sin 3^{\circ }=\cos 87^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}+1)({\sqrt {5}}-1)-2({\sqrt {3}}-1){\sqrt {5+{\sqrt {5}}}}}{16}},} \sin \frac{\pi}{60} = \cos \frac{29\,\pi}{60} = \sin 3^\circ = \cos 87^\circ = \frac{\sqrt{2}(\sqrt{3}+1)(\sqrt{5}-1)-2(\sqrt{3}-1)\sqrt{5+\sqrt{5}}}{16},

cos ⁡ π 60 = sin ⁡ 29 π 60 = cos ⁡ 3 ∘ = sin ⁡ 87 ∘ = 2 ( 3 − 1 ) ( 5 − 1 ) + 2 ( 3 + 1 ) 5 + 5 16, {\displaystyle \cos {\frac {\pi }{60}}=\sin {\frac {29\,\pi }{60}}=\cos 3^{\circ }=\sin 87^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}-1)({\sqrt {5}}-1)+2({\sqrt {3}}+1){\sqrt {5+{\sqrt {5}}}}}{16}},} \cos \frac{\pi}{60} = \sin \frac{29\,\pi}{60} = \cos 3^\circ = \sin 87^\circ = \frac{\sqrt{2}(\sqrt{3}-1)(\sqrt{5}-1)+2(\sqrt{3}+1)\sqrt{5+\sqrt{5}}}{16},

tg ⁡ π 60 = ctg ⁡ 29 π 60 = tg ⁡ 3 ∘ = ctg ⁡ 87 ∘ = 2 ( 5 + 2 ) − 3 ( 5 + 3 ) + ( 2 − 3 ) ( 3 ( 5 + 1 ) − 2 ) 5 − 2 5 2, {\displaystyle \operatorname {tg} {\frac {\pi }{60}}=\operatorname {ctg} {\frac {29\,\pi }{60}}=\operatorname {tg} 3^{\circ }=\operatorname {ctg} 87^{\circ }={\frac {2({\sqrt {5}}+2)-{\sqrt {3}}({\sqrt {5}}+3)+(2-{\sqrt {3}})({\sqrt {3}}({\sqrt {5}}+1)-2){\sqrt {5-2{\sqrt {5}}}}}{2}},} \operatorname{tg} \frac{\pi}{60} = \operatorname{ctg} \frac{29\,\pi}{60} = \operatorname{tg} 3^\circ = \operatorname{ctg} 87^\circ = \frac{2(\sqrt{5}+2)-\sqrt{3}(\sqrt{5}+3)+(2-\sqrt{3})(\sqrt{3}(\sqrt{5}+1)-2)\sqrt{5-2\sqrt{5}}}{2},

ctg ⁡ π 60 = tg ⁡ 29 π 60 = ctg ⁡ 3 ∘ = tg ⁡ 87 ∘ = 2 ( 2 ( 5 + 2 ) + 3 ( 5 + 3 ) ) + ( 3 ( 5 − 1 ) + 2 ) 2 ( 25 + 11 5 ) 4, {\displaystyle \operatorname {ctg} {\frac {\pi }{60}}=\operatorname {tg} {\frac {29\,\pi }{60}}=\operatorname {ctg} 3^{\circ }=\operatorname {tg} 87^{\circ }={\frac {2(2({\sqrt {5}}+2)+{\sqrt {3}}({\sqrt {5}}+3))+({\sqrt {3}}({\sqrt {5}}-1)+2){\sqrt {2(25+11{\sqrt {5}})}}}{4}},} \operatorname{ctg} \frac{\pi}{60} = \operatorname{tg} \frac{29\,\pi}{60} = \operatorname{ctg} 3^\circ = \operatorname{tg} 87^\circ = \frac{2(2(\sqrt{5}+2)+\sqrt{3}(\sqrt{5}+3))+(\sqrt{3}(\sqrt{5}-1)+2)\sqrt{2(25+11\sqrt{5})}}{4},

sin ⁡ π 30 = cos ⁡ 7 π 15 = sin ⁡ 6 ∘ = cos ⁡ 84 ∘ = 6 ( 5 − 5 ) − 5 − 1 8, {\displaystyle \sin {\frac {\pi }{30}}=\cos {\frac {7\,\pi }{15}}=\sin 6^{\circ }=\cos 84^{\circ }={\frac {{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1}{8}},} \sin \frac{\pi}{30} = \cos \frac{7\,\pi}{15} = \sin 6^\circ = \cos 84^\circ = \frac{\sqrt{6(5-\sqrt{5})}-\sqrt{5}-1}{8},

cos ⁡ π 30 = sin ⁡ 7 π 15 = cos ⁡ 6 ∘ = sin ⁡ 84 ∘ = 2 ( 5 − 5 ) + 3 ( 5 + 1 ) 8, {\displaystyle \cos {\frac {\pi }{30}}=\sin {\frac {7\,\pi }{15}}=\cos 6^{\circ }=\sin 84^{\circ }={\frac {{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)}{8}},} \cos \frac{\pi}{30} = \sin \frac{7\,\pi}{15} = \cos 6^\circ = \sin 84^\circ = \frac{\sqrt{2(5-\sqrt{5})}+\sqrt{3}(\sqrt{5}+1)}{8},

