三角函數精確值是利用三角函數的公式將特定的三角函數值加以化簡，並以數學根式或分數表示。 用根式或分數表達的精確三角函數有時很有用，主要用於簡化的解決某些方程式能進一步化簡。 根據尼雲定理，有理數度數的角的正弦值，其中的有理數僅有0， ± 1 2 {\displaystyle \pm {\frac {1}{2}}} ，±1。 更多資訊 , ... 相同角度的轉換表 角度單位 值 轉 0 {\displaystyle 0} 1 12 {\displaystyle {\frac {1}{12}}} 1 8 {\displaystyle {\frac {1}{8}}} 1 6 {\displaystyle {\frac {1}{6}}} 1 4 {\displaystyle {\frac {1}{4}}} 1 2 {\displaystyle {\frac {1}{2}}} 3 4 {\displaystyle {\frac {3}{4}}} 1 {\displaystyle 1} 角度 0 ∘ {\displaystyle 0^{\circ }} 30 ∘ {\displaystyle 30^{\circ }} 45 ∘ {\displaystyle 45^{\circ }} 60 ∘ {\displaystyle 60^{\circ }} 90 ∘ {\displaystyle 90^{\circ }} 180 ∘ {\displaystyle 180^{\circ }} 270 ∘ {\displaystyle 270^{\circ }} 360 ∘ {\displaystyle 360^{\circ }} 弧度 0 {\displaystyle 0} π 6 {\displaystyle {\frac {\pi }{6}}} π 4 {\displaystyle {\frac {\pi }{4}}} π 3 {\displaystyle {\frac {\pi }{3}}} π 2 {\displaystyle {\frac {\pi }{2}}} π {\displaystyle \pi } 3 π 2 {\displaystyle {\frac {3\pi }{2}}} 2 π {\displaystyle 2\pi } 梯度 0 g {\displaystyle 0^{g}} 33 1 3 g {\displaystyle 33{\frac {1}{3}}^{g}} 50 g {\displaystyle 50^{g}} 66 2 3 g {\displaystyle 66{\frac {2}{3}}^{g}} 100 g {\displaystyle 100^{g}} 200 g {\displaystyle 200^{g}} 300 g {\displaystyle 300^{g}} 400 g {\displaystyle 400^{g}} 關閉 計算方式 基於常識 例如：0°、30°、45° 單位圓 經由半角公式的計算 參見：半角公式 例如：15°、22.5° sin ( x 2 ) = ± 1 2 ( 1 − cos x ) {\displaystyle \sin \left({\frac {x}{2}}\right)=\pm \,{\sqrt {{\tfrac {1}{2}}(1-\cos x)}}} cos ( x 2 ) = ± 1 2 ( 1 + cos x ) {\displaystyle \cos \left({\frac {x}{2}}\right)=\pm \,{\sqrt {{\tfrac {1}{2}}(1+\cos x)}}} 利用三倍角公式求 1 3 {\displaystyle {\frac {1}{3}}\,} 角 參見：三倍角公式 例如：10°、20°、7°......等等，非三的倍數的角的精確值。 sin 3 θ = 3 sin θ − 4 sin 3 θ {\displaystyle \sin 3\theta =3\sin \theta -4\sin ^{3}\theta \,} cos 3 θ = 4 cos 3 θ − 3 cos θ {\displaystyle \cos 3\theta =4\cos ^{3}\theta -3\cos \theta \,} 把它改為 sin θ = 3 sin 1 3 θ − 4 sin 3 1 3 θ {\displaystyle \sin \theta =3\sin {\frac {1}{3}}\theta -4\sin ^{3}{\frac {1}{3}}\theta \,} cos θ = 4 cos 3 1 3 θ − 3 cos 1 3 θ {\displaystyle \cos \theta =4\cos ^{3}{\frac {1}{3}}\theta -3\cos {\frac {1}{3}}\theta \,} 把 cos 1 3 θ {\displaystyle \cos {\frac {1}{3}}\theta \,} 當成未知數， cos θ {\displaystyle \cos \theta \,} 當成常數項 解一元三次方程式即可求出 例如： sin π 9 = sin 20 ∘ = − 3 16 + − 1 256 3 + − 3 16 − − 1 256 3 {\displaystyle \sin {\frac {\pi }{9}}=\sin 20^{\circ }={\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}+{\sqrt {-{\frac {1}{256}}}}}}+{\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}-{\sqrt {-{\frac {1}{256}}}}}}} 同樣地，若角度代未知數，則會得到三分之一角公式。 經由歐拉公式的計算 更多資訊：歐拉公式 cos θ n = ℜ ( cos θ + i sin θ n ) = 1 2 ( cos θ + i sin θ n + cos θ − i sin θ n ) {\displaystyle \cos {\frac {\theta }{n}}=\Re \left({\sqrt[{n}]{\cos \theta +i\sin \theta }}\right)={\frac {1}{2}}\left({\sqrt[{n}]{\cos \theta +i\sin \theta }}+{\sqrt[{n}]{\cos \theta -i\sin \theta }}\right)} sin θ n = ℑ ( cos θ + i sin θ n ) = 1 2 i ( cos θ + i sin θ n − cos θ − i sin θ n ) {\displaystyle \sin {\frac {\theta }{n}}=\Im \left({\sqrt[{n}]{\cos \theta +i\sin \theta }}\right)={\frac {1}{2i}}\left({\sqrt[{n}]{\cos \theta +i\sin \theta }}-{\sqrt[{n}]{\cos \theta -i\sin \theta }}\right)} 例如： sin 1 ∘ = 1 2 i ( cos 3 ∘ + i sin 3 ∘ 3 − cos 3 ∘ − i sin 3 ∘ 3 ) {\displaystyle \sin {1^{\circ }}={\frac {1}{2i}}\left({\sqrt[{3}]{\cos {3^{\circ }}+i\sin {3^{\circ }}}}-{\sqrt[{3}]{\cos {3^{\circ }}-i\sin {3^{\circ }}}}\right)} = 1 4 2 3 i { [ 2 ( 1 + 3 ) 5 + 5 + 2 ( 5 − 1 ) ( 3 − 1 ) ] + i [ 2 ( 1 − 3 ) 5 + 5 + 2 ( 5 − 1 ) ( 3 + 1 ) ] 3 {\displaystyle ={\frac {1}{4{\sqrt[{3}]{2}}i}}{\Bigg \{}{\sqrt[{3}]{\left[2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)\right]+i\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]}}} − [ 2 ( 1 + 3 ) 5 + 5 + 2 ( 5 − 1 ) ( 3 − 1 ) ] − i [ 2 ( 1 − 3 ) 5 + 5 + 2 ( 5 − 1 ) ( 3 + 1 ) ] 3 } {\displaystyle -{\sqrt[{3}]{\left[2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)\right]-i\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]}}{\Bigg \}}} [1] 經由和角公式的計算 例如：21° = 9° + 12° sin ( x ± y ) = sin ( x ) cos ( y ) ± cos ( x ) sin ( y ) {\displaystyle \sin(x\pm y)=\sin(x)\cos(y)\pm \cos(x)\sin(y)\,} cos ( x ± y ) = cos ( x ) cos ( y ) ∓ sin ( x ) sin ( y ) {\displaystyle \cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)\,} 經由托勒密定理的計算 更多資訊：托勒密定理和弦函數 Chord(36°) = a/b = 1/φ, 根據托勒密定理 例如：18° 根據托勒密定理，在圓內接四邊形ABCD中， a 2 + a b = b 2 {\displaystyle a^{2}+ab=b^{2}} ( a b ) 2 + a b = 1 {\displaystyle \left({\frac {a}{b}}\right)^{2}+{\frac {a}{b}}=1} c r d 36 ∘ = c r d ( ∠ A D B ) = a b = 5 − 1 2 {\displaystyle \mathrm {crd} \ {36^{\circ }}=\mathrm {crd} \left(\angle \mathrm {ADB} \right)={\frac {a}{b}}={\frac {{\sqrt {5}}-1}{2}}} c r d θ = 2 sin θ 2 {\displaystyle \mathrm {crd} \ {\theta }=2\sin {\frac {\theta }{2}}\,} sin 18 ∘ = 5 − 1 4 {\displaystyle \sin {18^{\circ }}={\frac {{\sqrt {5}}-1}{4}}} 三角函數精確值列表 由於三角函數的特性，大於45°角度的三角函數值，可以經由自0°～45°的角度的三角函數值的相關的計算取得。 