tg ⁡ π 30 = ctg ⁡ 7 π 15 = tg ⁡ 6 ∘ = ctg ⁡ 84 ∘ = 2 ( 5 − 5 ) − 3 ( 5 − 1 ) 2, {\displaystyle \operatorname {tg} {\frac {\pi }{30}}=\operatorname {ctg} {\frac {7\,\pi }{15}}=\operatorname {tg} 6^{\circ }=\operatorname {ctg} 84^{\circ }={\frac {{\sqrt {2(5-{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)}{2}},} \operatorname{tg} \frac{\pi}{30} = \operatorname{ctg} \frac{7\,\pi}{15} = \operatorname{tg} 6^\circ = \operatorname{ctg} 84^\circ = \frac{\sqrt{2(5-\sqrt{5})}-\sqrt{3}(\sqrt{5}-1)}{2},

ctg ⁡ π 30 = tg ⁡ 7 π 15 = ctg ⁡ 6 ∘ = tg ⁡ 84 ∘ = 2 ( 25 + 11 5 ) + 3 ( 5 + 3 ) 2, {\displaystyle \operatorname {ctg} {\frac {\pi }{30}}=\operatorname {tg} {\frac {7\,\pi }{15}}=\operatorname {ctg} 6^{\circ }=\operatorname {tg} 84^{\circ }={\frac {{\sqrt {2(25+11{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+3)}{2}},} \operatorname{ctg} \frac{\pi}{30} = \operatorname{tg} \frac{7\,\pi}{15} = \operatorname{ctg} 6^\circ = \operatorname{tg} 84^\circ = \frac{\sqrt{2(25+11\sqrt{5})}+\sqrt{3}(\sqrt{5}+3)}{2},

sin ⁡ π 20 = cos ⁡ 9 π 20 = sin ⁡ 9 ∘ = cos ⁡ 81 ∘ = 2 ( 5 + 1 ) − 2 5 − 5 8, {\displaystyle \sin {\frac {\pi }{20}}=\cos {\frac {9\,\pi }{20}}=\sin 9^{\circ }=\cos 81^{\circ }={\frac {{\sqrt {2}}({\sqrt {5}}+1)-2{\sqrt {5-{\sqrt {5}}}}}{8}},} \sin \frac{\pi}{20} = \cos \frac{9\,\pi}{20} = \sin 9^\circ = \cos 81^\circ = \frac{\sqrt{2}(\sqrt{5}+1)-2\sqrt{5-\sqrt{5}}}{8},

cos ⁡ π 20 = sin ⁡ 9 π 20 = cos ⁡ 9 ∘ = sin ⁡ 81 ∘ = 2 ( 5 + 1 ) + 2 5 − 5 8, {\displaystyle \cos {\frac {\pi }{20}}=\sin {\frac {9\,\pi }{20}}=\cos 9^{\circ }=\sin 81^{\circ }={\frac {{\sqrt {2}}({\sqrt {5}}+1)+2{\sqrt {5-{\sqrt {5}}}}}{8}},} \cos \frac{\pi}{20} = \sin \frac{9\,\pi}{20} = \cos 9^\circ = \sin 81^\circ = \frac{\sqrt{2}(\sqrt{5}+1)+2\sqrt{5-\sqrt{5}}}{8},

tg ⁡ π 20 = ctg ⁡ 9 π 20 = tg ⁡ 9 ∘ = ctg ⁡ 81 ∘ = 5 + 1 − 5 + 2 5, {\displaystyle \operatorname {tg} {\frac {\pi }{20}}=\operatorname {ctg} {\frac {9\,\pi }{20}}=\operatorname {tg} 9^{\circ }=\operatorname {ctg} 81^{\circ }={{\sqrt {5}}+1-{\sqrt {5+2{\sqrt {5}}}}},} \operatorname{tg} \frac{\pi}{20} = \operatorname{ctg} \frac{9\,\pi}{20} = \operatorname{tg} 9^\circ = \operatorname{ctg} 81^\circ = {\sqrt{5}+1-\sqrt{5+2\sqrt{5}}},

ctg ⁡ π 20 = tg ⁡ 9 π 20 = ctg ⁡ 9 ∘ = tg ⁡ 81 ∘ = 5 + 1 + 5 + 2 5, {\displaystyle \operatorname {ctg} {\frac {\pi }{20}}=\operatorname {tg} {\frac {9\,\pi }{20}}=\operatorname {ctg} 9^{\circ }=\operatorname {tg} 81^{\circ }={{\sqrt {5}}+1+{\sqrt {5+2{\sqrt {5}}}}},} \operatorname{ctg} \frac{\pi}{20} = \operatorname{tg} \frac{9\,\pi}{20} = \operatorname{ctg} 9^\circ = \operatorname{tg} 81^\circ = {\sqrt{5}+1+\sqrt{5+2\sqrt{5}}},

sin ⁡ π 15 = cos ⁡ 13 π 30 = sin ⁡ 12 ∘ = cos ⁡ 78 ∘ = 2 ( 5 + 5 ) − 3 ( 5 − 1 ) 8, {\displaystyle \sin {\frac {\pi }{15}}=\cos {\frac {13\,\pi }{30}}=\sin 12^{\circ }=\cos 78^{\circ }={\frac {{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)}{8}},} \sin \frac{\pi}{15} = \cos \frac{13\,\pi}{30} = \sin 12^\circ = \cos 78^\circ = \frac{\sqrt{2(5+\sqrt{5})}-\sqrt{3}(\sqrt{5}-1)}{8},

cos ⁡ π 15 = sin ⁡ 13 π 30 = cos ⁡ 12 ∘ = sin ⁡ 78 ∘ = 6 ( 5 + 5 ) + 5 − 1 8, {\displaystyle \cos {\frac {\pi }{15}}=\sin {\frac {13\,\pi }{30}}=\cos 12^{\circ }=\sin 78^{\circ }={\frac {{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1}{8}},} \cos \frac{\pi}{15} = \sin \frac{13\,\pi}{30} = \cos 12^\circ = \sin 78^\circ = \frac{\sqrt{6(5+\sqrt{5})}+\sqrt{5}-1}{8},

tg ⁡ π 15 = ctg ⁡ 13 π 30 = tg ⁡ 12 ∘ = ctg ⁡ 78 ∘ = 3 ( 3 − 5 ) − 2 ( 25 − 11 5 ) 2, {\displaystyle \operatorname {tg} {\frac {\pi }{15}}=\operatorname {ctg} {\frac {13\,\pi }{30}}=\operatorname {tg} 12^{\circ }=\operatorname {ctg} 78^{\circ }={\frac {{\sqrt {3}}(3-{\sqrt {5}})-{\sqrt {2(25-11{\sqrt {5}})}}}{2}},} \operatorname{tg} \frac{\pi}{15} = \operatorname{ctg} \frac{13\,\pi}{30} = \operatorname{tg} 12^\circ = \operatorname{ctg} 78^\circ = \frac{\sqrt{3}(3-\sqrt{5})-\sqrt{2(25-11\sqrt{5})}}{2},

ctg ⁡ π 15 = tg ⁡ 13 π 30 = ctg ⁡ 12 ∘ = tg ⁡ 78 ∘ = 3 ( 5 + 1 ) + 2 ( 5 + 5 ) 2, {\displaystyle \operatorname {ctg} {\frac {\pi }{15}}=\operatorname {tg} {\frac {13\,\pi }{30}}=\operatorname {ctg} 12^{\circ }=\operatorname {tg} 78^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)+{\sqrt {2(5+{\sqrt {5}})}}}{2}},} \operatorname{ctg} \frac{\pi}{15} = \operatorname{tg} \frac{13\,\pi}{30} = \operatorname{ctg} 12^\circ = \operatorname{tg} 78^\circ = \frac{\sqrt{3}(\sqrt{5}+1)+\sqrt{2(5+\sqrt{5})}}{2},

sin ⁡ 7 π 60 = cos ⁡ 23 π 60 = sin ⁡ 21 ∘ = cos ⁡ 69 ∘ = − 2 ( 3 − 1 ) ( 5 + 1 ) + 2 ( 3 + 1 ) 5 − 5 16, {\displaystyle \sin {\frac {7\,\pi }{60}}=\cos {\frac {23\,\pi }{60}}=\sin 21^{\circ }=\cos 69^{\circ }={\frac {-{\sqrt {2}}({\sqrt {3}}-1)({\sqrt {5}}+1)+2({\sqrt {3}}+1){\sqrt {5-{\sqrt {5}}}}}{16}},} \sin \frac{7\,\pi}{60} = \cos \frac{23\,\pi}{60} = \sin 21^\circ = \cos 69^\circ = \frac{-\sqrt{2}(\sqrt{3}-1)(\sqrt{5}+1)+2(\sqrt{3}+1)\sqrt{5-\sqrt{5}}}{16},

cos ⁡ 7 π 60 = sin ⁡ 23 π 60 = cos ⁡ 21 ∘ = sin ⁡ 69 ∘ = 2 ( 3 + 1 ) ( 5 + 1 ) + 2 ( 3 − 1 ) 5 − 5 16, {\displaystyle \cos {\frac {7\,\pi }{60}}=\sin {\frac {23\,\pi }{60}}=\cos 21^{\circ }=\sin 69^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}+1)({\sqrt {5}}+1)+2({\sqrt {3}}-1){\sqrt {5-{\sqrt {5}}}}}{16}},} \cos \frac{7\,\pi}{60} = \sin \frac{23\,\pi}{60} = \cos 21^\circ = \sin 69^\circ = \frac{\sqrt{2}(\sqrt{3}+1)(\sqrt{5}+1)+2(\sqrt{3}-1)\sqrt{5-\sqrt{5}}}{16},

tg ⁡ 7 π 60 = ctg ⁡ 23 π 60 = tg ⁡ 21 ∘ = ctg ⁡ 69 ∘ = 2 ( 2 ( 5 − 2 ) − 3 ( 3 − 5 ) ) + ( 3 ( 5 + 1 ) − 2 ) 2 ( 25 − 11 5 ) 4, {\displaystyle \operatorname {tg} {\frac {7\,\pi }{60}}=\operatorname {ctg} {\frac {23\,\pi }{60}}=\operatorname {tg} 21^{\circ }=\operatorname {ctg} 69^{\circ }={\frac {2(2({\sqrt {5}}-2)-{\sqrt {3}}(3-{\sqrt {5}}))+({\sqrt {3}}({\sqrt {5}}+1)-2){\sqrt {2(25-11{\sqrt {5}})}}}{4}},} \operatorname{tg} \frac{7\,\pi}{60} = \operatorname{ctg} \frac{23\,\pi}{60} = \operatorname{tg} 21^\circ = \operatorname{ctg} 69^\circ = \frac{2(2(\sqrt{5}-2)-\sqrt{3}(3-\sqrt{5}))+(\sqrt{3}(\sqrt{5}+1)-2)\sqrt{2(25-11\sqrt{5})}}{4},