0°：根本 sin 0 = 0 {\displaystyle \sin 0=0\,} cos 0 = 1 {\displaystyle \cos 0=1\,} tan 0 = 0 {\displaystyle \tan 0=0\,} 1°：2°的一半 sin 1 ∘ = 1 + 3 i 16 4 30 − 8 15 + 3 5 + 8 5 + 5 + 4 10 − 4 6 − 4 2 + ( 4 30 + 8 15 + 3 5 + 8 5 + 5 − 4 10 − 4 6 + 4 2 ) i 3 + {\displaystyle \sin {1^{\circ }}={\frac {1+{\sqrt {3}}i}{16}}{\sqrt[{3}]{4{\sqrt {30}}-8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}+4{\sqrt {10}}-4{\sqrt {6}}-4{\sqrt {2}}+\left(4{\sqrt {30}}+8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}-4{\sqrt {10}}-4{\sqrt {6}}+4{\sqrt {2}}\right)i}}+} 1 − 3 i 16 4 30 − 8 15 + 3 5 + 8 5 + 5 + 4 10 − 4 6 − 4 2 − ( 4 30 + 8 15 + 3 5 + 8 5 + 5 − 4 10 − 4 6 + 4 2 ) i 3 {\displaystyle {\frac {1-{\sqrt {3}}i}{16}}{\sqrt[{3}]{4{\sqrt {30}}-8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}+4{\sqrt {10}}-4{\sqrt {6}}-4{\sqrt {2}}-\left(4{\sqrt {30}}+8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}-4{\sqrt {10}}-4{\sqrt {6}}+4{\sqrt {2}}\right)i}}} [2] 1.5°：正一百二十邊形 sin ( π 120 ) = sin ( 1.5 ∘ ) = ( 2 + 2 ) ( 15 + 3 − 10 − 2 5 ) − ( 2 − 2 ) ( 30 − 6 5 + 5 + 1 ) 16 {\displaystyle \sin \left({\frac {\pi }{120}}\right)=\sin \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)-\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)}{16}}} cos ( π 120 ) = cos ( 1.5 ∘ ) = ( 2 + 2 ) ( 30 − 6 5 + 5 + 1 ) + ( 2 − 2 ) ( 15 + 3 − 10 − 2 5 ) 16 {\displaystyle \cos \left({\frac {\pi }{120}}\right)=\cos \left(1.5^{\circ }\right)={\frac {\left({\sqrt {2+{\sqrt {2}}}}\right)\left({\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}+1\right)+\left({\sqrt {2-{\sqrt {2}}}}\right)\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}{16}}} 1.875°：正九十六邊形 sin ( π 96 ) = sin ( 1.875 ∘ ) = 1 2 2 − 2 + 2 + 2 + 3 {\displaystyle \sin \left({\frac {\pi }{96}}\right)=\sin \left(1.875^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}} cos ( π 96 ) = cos ( 1.875 ∘ ) = 1 2 2 + 2 + 2 + 2 + 3 {\displaystyle \cos \left({\frac {\pi }{96}}\right)=\cos \left(1.875^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}}}} tan ( π 96 ) = tan ( 1.875 ∘ ) = 2 − 3 + 2 + 2 + 2 3 + 2 + 2 + 2 + 2 {\displaystyle \tan \left({\frac {\pi }{96}}\right)=\tan \left(1.875^{\circ }\right)={\frac {\sqrt {2-{\sqrt {{\sqrt {{\sqrt {{\sqrt {3}}+2}}+2}}+2}}}}{\sqrt {{\sqrt {{\sqrt {{\sqrt {{\sqrt {3}}+2}}+2}}+2}}+2}}}} 2°：6°的三分之一 sin 2 ∘ = 1 2 i ( cos 6 ∘ + i sin 6 ∘ 3 − cos 6 ∘ − i sin 6 ∘ 3 ) {\displaystyle \sin {2^{\circ }}={\frac {1}{2i}}\left({\sqrt[{3}]{\cos {6^{\circ }}+i\sin {6^{\circ }}}}-{\sqrt[{3}]{\cos {6^{\circ }}-i\sin {6^{\circ }}}}\right)} = 1 4 i { [ 2 ( 5 − 5 ) + 3 ( 5 + 1 ) ] + i [ 6 ( 5 − 5 ) − 5 − 1 ] 3 {\displaystyle ={\frac {1}{4i}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]+i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}} − [ 2 ( 5 − 5 ) + 3 ( 5 + 1 ) ] − i [ 6 ( 5 − 5 ) − 5 − 1 ] 3 } {\displaystyle -{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]-i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}{\Bigg \}}} cos 2 ∘ = 1 2 ( cos 6 ∘ + i sin 6 ∘ 3 + cos 6 ∘ − i sin 6 ∘ 3 ) {\displaystyle \cos {2^{\circ }}={\frac {1}{2}}\left({\sqrt[{3}]{\cos {6^{\circ }}+i\sin {6^{\circ }}}}+{\sqrt[{3}]{\cos {6^{\circ }}-i\sin {6^{\circ }}}}\right)} = 1 4 { [ 2 ( 5 − 5 ) + 3 ( 5 + 1 ) ] + i [ 6 ( 5 − 5 ) − 5 − 1 ] 3 {\displaystyle ={\frac {1}{4}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]+i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}} + [ 2 ( 5 − 5 ) + 3 ( 5 + 1 ) ] − i [ 6 ( 5 − 5 ) − 5 − 1 ] 3 } {\displaystyle +{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]-i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}{\Bigg \}}} 2.25°：正八十邊形 sin ( π 80 ) = sin ( 2.25 ∘ ) = − 2 2 2 2 5 + 5 + 4 + 4 + 8 4 {\displaystyle \sin \left({\frac {\pi }{80}}\right)=\sin \left(2.25^{\circ }\right)={\frac {\sqrt {-2{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+8}}{4}}} cos ( π 80 ) = cos ( 2.25 ∘ ) = 2 2 2 2 5 + 5 + 4 + 4 + 8 4 {\displaystyle \cos \left({\frac {\pi }{80}}\right)=\cos \left(2.25^{\circ }\right)={\frac {\sqrt {2{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+8}}{4}}} tan ( π 80 ) = tan ( 2.25 ∘ ) = − 2 2 2 5 + 5 + 4 + 4 + 4 2 2 2 5 + 5 + 4 + 4 + 4 {\displaystyle \tan \left({\frac {\pi }{80}}\right)=\tan \left(2.25^{\circ }\right)={\frac {\sqrt {-{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}} cot ( π 80 ) = cot ( 2.25 ∘ ) = 2 2 2 5 + 5 + 4 + 4 + 4 − 2 2 2 5 + 5 + 4 + 4 + 4 {\displaystyle \cot \left({\frac {\pi }{80}}\right)=\cot \left(2.25^{\circ }\right)={\frac {\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}{\sqrt {-{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}} sec ( π 80 ) = sec ( 2.25 ∘ ) = 2 2 2 2 2 5 + 5 + 4 + 4 + 4 {\displaystyle \sec \left({\frac {\pi }{80}}\right)=\sec \left(2.25^{\circ }\right)={\frac {2{\sqrt {2}}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}} csc ( π 80 ) = csc ( 2.25 ∘ ) = 2 2 − 2 2 2 5 + 5 + 4 + 4 + 4 {\displaystyle \csc \left({\frac {\pi }{80}}\right)=\csc \left(2.25^{\circ }\right)={\frac {2{\sqrt {2}}}{\sqrt {-{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {2}}{\sqrt {{\sqrt {5}}+5}}+4}}+4}}+4}}}} 2.8125°：正六十四邊形 sin ( π 64 ) = sin ( 2.8125 ∘ ) = 1 2 2 − 2 + 2 + 2 + 2 {\displaystyle \sin \left({\frac {\pi }{64}}\right)=\sin \left(2.8125^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}} cos ( π 64 ) = cos ( 2.8125 ∘ ) = 1 2 2 + 2 + 2 + 2 + 2 {\displaystyle \cos \left({\frac {\pi }{64}}\right)=\cos \left(2.8125^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}}}} 3°：正六十邊形 sin π 60 = sin 3 ∘ = 1 4 8 − 3 − 15 − 10 − 2 5 {\displaystyle \sin {\frac {\pi }{60}}=\sin 3^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt {3}}-{\sqrt {15}}-{\sqrt {10-2{\sqrt {5}}}}}}\,} cos π 60 = cos 3 ∘ = 1 4 8 + 3 + 15 + 10 − 2 5 {\displaystyle \cos {\frac {\pi }{60}}=\cos 3^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt {3}}+{\sqrt {15}}+{\sqrt {10-2{\sqrt {5}}}}}}\,} tan π 60 = tan 3 ∘ = 1 4 [ ( 2 − 3 ) ( 3 + 5 ) − 2 ] [ 2 − 2 ( 5 − 5 ) ] {\displaystyle \tan {\frac {\pi }{60}}=\tan 3^{\circ }={\tfrac {1}{4}}\left[(2-{\sqrt {3}})(3+{\sqrt {5}})-2\right]\left[2-{\sqrt {2(5-{\sqrt {5}})}}\right]\,} 3.75°：正四十八邊形 sin ( π 48 ) = sin ( 3.75 ∘ ) = 1 2 2 − 2 + 2 + 3 {\displaystyle \sin \left({\frac {\pi }{48}}\right)=\sin \left(3.75^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}} cos ( π 48 ) = cos ( 3.75 ∘ ) = 1 2 2 + 2 + 2 + 3 {\displaystyle \cos \left({\frac {\pi }{48}}\right)=\cos \left(3.75^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}}}} 4°：12°的三分之一 sin 4 ∘ = 1 2 i ( cos 12 ∘ + i sin 12 ∘ 3 − cos 12 ∘ − i sin 12 ∘ 3 ) {\displaystyle \sin {4^{\circ }}={\frac {1}{2i}}\left({\sqrt[{3}]{\cos {12^{\circ }}+i\sin {12^{\circ }}}}-{\sqrt[{3}]{\cos {12^{\circ }}-i\sin {12^{\circ }}}}\right)} = 1 4 i { [ 6 ( 5 + 5 ) + 5 − 1 ] + i [ 2 ( 5 + 5 ) − 3 ( 5 − 1 ) ] 3 {\displaystyle ={\frac {1}{4i}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]+i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}} − [ 6 ( 5 + 5 ) + 5 − 1 ] − i [ 2 ( 5 + 5 ) − 3 ( 5 − 1 ) ] 3 } {\displaystyle -{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]-i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}{\Bigg \}}} cos 4 ∘ = 1 2 ( cos 12 ∘ + i sin 12 ∘ 3 + cos 12 ∘ − i sin 12 ∘ 3 ) {\displaystyle \cos {4^{\circ }}={\frac {1}{2}}\left({\sqrt[{3}]{\cos {12^{\circ }}+i\sin {12^{\circ }}}}+{\sqrt[{3}]{\cos {12^{\circ }}-i\sin {12^{\circ }}}}\right)} = 1 4 { [ 6 ( 5 + 5 ) + 5 − 1 ] + i [ 2 ( 5 + 5 ) − 3 ( 5 − 1 ) ] 3 {\displaystyle ={\frac {1}{4}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]+i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}} + [ 6 ( 5 + 5 ) + 5 − 1 ] − i [ 2 ( 5 + 5 ) − 3 ( 5 − 1 ) ] 3 } {\displaystyle +{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]-i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}{\Bigg \}}} 4.5°：正四十邊形 sin ( π 40 ) = sin ( 4.5 ∘ ) = 1 2 2 − 2 + 5 + 5 2 {\displaystyle \sin \left({\frac {\pi }{40}}\right)=\sin \left(4.5^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}}} cos ( π 40 ) = cos ( 4.5 ∘ ) = 1 2 2 + 2 + 5 + 5 2 {\displaystyle \cos \left({\frac {\pi }{40}}\right)=\cos \left(4.5^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {\frac {5+{\sqrt {5}}}{2}}}}}}}} 5°：15°的三分之一、正三十六邊形 sin π 36 = sin 5 ∘ = 2 − 2 3 i 2 2 ( 2 − 6 ) 3 − 2 − 3 − ( 1 + 3 i ) 2 ( 2 − 6 ) 3 − 2 − 3 8 {\displaystyle \sin {\frac {\pi }{36}}=\sin 5^{\circ }={\frac {2-2{\sqrt {3}}\mathrm {i} }{2{\sqrt[{3}]{2({\sqrt {2}}-{\sqrt {6}})}}-2-{\sqrt {3}}}}-{\frac {(1+{\sqrt {3}}\mathrm {i} ){\sqrt[{3}]{2({\sqrt {2}}-{\sqrt {6}})}}-2-{\sqrt {3}}}{8}}\,} 5.625°：正三十二邊形 sin ( π 32 ) = sin ( 5.625 ∘ ) = 1 2 2 − 2 + 2 + 2 {\displaystyle \sin \left({\frac {\pi }{32}}\right)=\sin \left(5.625^{\circ }\right)={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}} cos ( π 32 ) = cos ( 5.625 ∘ ) = 1 2 2 + 2 + 2 + 2 {\displaystyle \cos \left({\frac {\pi }{32}}\right)=\cos \left(5.625^{\circ }\right)={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}}}} 6°：正三十邊形 sin π 30 = sin 6 ∘ = 1 8 [ 6 ( 5 − 5 ) − 5 − 1 ] {\displaystyle \sin {\frac {\pi }{30}}=\sin 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]\,} cos π 30 = cos 6 ∘ = 1 8 [ 2 ( 5 − 5 ) + 3 ( 5 + 1 ) ] {\displaystyle \cos {\frac {\pi }{30}}=\cos 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]\,} tan π 30 = tan 6 ∘ = 1 2 [ 2 ( 5 − 5 ) − 3 ( 5 − 1 ) ] {\displaystyle \tan {\frac {\pi }{30}}=\tan 6^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(5-{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,} cot π 30 = cot 6 ∘ = 1 2 ( 50 + 22 5 + 3 3 + 15 ) {\displaystyle \cot {\frac {\pi }{30}}=\cot 6^{\circ }={\tfrac {1}{2}}\left({\sqrt {50+22{\sqrt {5}}}}+3{\sqrt {3}}+{\sqrt {15}}\right)\,} sec π 30 = sec 6 ∘ = 3 − 5 − 2 5 {\displaystyle \sec {\frac {\pi }{30}}=\sec 6^{\circ }={\sqrt {3}}-{\sqrt {5-2{\sqrt {5}}}}\,} csc π 30 = csc 6 ∘ = 2 + 5 + 15 + 6 5 {\displaystyle \csc {\frac {\pi }{30}}=\csc 6^{\circ }=2+{\sqrt {5}}+{\sqrt {15+6{\sqrt {5}}}}\,} 7.5°：正二十四邊形 sin π 24 = sin 7.5 ∘ = 1 4 8 − 2 6 − 2 2 {\displaystyle \sin {\frac {\pi }{24}}=\sin 7.5^{\circ }={\tfrac {1}{4}}{\sqrt {8-2{\sqrt {6}}-2{\sqrt {2}}}}\,} cos π 24 = cos 7.5 ∘ = 1 4 8 + 2 6 + 2 2 {\displaystyle \cos {\frac {\pi }{24}}=\cos 7.5^{\circ }={\tfrac {1}{4}}{\sqrt {8+2{\sqrt {6}}+2{\sqrt {2}}}}\,} tan π 24 = tan 7.5 ∘ = 6 + 2 − 2 − 3 {\displaystyle \tan {\frac {\pi }{24}}=\tan 7.5^{\circ }={\sqrt {6}}+{\sqrt {2}}-2-{\sqrt {3}}\,} cot π 24 = cot 7.5 ∘ = 6 + 2 + 2 + 3 {\displaystyle \cot {\frac {\pi }{24}}=\cot 7.5^{\circ }={\sqrt {6}}+{\sqrt {2}}+2+{\sqrt {3}}\,} sec π 24 = sec 7.5 ∘ = 16 − 6 6 − 10 2 + 8 3 {\displaystyle \sec {\frac {\pi }{24}}=\sec 7.5^{\circ }={\sqrt {16-6{\sqrt {6}}-10{\sqrt {2}}+8{\sqrt {3}}}}\,} csc π 24 = csc 7.5 ∘ = 16 + 6 6 + 10 2 + 8 3 {\displaystyle \csc {\frac {\pi }{24}}=\csc 7.5^{\circ }={\sqrt {16+6{\sqrt {6}}+10{\sqrt {2}}+8{\sqrt {3}}}}\,} 9°：正二十邊形 更多資訊：二十邊形 sin π 20 = sin 9 ∘ = 1 4 8 − 2 10 + 2 5 {\displaystyle \sin {\frac {\pi }{20}}=\sin 9^{\circ }={\tfrac {1}{4}}{\sqrt {8-2{\sqrt {10+2{\sqrt {5}}}}}}\,} cos π 20 = cos 9 ∘ = 1 4 8 + 2 10 + 2 5 {\displaystyle \cos {\frac {\pi }{20}}=\cos 9^{\circ }={\tfrac {1}{4}}{\sqrt {8+2{\sqrt {10+2{\sqrt {5}}}}}}\,} tan π 20 = tan 9 ∘ = 5 + 1 − 5 + 2 5 {\displaystyle \tan {\frac {\pi }{20}}=\tan 9^{\circ }={\sqrt {5}}+1-{\sqrt {5+2{\sqrt {5}}}}\,} cot π 20 = cot 9 ∘ = 5 + 1 + 5 + 2 5 {\displaystyle \cot {\frac {\pi }{20}}=\cot 9^{\circ }={\sqrt {5}}+1+{\sqrt {5+2{\sqrt {5}}}}\,} 10°：正十八邊形 更多資訊：十八邊形 tan 10 ∘ = − − 1 − 3 i 6 − 12 3 + 36 i 3 − − 1 + 3 i 6 − 12 3 − 36 i 3 + 1 3 {\displaystyle {\tan 10^{\circ }=-{\frac {-1-{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}+36{\rm {i}}}}-{\frac {-1+{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}-36{\rm {i}}}}+{\frac {1}{\sqrt {3}}}}\,} 11.25°：正十六邊形 sin π 16 = sin 11.25 ∘ = 1 2 2 − 2 + 2 {\displaystyle \sin {\frac {\pi }{16}}=\sin 11.25^{\circ }={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}} cos π 16 = cos 11.25 ∘ = 1 2 2 + 2 + 2 {\displaystyle \cos {\frac {\pi }{16}}=\cos 11.25^{\circ }={\frac {1}{2}}{\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}} tan π 16 = tan 11.25 ∘ = 4 + 2 2 − 2 − 1 {\displaystyle \tan {\frac {\pi }{16}}=\tan 11.25^{\circ }={\sqrt {4+2{\sqrt {2}}}}-{\sqrt {2}}-1} cot π 16 = cot 11.25 ∘ = 4 + 2 2 + 2 + 1 {\displaystyle \cot {\frac {\pi }{16}}=\cot 11.25^{\circ }={\sqrt {4+2{\sqrt {2}}}}+{\sqrt {2}}+1} 12°：正十五邊形 更多資訊：十五邊形 sin π 15 = sin 12 ∘ = 1 8 [ 2 ( 5 + 5 ) − 3 ( 5 − 1 ) ] {\displaystyle \sin {\frac {\pi }{15}}=\sin 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,} cos π 15 = cos 12 ∘ = 1 8 [ 6 ( 5 + 5 ) + 5 − 1 ] {\displaystyle \cos {\frac {\pi }{15}}=\cos 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]\,} tan π 15 = tan 12 ∘ = 1 2 [ 3 ( 3 − 5 ) − 2 ( 25 − 11 5 ) ] {\displaystyle \tan {\frac {\pi }{15}}=\tan 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt {3}}(3-{\sqrt {5}})-{\sqrt {2(25-11{\sqrt {5}})}}\right]\,} 15°：正十二邊形 更多資訊：十二邊形 sin π 12 = sin 15 ∘ = 1 4 2 ( 3 − 1 ) {\displaystyle \sin {\frac {\pi }{12}}=\sin 15^{\circ }={\frac {1}{4}}{\sqrt {2}}\left({\sqrt {3}}-1\right)\,} cos π 12 = cos 15 ∘ = 1 4 2 ( 3 + 1 ) {\displaystyle \cos {\frac {\pi }{12}}=\cos 15^{\circ }={\frac {1}{4}}{\sqrt {2}}\left({\sqrt {3}}+1\right)\,} tan π 12 = tan 15 ∘ = 2 − 3 {\displaystyle \tan {\frac {\pi }{12}}=\tan 15^{\circ }=2-{\sqrt {3}}\,} cot π 12 = cot 15 ∘ = 2 + 3 {\displaystyle \cot {\frac {\pi }{12}}=\cot 15^{\circ }=2+{\sqrt {3}}\,} 18°：正十邊形 更多資訊：十邊形 sin π 10 = sin 18 ∘ = 1 4 ( 5 − 1 ) = 1 2 φ − 1 {\displaystyle \sin {\frac {\pi }{10}}=\sin 18^{\circ }={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)={\tfrac {1}{2}}\varphi ^{-1}\,} cos π 10 = cos 18 ∘ = 1 4 2 ( 5 + 5 ) {\displaystyle \cos {\frac {\pi }{10}}=\cos 18^{\circ }={\tfrac {1}{4}}{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,} tan π 10 = tan 18 ∘ = 1 5 5 ( 5 − 2 5 ) {\displaystyle \tan {\frac {\pi }{10}}=\tan 18^{\circ }={\tfrac {1}{5}}{\sqrt {5\left(5-2{\sqrt {5}}\right)}}\,} 20°：正九邊形、60°的三分之一 更多資訊：九邊形 sin π 9 = sin 20 ∘ = − 3 16 + − 1 256 3 + − 3 16 − − 1 256 3 = {\displaystyle \sin {\frac {\pi }{9}}=\sin 20^{\circ }={\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}+{\sqrt {-{\frac {1}{256}}}}}}+{\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}-{\sqrt {-{\frac {1}{256}}}}}}=} 2 − 4 3 ( i − 3 3 − i + 3 3 ) {\displaystyle 2^{-{\frac {4}{3}}}\left({\sqrt[{3}]{i-{\sqrt {3}}}}-{\sqrt[{3}]{i+{\sqrt {3}}}}\right)} cos π 9 = cos 20 ∘ = {\displaystyle \cos {\frac {\pi }{9}}=\cos 20^{\circ }=} 2 − 4 3 ( 1 + i 3 3 + 1 − i 3 3 ) {\displaystyle 2^{-{\frac {4}{3}}}\left({\sqrt[{3}]{1+i{\sqrt {3}}}}+{\sqrt[{3}]{1-i{\sqrt {3}}}}\right)} 21°：9°與12°的和 sin 7 π 60 = sin 21 ∘ = 1 4 8 + 3 − 15 − 10 + 2 5 {\displaystyle \sin {\frac {7\pi }{60}}=\sin 21^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt {3}}-{\sqrt {15}}-{\sqrt {10+2{\sqrt {5}}}}}}\,} cos 7 π 60 = cos 21 ∘ = 1 4 8 − 3 + 15 + 10 + 2 5 {\displaystyle \cos {\frac {7\pi }{60}}=\cos 21^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt {3}}+{\sqrt {15}}+{\sqrt {10+2{\sqrt {5}}}}}}\,} tan 7 π 60 = tan 21 ∘ = 1 4 [ 2 − ( 2 + 3 ) ( 3 − 5 ) ] [ 2 − 2 ( 5 + 5 ) ] {\displaystyle \tan {\frac {7\pi }{60}}=\tan 21^{\circ }={\tfrac {1}{4}}\left[2-\left(2+{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]\,} 360/17°， ( 21 3 17 ) ∘ {\displaystyle \mathbf {\left(21{\frac {3}{17}}\right)^{\circ }} } ， ( 360 17 ) ∘ {\displaystyle \mathbf {\left({\frac {360}{17}}\right)^{\circ }} } ：正十七邊形 更多資訊：十七邊形 cos 2 π 17 = − 1 + 17 + 34 − 2 17 + 2 17 + 3 17 − 34 − 2 17 − 2 34 + 2 17 16 {\displaystyle \operatorname {cos} {2\pi \over 17}={\frac {-1+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}+2{\sqrt {17+3{\sqrt {17}}-{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {34+2{\sqrt {17}}}}}}}{16}}} 22.5°：正八邊形 更多資訊：八邊形 sin π 8 = sin 22.5 ∘ = 1 2 ( 2 − 2 ) {\displaystyle \sin {\frac {\pi }{8}}=\sin 22.5^{\circ }={\tfrac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\right)} cos π 8 = cos 22.5 ∘ = 1 2 ( 2 + 2 ) {\displaystyle \cos {\frac {\pi }{8}}=\cos 22.5^{\circ }={\tfrac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\right)\,} tan π 8 = tan 22.5 ∘ = 2 − 1 {\displaystyle \tan {\frac {\pi }{8}}=\tan 22.5^{\circ }={\sqrt {2}}-1\,} 24°：12°的二倍 sin 2 π 15 = sin 24 ∘ = 1 8 [ 3 ( 5 + 1 ) − 2 5 − 5 ] {\displaystyle \sin {\frac {2\pi }{15}}=\sin 24^{\circ }={\tfrac {1}{8}}\left[{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}\right]\,} cos 2 π 15 = cos 24 ∘ = 1 8 ( 6 5 − 5 + 5 + 1 ) {\displaystyle \cos {\frac {2\pi }{15}}=\cos 24^{\circ }={\tfrac {1}{8}}\left({\sqrt {6}}{\sqrt {5-{\sqrt {5}}}}+{\sqrt {5}}+1\right)\,} tan 2 π 15 = tan 24 ∘ = 1 2 [ 2 ( 25 + 11 5 ) − 3 ( 3 + 5 ) ] {\displaystyle \tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(25+11{\sqrt {5}})}}-{\sqrt {3}}(3+{\sqrt {5}})\right]\,} 180/7°， ( 25 5 7 ) ∘ {\displaystyle \mathbf {\left(25{\frac {5}{7}}\right)^{\circ }} } ， ( 180 7 ) ∘ {\displaystyle \mathbf {\left({\frac {180}{7}}\right)^{\circ }} } ：正七邊形 更多資訊：七邊形 cos π 7 = cos 180 7 ∘ = cos 25 5 7 ∘ = 1 6 + 1 − 3 i 24 28 − 84 3 i 3 + 1 + 3 i 24 28 − 84 3 i 3 {\displaystyle \cos {\frac {\pi }{7}}=\cos {\frac {180}{7}}^{\circ }=\cos 25{\frac {5}{7}}^{\circ }={\frac {1}{6}}+{\frac {1-{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}+{\frac {1+{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}} 27°：12°與15°的和 sin 3 π 20 = sin 27 ∘ = 1 8 [ 2 5 + 5 − 2 ( 5 − 1 ) ] {\displaystyle \sin {\frac {3\pi }{20}}=\sin 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\;({\sqrt {5}}-1)\right]\,} cos 3 π 20 = cos 27 ∘ = 1 8 [ 2 5 + 5 + 2 ( 5 − 1 ) ] {\displaystyle \cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\;\left({\sqrt {5}}-1\right)\right]\,} tan 3 π 20 = tan 27 ∘ = 5 − 1 − 5 − 2 5 {\displaystyle \tan {\frac {3\pi }{20}}=\tan 27^{\circ }={\sqrt {5}}-1-{\sqrt {5-2{\sqrt {5}}}}\,} 30°：正六邊形 更多資訊：六邊形 sin π 6 = sin 30 ∘ = 1 2 {\displaystyle \sin {\frac {\pi }{6}}=\sin 30^{\circ }={\tfrac {1}{2}}\,} cos π 6 = cos 30 ∘ = 1 2 3 {\displaystyle \cos {\frac {\pi }{6}}=\cos 30^{\circ }={\tfrac {1}{2}}{\sqrt {3}}\,} tan π 6 = tan 30 ∘ = 1 3 3 {\displaystyle \tan {\frac {\pi }{6}}=\tan 30^{\circ }={\tfrac {1}{3}}{\sqrt {3}}\,} 33°：15°與18°的和 sin 11 π 60 = sin 33 ∘ = 1 4 8 − 3 − 15 + 10 − 2 5 {\displaystyle \sin {\frac {11\pi }{60}}=\sin 33^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt {3}}-{\sqrt {15}}+{\sqrt {10-2{\sqrt {5}}}}}}\,} cos 11 π 60 = cos 33 ∘ = 1 4 8 + 3 + 15 − 10 − 2 5 {\displaystyle \cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt {3}}+{\sqrt {15}}-{\sqrt {10-2{\sqrt {5}}}}}}\,} tan 11 π 60 = tan 33 ∘ = 1 4 ( 2 3 − 5 − 1 ) ( 2 5 + 2 5 + 3 + 5 ) {\displaystyle \tan {\frac {11\pi }{60}}=\tan 33^{\circ }={\tfrac {1}{4}}\left(2{\sqrt {3}}-{\sqrt {5}}-1\right)\left(2{\sqrt {5+2{\sqrt {5}}}}+3+{\sqrt {5}}\right)\,} cot 11 π 60 = cot 33 ∘ = 1 4 ( 2 3 + 5 + 1 ) ( 2 5 + 2 5 − 3 − 5 ) {\displaystyle \cot {\frac {11\pi }{60}}=\cot 33^{\circ }={\tfrac {1}{4}}\left(2{\sqrt {3}}+{\sqrt {5}}+1\right)\left(2{\sqrt {5+2{\sqrt {5}}}}-3-{\sqrt {5}}\right)\,} 36°：正五邊形 更多資訊：五邊形 sin π 5 = sin 36 ∘ = 1 4 [ 2 ( 5 − 5 ) ] {\displaystyle \sin {\frac {\pi }{5}}=\sin 36^{\circ }={\tfrac {1}{4}}\left[{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]\,} cos π 5 = cos 36 ∘ = 1 + 5 4 = 1 2 φ {\displaystyle \cos {\frac {\pi }{5}}=\cos 36^{\circ }={\frac {1+{\sqrt {5}}}{4}}={\tfrac {1}{2}}\varphi \,} tan π 5 = tan 36 ∘ = 5 − 2 5 {\displaystyle \tan {\frac {\pi }{5}}=\tan 36^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,} 39°：18°與21°的和 sin 13 π 60 = sin 39 ∘ = 1 4 8 − 3 + 15 + 10 + 2 5 {\displaystyle \sin {\frac {13\pi }{60}}=\sin 39^{\circ }={\tfrac {1}{4}}{\sqrt {8-{\sqrt {3}}+{\sqrt {15}}+{\sqrt {10+2{\sqrt {5}}}}}}\,} cos 13 π 60 = cos 39 ∘ = 1 4 8 + 3 − 15 + 10 + 2 5 {\displaystyle \cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\tfrac {1}{4}}{\sqrt {8+{\sqrt {3}}-{\sqrt {15}}+{\sqrt {10+2{\sqrt {5}}}}}}\,} tan 13 π 60 = tan 39 ∘ = 1 4 [ ( 2 − 3 ) ( 3 − 5 ) − 2 ] [ 2 − 2 ( 5 + 5 ) ] {\displaystyle \tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\tfrac {1}{4}}\left[\left(2-{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)-2\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]\,} 42°：21°的2倍 sin 7 π 30 = sin 42 ∘ = 6 5 + 5 − 5 + 1 8 {\displaystyle \sin {\frac {7\pi }{30}}=\sin 42^{\circ }={\frac {{\sqrt {6}}{\sqrt {5+{\sqrt {5}}}}-{\sqrt {5}}+1}{8}}\,} cos 7 π 30 = cos 42 ∘ = 2 5 + 5 + 3 ( 5 − 1 ) 8 {\displaystyle \cos {\frac {7\pi }{30}}=\cos 42^{\circ }={\frac {{\sqrt {2}}{\sqrt {5+{\sqrt {5}}}}+{\sqrt {3}}\left({\sqrt {5}}-1\right)}{8}}\,} tan 7 π 30 = tan 42 ∘ = 1 2 ( 3 + 15 − 10 + 2 5 ) {\displaystyle \tan {\frac {7\pi }{30}}=\tan 42^{\circ }={\frac {1}{2}}\left({\sqrt {3}}+{\sqrt {15}}-{\sqrt {10+2{\sqrt {5}}}}\right)\,} cot 7 π 30 = cot 42 ∘ = 1 2 ( 3 3 − 15 + 50 − 22 5 ) {\displaystyle \cot {\frac {7\pi }{30}}=\cot 42^{\circ }={\frac {1}{2}}\left(3{\sqrt {3}}-{\sqrt {15}}+{\sqrt {50-22{\sqrt {5}}}}\right)\,} sec 7 π 30 = sec 42 ∘ = 5 + 2 5 − 3 {\displaystyle \sec {\frac {7\pi }{30}}=\sec 42^{\circ }={\sqrt {5+2{\sqrt {5}}}}-{\sqrt {3}}\,} sec 7 π 30 = sec 42 ∘ = 15 − 6 5 + 5 − 2 {\displaystyle \sec {\frac {7\pi }{30}}=\sec 42^{\circ }={\sqrt {15-6{\sqrt {5}}}}+{\sqrt {5}}-2\,} 45°：正方形 更多資訊：正方形 sin π 4 = sin 45 ∘ = 2 2 = 1 2 {\displaystyle \sin {\frac {\pi }{4}}=\sin 45^{\circ }={\frac {\sqrt {2}}{2}}={\frac {1}{\sqrt {2}}}\,} cos π 4 = cos 45 ∘ = 2 2 = 1 2 {\displaystyle \cos {\frac {\pi }{4}}=\cos 45^{\circ }={\frac {\sqrt {2}}{2}}={\frac {1}{\sqrt {2}}}\,} tan π 4 = tan 45 ∘ = 1 {\displaystyle \tan {\frac {\pi }{4}}=\tan 45^{\circ }=1} 48° sin 48 ∘ = 1 4 7 − 5 + 6 ( 5 − 5 ) {\displaystyle \sin 48^{\circ }={\frac {1}{4}}{\sqrt {7-{\sqrt {5}}+{\sqrt {6(5-{\sqrt {5}})}}}}} 54°：27°與27°的和 sin 3 π 10 = sin 54 ∘ = 5 + 1 4 {\displaystyle \sin {\frac {3\pi }{10}}=\sin 54^{\circ }={\frac {{\sqrt {5}}+1}{4}}\,\!