ctg ⁡ 7 π 60 = tg ⁡ 23 π 60 = ctg ⁡ 21 ∘ = tg ⁡ 69 ∘ = 2 ( 2 ( 5 − 2 ) + 3 ( 3 − 5 ) ) + ( 3 ( 5 + 1 ) + 2 ) 2 ( 25 − 11 5 ) 4, {\displaystyle \operatorname {ctg} {\frac {7\,\pi }{60}}=\operatorname {tg} {\frac {23\,\pi }{60}}=\operatorname {ctg} 21^{\circ }=\operatorname {tg} 69^{\circ }={\frac {2(2({\sqrt {5}}-2)+{\sqrt {3}}(3-{\sqrt {5}}))+({\sqrt {3}}({\sqrt {5}}+1)+2){\sqrt {2(25-11{\sqrt {5}})}}}{4}},} \operatorname{ctg} \frac{7\,\pi}{60} = \operatorname{tg} \frac{23\,\pi}{60} = \operatorname{ctg} 21^\circ = \operatorname{tg} 69^\circ = \frac{2(2(\sqrt{5}-2)+\sqrt{3}(3-\sqrt{5}))+(\sqrt{3}(\sqrt{5}+1)+2)\sqrt{2(25-11\sqrt{5})}}{4},

sin ⁡ 2 π 15 = cos ⁡ 11 π 30 = sin ⁡ 24 ∘ = cos ⁡ 66 ∘ = 3 ( 5 + 1 ) − 2 ( 5 − 5 ) 8, {\displaystyle \sin {\frac {2\,\pi }{15}}=\cos {\frac {11\,\pi }{30}}=\sin 24^{\circ }=\cos 66^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2(5-{\sqrt {5}})}}}{8}},} \sin \frac{2\,\pi}{15} = \cos \frac{11\,\pi}{30} = \sin 24^\circ = \cos 66^\circ = \frac{\sqrt{3}(\sqrt{5}+1)-\sqrt{2(5-\sqrt{5})}}{8},

cos ⁡ 2 π 15 = sin ⁡ 11 π 30 = cos ⁡ 24 ∘ = sin ⁡ 66 ∘ = 5 + 1 + 6 ( 5 − 5 ) 8, {\displaystyle \cos {\frac {2\,\pi }{15}}=\sin {\frac {11\,\pi }{30}}=\cos 24^{\circ }=\sin 66^{\circ }={\frac {{\sqrt {5}}+1+{\sqrt {6(5-{\sqrt {5}})}}}{8}},} \cos \frac{2\,\pi}{15} = \sin \frac{11\,\pi}{30} = \cos 24^\circ = \sin 66^\circ = \frac{\sqrt{5}+1+\sqrt{6(5-\sqrt{5})}}{8},

tg ⁡ 2 π 15 = ctg ⁡ 11 π 30 = tg ⁡ 24 ∘ = ctg ⁡ 66 ∘ = − 3 ( 3 + 5 ) + 2 ( 25 + 11 5 ) 2, {\displaystyle \operatorname {tg} {\frac {2\,\pi }{15}}=\operatorname {ctg} {\frac {11\,\pi }{30}}=\operatorname {tg} 24^{\circ }=\operatorname {ctg} 66^{\circ }={\frac {-{\sqrt {3}}(3+{\sqrt {5}})+{\sqrt {2(25+11{\sqrt {5}})}}}{2}},} \operatorname{tg} \frac{2\,\pi}{15} = \operatorname{ctg} \frac{11\,\pi}{30} = \operatorname{tg} 24^\circ = \operatorname{ctg} 66^\circ = \frac{-\sqrt{3}(3+\sqrt{5})+\sqrt{2(25+11\sqrt{5})}}{2},

ctg ⁡ 2 π 15 = tg ⁡ 11 π 30 = ctg ⁡ 24 ∘ = tg ⁡ 66 ∘ = 3 ( 5 − 1 ) + 2 ( 5 − 5 ) 2, {\displaystyle \operatorname {ctg} {\frac {2\,\pi }{15}}=\operatorname {tg} {\frac {11\,\pi }{30}}=\operatorname {ctg} 24^{\circ }=\operatorname {tg} 66^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}-1)+{\sqrt {2(5-{\sqrt {5}})}}}{2}},} \operatorname{ctg} \frac{2\,\pi}{15} = \operatorname{tg} \frac{11\,\pi}{30} = \operatorname{ctg} 24^\circ = \operatorname{tg} 66^\circ = \frac{\sqrt{3}(\sqrt{5}-1)+\sqrt{2(5-\sqrt{5})}}{2},

sin ⁡ 3 π 20 = cos ⁡ 7 π 20 = sin ⁡ 27 ∘ = cos ⁡ 63 ∘ = − 2 ( 5 − 1 ) + 2 5 + 5 8, {\displaystyle \sin {\frac {3\,\pi }{20}}=\cos {\frac {7\,\pi }{20}}=\sin 27^{\circ }=\cos 63^{\circ }={\frac {-{\sqrt {2}}({\sqrt {5}}-1)+2{\sqrt {5+{\sqrt {5}}}}}{8}},} \sin \frac{3\,\pi}{20} = \cos \frac{7\,\pi}{20} = \sin 27^\circ = \cos 63^\circ = \frac{-\sqrt{2}(\sqrt{5}-1)+2\sqrt{5+\sqrt{5}}}{8},