} cos 3 π 10 = cos 54 ∘ = 10 − 2 5 4 {\displaystyle \cos {\frac {3\pi }{10}}=\cos 54^{\circ }={\frac {\sqrt {10-2{\sqrt {5}}}}{4}}} tan 3 π 10 = tan 54 ∘ = 25 + 10 5 5 {\displaystyle \tan {\frac {3\pi }{10}}=\tan 54^{\circ }={\frac {\sqrt {25+10{\sqrt {5}}}}{5}}\,} cot 3 π 10 = cot 54 ∘ = 5 − 2 5 {\displaystyle \cot {\frac {3\pi }{10}}=\cot 54^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,} 60°：等邊三角形 sin π 3 = sin 60 ∘ = 3 2 {\displaystyle \sin {\frac {\pi }{3}}=\sin 60^{\circ }={\frac {\sqrt {3}}{2}}\,} cos π 3 = cos 60 ∘ = 1 2 {\displaystyle \cos {\frac {\pi }{3}}=\cos 60^{\circ }={\frac {1}{2}}\,} tan π 3 = tan 60 ∘ = 3 {\displaystyle \tan {\frac {\pi }{3}}=\tan 60^{\circ }={\sqrt {3}}\,} cot π 3 = cot 60 ∘ = 3 3 = 1 3 {\displaystyle \cot {\frac {\pi }{3}}=\cot 60^{\circ }={\frac {\sqrt {3}}{3}}={\frac {1}{\sqrt {3}}}\,} 67.5°：7.5°與60°的和 sin 3 π 8 = sin 67.5 ∘ = 1 2 2 + 2 {\displaystyle \sin {\frac {3\pi }{8}}=\sin 67.5^{\circ }={\tfrac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\,} cos 3 π 8 = cos 67.5 ∘ = 1 2 2 − 2 {\displaystyle \cos {\frac {3\pi }{8}}=\cos 67.5^{\circ }={\tfrac {1}{2}}{\sqrt {2-{\sqrt {2}}}}\,} tan 3 π 8 = tan 67.5 ∘ = 2 + 1 {\displaystyle \tan {\frac {3\pi }{8}}=\tan 67.5^{\circ }={\sqrt {2}}+1\,} cot 3 π 8 = cot 67.5 ∘ = 2 − 1 {\displaystyle \cot {\frac {3\pi }{8}}=\cot 67.5^{\circ }={\sqrt {2}}-1\,} 72°：36°的二倍 sin 2 π 5 = sin 72 ∘ = 1 4 2 ( 5 + 5 ) {\displaystyle \sin {\frac {2\pi }{5}}=\sin 72^{\circ }={\tfrac {1}{4}}{\sqrt {2\left(5+{\sqrt {5}}\right)}}\,} cos 2 π 5 = cos 72 ∘ = 1 4 ( 5 − 1 ) {\displaystyle \cos {\frac {2\pi }{5}}=\cos 72^{\circ }={\tfrac {1}{4}}\left({\sqrt {5}}-1\right)\,} tan 2 π 5 = tan 72 ∘ = 5 + 2 5 {\displaystyle \tan {\frac {2\pi }{5}}=\tan 72^{\circ }={\sqrt {5+2{\sqrt {5}}}}\,} cot 2 π 5 = cot 72 ∘ = 1 5 5 ( 5 − 2 5 ) {\displaystyle \cot {\frac {2\pi }{5}}=\cot 72^{\circ }={\tfrac {1}{5}}{\sqrt {5\left(5-2{\sqrt {5}}\right)}}\,} 75°: 30°與45°的和 sin 5 π 12 = sin 75 ∘ = 1 4 ( 6 + 2 ) {\displaystyle \sin {\frac {5\pi }{12}}=\sin 75^{\circ }={\tfrac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)\,} cos 5 π 12 = cos 75 ∘ = 1 4 ( 6 − 2 ) {\displaystyle \cos {\frac {5\pi }{12}}=\cos 75^{\circ }={\tfrac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)\,} tan 5 π 12 = tan 75 ∘ = 2 + 3 {\displaystyle \tan {\frac {5\pi }{12}}=\tan 75^{\circ }=2+{\sqrt {3}}\,} cot 5 π 12 = cot 75 ∘ = 2 − 3 {\displaystyle \cot {\frac {5\pi }{12}}=\cot 75^{\circ }=2-{\sqrt {3}}\,} 81° sin 81 ∘ = 1 2 1 2 ( 4 + 2 ( 5 + 5 ) ) {\displaystyle \sin 81^{\circ }={\frac {1}{2}}{\sqrt {{\frac {1}{2}}{\Big (}4+{\sqrt {2(5+{\sqrt {5}})}}{\Big )}}}} 90°：根本 sin π 2 = sin 90 ∘ = 1 {\displaystyle \sin {\frac {\pi }{2}}=\sin 90^{\circ }=1\,} cos π 2 = cos 90 ∘ = 0 {\displaystyle \cos {\frac {\pi }{2}}=\cos 90^{\circ }=0\,} cot π 2 = cot 90 ∘ = 0 {\displaystyle \cot {\frac {\pi }{2}}=\cot 90^{\circ }=0\,} 列表 在下表中， i {\displaystyle i} 為虛數單位， ω = exp ( π i 3 ) = − 1 2 + 1 2 i 3 {\displaystyle \omega =\exp({\frac {\pi i}{3}})=-{\frac {1}{2}}+{\frac {1}{2}}i{\sqrt {3}}} 。 n {\displaystyle n} sin ( 2 π n ) {\displaystyle \sin \left({\frac {2\pi }{n}}\right)} cos ( 2 π n ) {\displaystyle \cos \left({\frac {2\pi }{n}}\right)} tan ( 2 π n ) {\displaystyle \tan \left({\frac {2\pi }{n}}\right)} 1 0 {\displaystyle 0} 1 {\displaystyle 1} 0 {\displaystyle 0} 2 0 {\displaystyle 0} − 1 {\displaystyle -1} 0 {\displaystyle 0} 3 1 2 3 {\displaystyle {\frac {1}{2}}{\sqrt {3}}} − 1 2 {\displaystyle -{\frac {1}{2}}} − 3 {\displaystyle -{\sqrt {3}}} 4 1 {\displaystyle 1} 0 {\displaystyle 0} ± ∞ {\displaystyle \pm \infty } 5 1 4 ( 10 + 2 5 ) {\displaystyle {\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)} 1 4 ( 5 − 1 ) {\displaystyle {\frac {1}{4}}\left({\sqrt {5}}-1\right)} 5 + 2 5 {\displaystyle {\sqrt {5+2{\sqrt {5}}}}} 6 1 2 3 {\displaystyle {\frac {1}{2}}{\sqrt {3}}} 1 2 {\displaystyle {\frac {1}{2}}} 3 {\displaystyle {\sqrt {3}}} 7 1 2 1 3 ( 7 − ω 2 7 + 21 − 3 2 3 − ω 7 − 21 − 3 2 3 ) {\displaystyle {\frac {1}{2}}{\sqrt {{\frac {1}{3}}\left(7-\omega ^{2}{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}-\omega {\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)}}} 1 6 ( − 1 + 7 + 21 − 3 2 3 + 7 − 21 − 3 2 3 ) {\displaystyle {\frac {1}{6}}\left(-1+{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}+{\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)} 8 1 2 2 {\displaystyle {\frac {1}{2}}{\sqrt {2}}} 1 2 2 {\displaystyle {\frac {1}{2}}{\sqrt {2}}} 1 {\displaystyle 1} 9 1 4 ( 4 ( − 1 + − 3 ) 3 + 4 ( − 1 − − 3 ) 3 ) {\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4(-1+{\sqrt {-3}})}}+{\sqrt[{3}]{4(-1-{\sqrt {-3}})}}\right)} 10 1 4 ( 10 − 2 5 ) {\displaystyle {\frac {1}{4}}\left({\sqrt {10-2{\sqrt {5}}}}\right)} 1 4 ( 5 + 1 ) {\displaystyle {\frac {1}{4}}\left({\sqrt {5}}+1\right)} 5 − 2 5 {\displaystyle {\sqrt {5-2{\sqrt {5}}}}} 11 12 1 2 {\displaystyle {\frac {1}{2}}} 1 2 3 {\displaystyle {\frac {1}{2}}{\sqrt {3}}} 1 3 3 {\displaystyle {\frac {1}{3}}{\sqrt {3}}} 13 14 1 24 3 ( 112 − 14336 + − 5549064193 3 − 14336 − − 5549064193 3 ) {\displaystyle {\frac {1}{24}}{\sqrt {3\left(112-{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}\right)}}} 1 24 3 ( 80 + 14336 + − 5549064193 3 + 14336 − − 5549064193 3 ) {\displaystyle {\frac {1}{24}}{\sqrt {3\left(80+{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}\right)}}} 112 − 14336 + − 5549064193 3 − 14336 − − 5549064193 3 80 + 14336 + − 5549064193 3 + 14336 − − 5549064193 3 {\displaystyle {\sqrt {\frac {112-{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}}{80+{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}}}}} 15 1 8 ( 15 + 3 − 10 − 2 5 ) {\displaystyle {\frac {1}{8}}\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)} 1 8 ( 1 + 5 + 30 − 6 5 ) {\displaystyle {\frac {1}{8}}\left(1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}\right)} 1 2 ( − 3 3 − 15 + 50 + 22 5 ) {\displaystyle {\frac {1}{2}}\left(-3{\sqrt {3}}-{\sqrt {15}}+{\sqrt {50+22{\sqrt {5}}}}\right)} 16 1 2 ( 2 − 2 ) {\displaystyle {\frac {1}{2}}\left({\sqrt {2-{\sqrt {2}}}}\right)} 1 2 ( 2 + 2 ) {\displaystyle {\frac {1}{2}}\left({\sqrt {2+{\sqrt {2}}}}\right)} 2 − 1 {\displaystyle {\sqrt {2}}-1} 17 1 4 8 − 2 ( 15 + 17 + 34 − 2 17 − 2 17 + 3 17 − 170 + 38 17 ) {\displaystyle {\frac {1}{4}}{\sqrt {8-{\sqrt {2\left(15+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {17+3{\sqrt {17}}-{\sqrt {170+38{\sqrt {17}}}}}}\right)}}}}} 1 16 ( − 1 + 17 + 34 − 2 17 + 2 17 + 3 17 − 34 − 2 17 − 2 34 + 2 17 ) {\displaystyle {\frac {1}{16}}\left(-1+{\sqrt {17}}+{\sqrt {34-2{\sqrt {17}}}}+2{\sqrt {17+3{\sqrt {17}}-{\sqrt {34-2{\sqrt {17}}}}-2{\sqrt {34+2{\sqrt {17}}}}}}\right)} 18 1 4 ( 4 − 1 − 4 3 3 − 4 − 1 + 4 3 3 ) {\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4{\sqrt {-1}}-4{\sqrt {3}}}}-{\sqrt[{3}]{4{\sqrt {-1}}+4{\sqrt {3}}}}\right)} 1 4 ( 4 + 4 − 3 3 + 4 − 4 − 3 3 ) {\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4+4{\sqrt {-3}}}}+{\sqrt[{3}]{4-4{\sqrt {-3}}}}\right)} 19 20 1 4 ( 5 − 1 ) {\displaystyle {\frac {1}{4}}\left({\sqrt {5}}-1\right)} 1 4 ( 10 + 2 5 ) {\displaystyle {\frac {1}{4}}\left({\sqrt {10+2{\sqrt {5}}}}\right)} 1 5 ( 25 − 10 5 ) {\displaystyle {\frac {1}{5}}\left({\sqrt {25-10{\sqrt {5}}}}\right)} 21 22 23 24 1 4 ( 6 − 2 ) {\displaystyle {\frac {1}{4}}\left({\sqrt {6}}-{\sqrt {2}}\right)} 1 4 ( 6 + 2 ) {\displaystyle {\frac {1}{4}}\left({\sqrt {6}}+{\sqrt {2}}\right)} 2 − 3 {\displaystyle 2-{\sqrt {3}}} 相關 更多資訊：三角函數和三角恆等式 更多資訊：en:Generating trigonometric tables 參見 可作圖多邊形 三角數 十七邊形 參考文獻 埃里克·韋斯坦因. Constructible polygon. MathWorld. 埃里克·韋斯坦因. Trigonometry angles. MathWorld. π/3 (60°)（頁面存檔備份，存於網際網路檔案館）—π/6 (30°)（頁面存檔備份，存於網際網路檔案館）—π/12 (15°)（頁面存檔備份，存於網際網路檔案館）—π/24 (7.5°)（頁面存檔備份，存於網際網路檔案館） π/4 (45°)（頁面存檔備份，存於網際網路檔案館）—π/8 (22.5°)（頁面存檔備份，存於網際網路檔案館）—π/16 (11.25°)（頁面存檔備份，存於網際網路檔案館）—π/32 (5.625°)（頁面存檔備份，存於網際網路檔案館） π/5 (36°)（頁面存檔備份，存於網際網路檔案館）—π/10 (18°)（頁面存檔備份，存於網際網路檔案館）—π/20 (9°)（頁面存檔備份，存於網際網路檔案館） π/7（頁面存檔備份，存於網際網路檔案館）—π/14 π/9 (20°)（頁面存檔備份，存於網際網路檔案館）—π/18 (10°)（頁面存檔備份，存於網際網路檔案館） π/11（頁面存檔備份，存於網際網路檔案館） π/13（頁面存檔備份，存於網際網路檔案館） π/15 (12°)（頁面存檔備份，存於網際網路檔案館）—π/30 (6°)（頁面存檔備份，存於網際網路檔案館） π/17（頁面存檔備份，存於網際網路檔案館） π/19 π/23（頁面存檔備份，存於網際網路檔案館） Bracken, Paul; Cizek, Jiri. Evaluation of quantum mechanical perturbation sums in terms of quadratic surds and their use in approximation of zeta(3)/pi^3. Int. J. Quantum Chemistry. 2002, 90 (1): 42–53. doi:10.1002/qua.1803. Conway, John H.; Radin, Charles; Radun, Lorenzo. On angles whose squared trigonometric functions are rational. 1998. arXiv:math-ph/9812019 . Conway, John H.; Radin, Charles; Radun, Lorenzo. On angles whose squared trigonometric functions are rational. Disc. Comput. Geom. 1999, 22 (3): 321–332. doi:10.1007/PL00009463. MR1706614 Girstmair, Kurt. Some linear relations between values of trigonometric functions at k*pi/n. Acta Arithmetica. 1997, 81: 387–398. MR1472818 Gurak, S. On the minimal polynomial of gauss periods for prime powers. Mathematics of Computation. 2006, 75 (256): 2021–2035. Bibcode:2006MaCom..75.2021G. doi:10.1090/S0025-5718-06-01885-0. MR2240647 Servi, L. D. Nested square roots of 2. Am. Math. Monthly. 2003, 110 (4): 326–330. doi:10.2307/3647881. MR1984573 JSTOR 3647881 注釋 [1]由Wolfram Alpha驗算：[1] （頁面存檔備份，存於網際網路檔案館） [2]使用Mathematica驗算，代碼為N[ArcSin[(1 + Sqrt[3] I)/16 Power[4 Sqrt[30] - 8 Sqrt[15 + 3 Sqrt[5]] + 8 Sqrt[5 + Sqrt[5]] + 4 Sqrt[10] - 4 Sqrt[6] - 4 Sqrt[2] + (4 Sqrt[30] + 8 Sqrt[15 + 3 Sqrt[5]] + 8 Sqrt[5 + Sqrt[5]] - 4 Sqrt[10] - 4 Sqrt[6] + 4 Sqrt[2]) I, (3)^-1] + (1 - Sqrt[3] I)/16 Power[4 Sqrt[30] - 8 Sqrt[15 + 3 Sqrt[5]] + 8 Sqrt[5 + Sqrt[5]] + 4 Sqrt[10] - 4 Sqrt[6] - 4 Sqrt[2] - (4 Sqrt[30] + 8 Sqrt[15 + 3 Sqrt[5]] + 8 Sqrt[5 + Sqrt[5]] - 4 Sqrt[10] - 4 Sqrt[6] + 4 Sqrt[2]) I, (3)^-1]], 100]/Degree結果為1與原角度無誤差 Wikiwand - on Seamless Wikipedia browsing. On steroids.