cos ⁡ 3 π 20 = sin ⁡ 7 π 20 = cos ⁡ 27 ∘ = sin ⁡ 63 ∘ = 2 ( 5 − 1 ) + 2 5 + 5 8, {\displaystyle \cos {\frac {3\,\pi }{20}}=\sin {\frac {7\,\pi }{20}}=\cos 27^{\circ }=\sin 63^{\circ }={\frac {{\sqrt {2}}({\sqrt {5}}-1)+2{\sqrt {5+{\sqrt {5}}}}}{8}},} \cos \frac{3\,\pi}{20} = \sin \frac{7\,\pi}{20} = \cos 27^\circ = \sin 63^\circ = \frac{\sqrt{2}(\sqrt{5}-1)+2\sqrt{5+\sqrt{5}}}{8},

tg ⁡ 3 π 20 = ctg ⁡ 7 π 20 = tg ⁡ 27 ∘ = ctg ⁡ 63 ∘ = 5 − 1 − 5 − 2 5, {\displaystyle \operatorname {tg} {\frac {3\,\pi }{20}}=\operatorname {ctg} {\frac {7\,\pi }{20}}=\operatorname {tg} 27^{\circ }=\operatorname {ctg} 63^{\circ }={{\sqrt {5}}-1-{\sqrt {5-2{\sqrt {5}}}}},} \operatorname{tg} \frac{3\,\pi}{20} = \operatorname{ctg} \frac{7\,\pi}{20} = \operatorname{tg} 27^\circ = \operatorname{ctg} 63^\circ = {\sqrt{5}-1-\sqrt{5-2\sqrt{5}}},

ctg ⁡ 3 π 20 = tg ⁡ 7 π 20 = ctg ⁡ 27 ∘ = tg ⁡ 63 ∘ = 5 − 1 + 5 − 2 5, {\displaystyle \operatorname {ctg} {\frac {3\,\pi }{20}}=\operatorname {tg} {\frac {7\,\pi }{20}}=\operatorname {ctg} 27^{\circ }=\operatorname {tg} 63^{\circ }={{\sqrt {5}}-1+{\sqrt {5-2{\sqrt {5}}}}},} \operatorname{ctg} \frac{3\,\pi}{20} = \operatorname{tg} \frac{7\,\pi}{20} = \operatorname{ctg} 27^\circ = \operatorname{tg} 63^\circ = {\sqrt{5}-1+\sqrt{5-2\sqrt{5}}},

sin ⁡ 11 π 60 = cos ⁡ 19 π 60 = sin ⁡ 33 ∘ = cos ⁡ 57 ∘ = 2 ( 3 + 1 ) ( 5 − 1 ) + 2 ( 3 − 1 ) 5 + 5 16, {\displaystyle \sin {\frac {11\,\pi }{60}}=\cos {\frac {19\,\pi }{60}}=\sin 33^{\circ }=\cos 57^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}+1)({\sqrt {5}}-1)+2({\sqrt {3}}-1){\sqrt {5+{\sqrt {5}}}}}{16}},} \sin \frac{11\,\pi}{60} = \cos \frac{19\,\pi}{60} = \sin 33^\circ = \cos 57^\circ = \frac{\sqrt{2}(\sqrt{3}+1)(\sqrt{5}-1)+2(\sqrt{3}-1)\sqrt{5+\sqrt{5}}}{16},

cos ⁡ 11 π 60 = sin ⁡ 19 π 60 = cos ⁡ 33 ∘ = sin ⁡ 57 ∘ = − 2 ( 3 − 1 ) ( 5 − 1 ) + 2 ( 3 + 1 ) 5 + 5 16, {\displaystyle \cos {\frac {11\,\pi }{60}}=\sin {\frac {19\,\pi }{60}}=\cos 33^{\circ }=\sin 57^{\circ }={\frac {-{\sqrt {2}}({\sqrt {3}}-1)({\sqrt {5}}-1)+2({\sqrt {3}}+1){\sqrt {5+{\sqrt {5}}}}}{16}},} \cos \frac{11\,\pi}{60} = \sin \frac{19\,\pi}{60} = \cos 33^\circ = \sin 57^\circ = \frac{-\sqrt{2}(\sqrt{3}-1)(\sqrt{5}-1)+2(\sqrt{3}+1)\sqrt{5+\sqrt{5}}}{16},

tg ⁡ 11 π 60 = ctg ⁡ 19 π 60 = tg ⁡ 33 ∘ = ctg ⁡ 57 ∘ = − 2 ( 5 + 2 ) + 3 ( 3 + 5 ) + ( 2 − 3 ) ( 3 ( 5 + 1 ) − 2 ) 5 − 2 5 2, {\displaystyle \operatorname {tg} {\frac {11\,\pi }{60}}=\operatorname {ctg} {\frac {19\,\pi }{60}}=\operatorname {tg} 33^{\circ }=\operatorname {ctg} 57^{\circ }={\frac {-2({\sqrt {5}}+2)+{\sqrt {3}}(3+{\sqrt {5}})+(2-{\sqrt {3}})({\sqrt {3}}({\sqrt {5}}+1)-2){\sqrt {5-2{\sqrt {5}}}}}{2}},} \operatorname{tg} \frac{11\,\pi}{60} = \operatorname{ctg} \frac{19\,\pi}{60} = \operatorname{tg} 33^\circ = \operatorname{ctg} 57^\circ = \frac{-2(\sqrt{5}+2)+\sqrt{3}(3+\sqrt{5})+(2-\sqrt{3})(\sqrt{3}(\sqrt{5}+1)-2)\sqrt{5-2\sqrt{5}}}{2},

ctg ⁡ 11 π 60 = tg ⁡ 19 π 60 = ctg ⁡ 33 ∘ = tg ⁡ 57 ∘ = − 2 ( 2 ( 5 + 2 ) + 3 ( 3 + 5 ) ) + ( 3 ( 5 − 1 ) + 2 ) 2 ( 25 + 11 5 ) 4, {\displaystyle \operatorname {ctg} {\frac {11\,\pi }{60}}=\operatorname {tg} {\frac {19\,\pi }{60}}=\operatorname {ctg} 33^{\circ }=\operatorname {tg} 57^{\circ }={\frac {-2(2({\sqrt {5}}+2)+{\sqrt {3}}(3+{\sqrt {5}}))+({\sqrt {3}}({\sqrt {5}}-1)+2){\sqrt {2(25+11{\sqrt {5}})}}}{4}},} \operatorname{ctg} \frac{11\,\pi}{60} = \operatorname{tg} \frac{19\,\pi}{60} = \operatorname{ctg} 33^\circ = \operatorname{tg} 57^\circ = \frac{-2(2(\sqrt{5}+2)+\sqrt{3}(3+\sqrt{5}))+(\sqrt{3}(\sqrt{5}-1)+2)\sqrt{2(25+11\sqrt{5})}}{4},

sin ⁡ 13 π 60 = cos ⁡ 17 π 60 = sin ⁡ 39 ∘ = cos ⁡ 51 ∘ = 2 ( 3 + 1 ) ( 5 + 1 ) − 2 ( 3 − 1 ) 5 − 5 16, {\displaystyle \sin {\frac {13\,\pi }{60}}=\cos {\frac {17\,\pi }{60}}=\sin 39^{\circ }=\cos 51^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}+1)({\sqrt {5}}+1)-2({\sqrt {3}}-1){\sqrt {5-{\sqrt {5}}}}}{16}},} \sin \frac{13\,\pi}{60} = \cos \frac{17\,\pi}{60} = \sin 39^\circ = \cos 51^\circ = \frac{\sqrt{2}(\sqrt{3}+1)(\sqrt{5}+1)-2(\sqrt{3}-1)\sqrt{5-\sqrt{5}}}{16},

cos ⁡ 13 π 60 = sin ⁡ 17 π 60 = cos ⁡ 39 ∘ = sin ⁡ 51 ∘ = 2 ( 3 − 1 ) ( 5 + 1 ) + 2 ( 3 + 1 ) 5 − 5 16, {\displaystyle \cos {\frac {13\,\pi }{60}}=\sin {\frac {17\,\pi }{60}}=\cos 39^{\circ }=\sin 51^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}-1)({\sqrt {5}}+1)+2({\sqrt {3}}+1){\sqrt {5-{\sqrt {5}}}}}{16}},} \cos \frac{13\,\pi}{60} = \sin \frac{17\,\pi}{60} = \cos 39^\circ = \sin 51^\circ = \frac{\sqrt{2}(\sqrt{3}-1)(\sqrt{5}+1)+2(\sqrt{3}+1)\sqrt{5-\sqrt{5}}}{16},

tg ⁡ 13 π 60 = ctg ⁡ 17 π 60 = tg ⁡ 39 ∘ = ctg ⁡ 51 ∘ = − 2 ( 2 ( 5 − 2 ) + 3 ( 3 − 5 ) ) + ( 3 ( 5 + 1 ) + 2 ) 2 ( 25 − 11 5 ) 4, {\displaystyle \operatorname {tg} {\frac {13\,\pi }{60}}=\operatorname {ctg} {\frac {17\,\pi }{60}}=\operatorname {tg} 39^{\circ }=\operatorname {ctg} 51^{\circ }={\frac {-2(2({\sqrt {5}}-2)+{\sqrt {3}}(3-{\sqrt {5}}))+({\sqrt {3}}({\sqrt {5}}+1)+2){\sqrt {2(25-11{\sqrt {5}})}}}{4}},} \operatorname{tg} \frac{13\,\pi}{60} = \operatorname{ctg} \frac{17\,\pi}{60} = \operatorname{tg} 39^\circ = \operatorname{ctg} 51^\circ = \frac{-2(2(\sqrt{5}-2)+\sqrt{3}(3-\sqrt{5}))+(\sqrt{3}(\sqrt{5}+1)+2)\sqrt{2(25-11\sqrt{5})}}{4},

ctg ⁡ 13 π 60 = tg ⁡ 17 π 60 = ctg ⁡ 39 ∘ = tg ⁡ 51 ∘ = − 2 ( 2 ( 5 − 2 ) − 3 ( 3 − 5 ) ) + ( 3 ( 5 + 1 ) − 2 ) 2 ( 25 − 11 5 ) 4, {\displaystyle \operatorname {ctg} {\frac {13\,\pi }{60}}=\operatorname {tg} {\frac {17\,\pi }{60}}=\operatorname {ctg} 39^{\circ }=\operatorname {tg} 51^{\circ }={\frac {-2(2({\sqrt {5}}-2)-{\sqrt {3}}(3-{\sqrt {5}}))+({\sqrt {3}}({\sqrt {5}}+1)-2){\sqrt {2(25-11{\sqrt {5}})}}}{4}},} \operatorname{ctg} \frac{13\,\pi}{60} = \operatorname{tg} \frac{17\,\pi}{60} = \operatorname{ctg} 39^\circ = \operatorname{tg} 51^\circ = \frac{-2(2(\sqrt{5}-2)-\sqrt{3}(3-\sqrt{5}))+(\sqrt{3}(\sqrt{5}+1)-2)\sqrt{2(25-11\sqrt{5})}}{4},

sin ⁡ 7 π 30 = cos ⁡ 8 π 30 = sin ⁡ 42 ∘ = cos ⁡ 48 ∘ = − ( 5 − 1 ) + 6 ( 5 + 5 ) 8, {\displaystyle \sin {\frac {7\,\pi }{30}}=\cos {\frac {8\,\pi }{30}}=\sin 42^{\circ }=\cos 48^{\circ }={\frac {-({\sqrt {5}}-1)+{\sqrt {6(5+{\sqrt {5}})}}}{8}},} \sin \frac{7\,\pi}{30} = \cos \frac{8\,\pi}{30} = \sin 42^\circ = \cos 48^\circ = \frac{-(\sqrt{5}-1)+\sqrt{6(5+\sqrt{5})}}{8},

cos ⁡ 7 π 30 = sin ⁡ 8 π 30 = cos ⁡ 42 ∘ = sin ⁡ 48 ∘ = 3 ( 5 − 1 ) + 2 ( 5 + 5 ) 8, {\displaystyle \cos {\frac {7\,\pi }{30}}=\sin {\frac {8\,\pi }{30}}=\cos 42^{\circ }=\sin 48^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}-1)+{\sqrt {2(5+{\sqrt {5}})}}}{8}},} \cos \frac{7\,\pi}{30} = \sin \frac{8\,\pi}{30} = \cos 42^\circ = \sin 48^\circ = \frac{\sqrt{3}(\sqrt{5}-1)+\sqrt{2(5+\sqrt{5})}}{8},

tg ⁡ 7 π 30 = ctg ⁡ 8 π 30 = tg ⁡ 42 ∘ = ctg ⁡ 48 ∘ = 3 ( 5 + 1 ) − 2 ( 5 + 5 ) 2, {\displaystyle \operatorname {tg} {\frac {7\,\pi }{30}}=\operatorname {ctg} {\frac {8\,\pi }{30}}=\operatorname {tg} 42^{\circ }=\operatorname {ctg} 48^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2(5+{\sqrt {5}})}}}{2}},} \operatorname{tg} \frac{7\,\pi}{30} = \operatorname{ctg} \frac{8\,\pi}{30} = \operatorname{tg} 42^\circ = \operatorname{ctg} 48^\circ = \frac{\sqrt{3}(\sqrt{5}+1)-\sqrt{2(5+\sqrt{5})}}{2},

ctg ⁡ 7 π 30 = tg ⁡ 8 π 30 = ctg ⁡ 42 ∘ = tg ⁡ 48 ∘ = 3 ( 3 − 5 ) + 2 ( 25 − 11 5 ) 2, {\displaystyle \operatorname {ctg} {\frac {7\,\pi }{30}}=\operatorname {tg} {\frac {8\,\pi }{30}}=\operatorname {ctg} 42^{\circ }=\operatorname {tg} 48^{\circ }={\frac {{\sqrt {3}}(3-{\sqrt {5}})+{\sqrt {2(25-11{\sqrt {5}})}}}{2}},} \operatorname{ctg} \frac{7\,\pi}{30} = \operatorname{tg} \frac{8\,\pi}{30} = \operatorname{ctg} 42^\circ = \operatorname{tg} 48^\circ = \frac{\sqrt{3}(3-\sqrt{5})+\sqrt{2(25-11\sqrt{5})}}{2},

tg ⁡ π 120 = ctg ⁡ 59 π 120 = tg ⁡ 1.5 ∘ = ctg ⁡ 88.5 ∘ = 8 − 2 ( 2 − 3 ) ( 3 − 5 ) − 2 ( 2 + 3 ) ( 5 + 5 ) 8 + 2 ( 2 − 3 ) ( 3 − 5 ) + 2 ( 2 + 3 ) ( 5 + 5 ), {\displaystyle \operatorname {tg} {\frac {\pi }{120}}=\operatorname {ctg} {\frac {59\,\pi }{120}}=\operatorname {tg} 1.5^{\circ }=\operatorname {ctg} 88.5^{\circ }={\sqrt {\frac {8-{\sqrt {2(2-{\sqrt {3}})(3-{\sqrt {5}})}}-{\sqrt {2(2+{\sqrt {3}})(5+{\sqrt {5}})}}}{8+{\sqrt {2(2-{\sqrt {3}})(3-{\sqrt {5}})}}+{\sqrt {2(2+{\sqrt {3}})(5+{\sqrt {5}})}}}}},} \operatorname{tg} \frac{\pi}{120} = \operatorname{ctg} \frac{59\,\pi}{120} = \operatorname{tg} 1.5^\circ = \operatorname{ctg} 88.5^\circ = \sqrt{\frac{8-\sqrt{2(2-\sqrt{3})(3-\sqrt{5})} - \sqrt{ 2(2+\sqrt{3})(5+\sqrt{5})}}{8+\sqrt{2(2-\sqrt{3})(3-\sqrt{5})}+\sqrt{2(2+\sqrt{3})(5+\sqrt{5})} }},

cos ⁡ π 240 = sin ⁡ 119 π 240 = cos ⁡ 0.75 ∘ = sin ⁡ 89.25 ∘ = 1 16 ( 2 − 2 + 2 ( 2 ( 5 + 5 ) + 3 ( 1 − 5 ) ) + {\displaystyle \cos {\frac {\pi }{240}}=\sin {\frac {119\,\pi }{240}}=\cos 0.75^{\circ }=\sin 89.25^{\circ }={\frac {1}{16}}\left({\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}\left({\sqrt {2(5+{\sqrt {5}})}}+{\sqrt {3}}(1-{\sqrt {5}})\right)+\right.} \cos \frac{\pi}{240} = \sin \frac{119\,\pi}{240} = \cos 0.75^\circ = \sin 89.25^\circ = \frac{1}{16} \left( \sqrt{2-\sqrt{2+\sqrt{2}}} \left(\sqrt{2(5+\sqrt{5})}+\sqrt{3}(1-\sqrt{5}) \right) + \right. + 2 + 2 + 2 ( 6 ( 5 + 5 ) + 5 − 1 ) ), {\displaystyle \left.+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}\left({\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right)\right),}  \left. + \sqrt{2+\sqrt{2+\sqrt{2}}} \left (\sqrt{6(5+\sqrt{5})}+\sqrt{5} - 1 \right) \right),

cos ⁡ π 17 = sin ⁡ 15 π 34 = 1 8 2 ( 2 3 17 − 2 ( 85 + 19 17 ) + 17 + 2 ( 17 − 17 ) + 17 + 15 ). {\displaystyle \cos {\frac {\pi }{17}}=\sin {\frac {15\,\pi }{34}}={\frac {1}{8}}{\sqrt {2\left(2{\sqrt {3{\sqrt {17}}-{\sqrt {2(85+19{\sqrt {17}})}}+17}}+{\sqrt {2(17-{\sqrt {17}})}}+{\sqrt {17}}+15\right)}}.} \cos \frac{\pi}{17} = \sin \frac{15\,\pi}{34} = \frac{1}{8}\sqrt{2 \left(2\sqrt{3\sqrt{17}-\sqrt{2(85+19\sqrt{17})} +17}+\sqrt{2(17-\sqrt{17})}+\sqrt{17}+15 \right)}.

sin ⁡ π 2 n + 1 = 1 2 2 − 2 + ⋯ + 2 ⏟ n, n ∈ N {\displaystyle \sin {\pi \over 2^{n+1}}={1 \over 2}\underbrace {\sqrt {2-{\sqrt {2+\dots +{\sqrt {2}}}}}} _{n},n\in \mathbb {N} } {\displaystyle \sin {\pi \over 2^{n+1}}={1 \over 2}\underbrace {\sqrt {2-{\sqrt {2+\dots +{\sqrt {2}}}}}} _{n},n\in \mathbb {N} }

cos ⁡ π 2 n + 1 = 1 2 2 + 2 + ⋯ + 2 ⏟ n, n ∈ N {\displaystyle \cos {\pi \over 2^{n+1}}={1 \over 2}\underbrace {\sqrt {2+{\sqrt {2+\dots +{\sqrt {2}}}}}} _{n},n\in \mathbb {N} } {\displaystyle \cos {\pi \over 2^{n+1}}={1 \over 2}\underbrace {\sqrt {2+{\sqrt {2+\dots +{\sqrt {2}}}}}} _{n},n\in \mathbb {N} }

sin ⁡ π 3 ⋅ 2 n = 1 2 2 − 2 + ⋯ + 3 ⏟ n, n ≥ 2 {\displaystyle \sin {\pi \over 3\cdot 2^{n}}={1 \over 2}\underbrace {\sqrt {2-{\sqrt {2+\dots +{\sqrt {3}}}}}} _{n},n\geq 2} {\displaystyle \sin {\pi \over 3\cdot 2^{n}}={1 \over 2}\underbrace {\sqrt {2-{\sqrt {2+\dots +{\sqrt {3}}}}}} _{n},n\geq 2}

cos ⁡ π 3 ⋅ 2 n = 1 2 2 + 2 + ⋯ + 3 ⏟ n, n ≥ 2 {\displaystyle \cos {\pi \over 3\cdot 2^{n}}={1 \over 2}\underbrace {\sqrt {2+{\sqrt {2+\dots +{\sqrt {3}}}}}} _{n},n\geq 2} {\displaystyle \cos {\pi \over 3\cdot 2^{n}}={1 \over 2}\underbrace {\sqrt {2+{\sqrt {2+\dots +{\sqrt {3}}}}}} _{n},n\geq 2}


Источник: https://ru.wikipedia.org/wiki/%D0%A2%D1%80%D0%B8%D0%B3%D0%BE%D0%BD%D0%BE%D0%BC%D0%B5%D1%82%D1%80%D0%B8%D1%87%D0%B5%D1%81%D0%BA%D0%B8%D0%B5_%D1%84%D1%83%D0%BD%D0%BA%D1%86%D0%B8%D0%B8



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