![{\displaystyle {\frac {1}{2}}{\sqrt {{\frac {1}{3}}\left(7-\omega ^{2}{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}-\omega {\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/8cc7dc542d7b26bf260027485f6a30538ffeffed)
![{\displaystyle {\frac {1}{6}}\left(-1+{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}+{\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)}](//wikimedia.org/api/rest_v1/media/math/render/svg/b9770256721ca5edf98ee18d025cdcea444febf4)
![{\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4(-1+{\sqrt {-3}})}}+{\sqrt[{3}]{4(-1-{\sqrt {-3}})}}\right)}](//wikimedia.org/api/rest_v1/media/math/render/svg/559fcd1611f8dce6e41277a61ed398ae11b67dad)
![{\displaystyle {\frac {1}{24}}{\sqrt {3\left(112-{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}\right)}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/6bcd3c3b7acbdcdc966c3f52fdb381166dd64a02)
![{\displaystyle {\frac {1}{24}}{\sqrt {3\left(80+{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}\right)}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/7b8523974b4edd2a69791141d1fbcd23955afabe)
![{\displaystyle {\sqrt {\frac {112-{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}-{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}}{80+{\sqrt[{3}]{14336+{\sqrt {-5549064193}}}}+{\sqrt[{3}]{14336-{\sqrt {-5549064193}}}}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/aac315bbf180a0a43ce39e951361a10dfb9e7d8f)
![{\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4{\sqrt {-1}}-4{\sqrt {3}}}}-{\sqrt[{3}]{4{\sqrt {-1}}+4{\sqrt {3}}}}\right)}](//wikimedia.org/api/rest_v1/media/math/render/svg/4af28bc5cf3d141c3815a9ca2f3ff7c0a10b3a57)
![{\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4+4{\sqrt {-3}}}}+{\sqrt[{3}]{4-4{\sqrt {-3}}}}\right)}](//wikimedia.org/api/rest_v1/media/math/render/svg/a3ac79b40df411129942e7df1ac3e9d2b805240f)
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![{\displaystyle \sin {\frac {\pi }{9}}=\sin 20^{\circ }={\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}+{\sqrt {-{\frac {1}{256}}}}}}+{\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}-{\sqrt {-{\frac {1}{256}}}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/6019908a9d7a84324bc27625b5e5213830d24d8a)
![{\displaystyle \cos {\frac {\theta }{n}}=\Re \left({\sqrt[{n}]{\cos \theta +i\sin \theta }}\right)={\frac {1}{2}}\left({\sqrt[{n}]{\cos \theta +i\sin \theta }}+{\sqrt[{n}]{\cos \theta -i\sin \theta }}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/71c3d32e31b02c92e787505c9eafa8fa4e8cd828)
![{\displaystyle \sin {\frac {\theta }{n}}=\Im \left({\sqrt[{n}]{\cos \theta +i\sin \theta }}\right)={\frac {1}{2i}}\left({\sqrt[{n}]{\cos \theta +i\sin \theta }}-{\sqrt[{n}]{\cos \theta -i\sin \theta }}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/2ea663e7086395418f2d54876843352a2d0bf147)
![{\displaystyle \sin {1^{\circ }}={\frac {1}{2i}}\left({\sqrt[{3}]{\cos {3^{\circ }}+i\sin {3^{\circ }}}}-{\sqrt[{3}]{\cos {3^{\circ }}-i\sin {3^{\circ }}}}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1c521d6e55fec7801936d21449aaaeb321b28e58)
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![{\displaystyle -{\sqrt[{3}]{\left[2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)\right]-i\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]}}{\Bigg \}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/066614d2502295fa13145c41b345358bf2238739)
![{\displaystyle \sin {1^{\circ }}={\frac {1+{\sqrt {3}}i}{16}}{\sqrt[{3}]{4{\sqrt {30}}-8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}+4{\sqrt {10}}-4{\sqrt {6}}-4{\sqrt {2}}+\left(4{\sqrt {30}}+8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}-4{\sqrt {10}}-4{\sqrt {6}}+4{\sqrt {2}}\right)i}}+}](http://wikimedia.org/api/rest_v1/media/math/render/svg/b9a7f8cda0d69d750a416bca1207841bab26cd6f)
![{\displaystyle {\frac {1-{\sqrt {3}}i}{16}}{\sqrt[{3}]{4{\sqrt {30}}-8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}+4{\sqrt {10}}-4{\sqrt {6}}-4{\sqrt {2}}-\left(4{\sqrt {30}}+8{\sqrt {15+3{\sqrt {5}}}}+8{\sqrt {5+{\sqrt {5}}}}-4{\sqrt {10}}-4{\sqrt {6}}+4{\sqrt {2}}\right)i}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/36c528db784d06bd39b38158572c4ae931517615)
![{\displaystyle \sin {2^{\circ }}={\frac {1}{2i}}\left({\sqrt[{3}]{\cos {6^{\circ }}+i\sin {6^{\circ }}}}-{\sqrt[{3}]{\cos {6^{\circ }}-i\sin {6^{\circ }}}}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/9cb6011d35cbc8d89b146f7fa0a014321c9f1d09)
![{\displaystyle ={\frac {1}{4i}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]+i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/76088f823ff50097d119f0df28cff18ea03e6c6e)
![{\displaystyle -{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]-i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}{\Bigg \}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/bc8b004b90010622da546669426b9a57564c5903)
![{\displaystyle \cos {2^{\circ }}={\frac {1}{2}}\left({\sqrt[{3}]{\cos {6^{\circ }}+i\sin {6^{\circ }}}}+{\sqrt[{3}]{\cos {6^{\circ }}-i\sin {6^{\circ }}}}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/6e40bf7865e0a5ceef13869ffa29c74560bced35)
![{\displaystyle ={\frac {1}{4}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]+i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/0ed4dc0c22095244d2f44b1a13799f3d249ebca5)
![{\displaystyle +{\sqrt[{3}]{\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]-i\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]}}{\Bigg \}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/57b380930ccaf8f99838d8352bcfc963be7680bd)
![{\displaystyle \tan {\frac {\pi }{60}}=\tan 3^{\circ }={\tfrac {1}{4}}\left[(2-{\sqrt {3}})(3+{\sqrt {5}})-2\right]\left[2-{\sqrt {2(5-{\sqrt {5}})}}\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/9afc8cae3b0181ce3ca0fa38cd9737a9096fc5de)
![{\displaystyle \sin {4^{\circ }}={\frac {1}{2i}}\left({\sqrt[{3}]{\cos {12^{\circ }}+i\sin {12^{\circ }}}}-{\sqrt[{3}]{\cos {12^{\circ }}-i\sin {12^{\circ }}}}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/622ba2d2d653fa9a61ba26ef726e4a46cc2acafb)
![{\displaystyle ={\frac {1}{4i}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]+i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/896906922b19b21a780a055f38801474ecafd495)
![{\displaystyle -{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]-i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}{\Bigg \}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/e2d8e586d6453abd5ac515018fcbfd022fa81011)
![{\displaystyle \cos {4^{\circ }}={\frac {1}{2}}\left({\sqrt[{3}]{\cos {12^{\circ }}+i\sin {12^{\circ }}}}+{\sqrt[{3}]{\cos {12^{\circ }}-i\sin {12^{\circ }}}}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/2100bbeb5c6a431e34e47bf0f26e4c67bddd1cbb)
![{\displaystyle ={\frac {1}{4}}{\Bigg \{}{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]+i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/d98084504214bdee5478f5af195113cb7476e2db)
![{\displaystyle +{\sqrt[{3}]{\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]-i\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]}}{\Bigg \}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/cbd37da55c50e7c12ca7ba6ebd337630973cf60c)
![{\displaystyle \sin {\frac {\pi }{36}}=\sin 5^{\circ }={\frac {2-2{\sqrt {3}}\mathrm {i} }{2{\sqrt[{3}]{2({\sqrt {2}}-{\sqrt {6}})}}-2-{\sqrt {3}}}}-{\frac {(1+{\sqrt {3}}\mathrm {i} ){\sqrt[{3}]{2({\sqrt {2}}-{\sqrt {6}})}}-2-{\sqrt {3}}}{8}}\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/74aa3cf4e94e893298f0dd9edba498d11b4a2249)
![{\displaystyle \sin {\frac {\pi }{30}}=\sin 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/e32eed64add3521d7d7c9bf695482e8c3add4c36)
![{\displaystyle \cos {\frac {\pi }{30}}=\cos 6^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/f90aff5eb92b8553f8dfd11cd76f3f30bd223560)
![{\displaystyle \tan {\frac {\pi }{30}}=\tan 6^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(5-{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/ca95e17f7bfec511a0d9029c1aa1a0c8e5cc5630)
![{\displaystyle {\tan 10^{\circ }=-{\frac {-1-{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}+36{\rm {i}}}}-{\frac {-1+{\sqrt {3}}{\rm {i}}}{6}}{\sqrt[{3}]{-12{\sqrt {3}}-36{\rm {i}}}}+{\frac {1}{\sqrt {3}}}}\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/bff3bab5cf2730334855083f4576b03c09c9dd07)
![{\displaystyle \sin {\frac {\pi }{15}}=\sin 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/55fb1a72c84d5dec8bf231d619c534df13c261ef)
![{\displaystyle \cos {\frac {\pi }{15}}=\cos 12^{\circ }={\tfrac {1}{8}}\left[{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/4923130317e1553f2cfd21ca42c4e29cd2f1f836)
![{\displaystyle \tan {\frac {\pi }{15}}=\tan 12^{\circ }={\tfrac {1}{2}}\left[{\sqrt {3}}(3-{\sqrt {5}})-{\sqrt {2(25-11{\sqrt {5}})}}\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/a207e48f104ece1821014496f7b32e371956076d)
![{\displaystyle \sin {\frac {\pi }{9}}=\sin 20^{\circ }={\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}+{\sqrt {-{\frac {1}{256}}}}}}+{\sqrt[{3}]{-{\frac {\sqrt {3}}{16}}-{\sqrt {-{\frac {1}{256}}}}}}=}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1e39576a98052e848f2f9291c5d5da2437d6d056)
![{\displaystyle 2^{-{\frac {4}{3}}}\left({\sqrt[{3}]{i-{\sqrt {3}}}}-{\sqrt[{3}]{i+{\sqrt {3}}}}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/4fb7a6cadc9f67c9fc96b0b6bb7165af68824e9d)
![{\displaystyle 2^{-{\frac {4}{3}}}\left({\sqrt[{3}]{1+i{\sqrt {3}}}}+{\sqrt[{3}]{1-i{\sqrt {3}}}}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/c1fd794ae0ddf0128bdcedcca67ead75c8174e89)
![{\displaystyle \tan {\frac {7\pi }{60}}=\tan 21^{\circ }={\tfrac {1}{4}}\left[2-\left(2+{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/92f16397799bd47711ac25e404f6de198c9bb28c)
![{\displaystyle \sin {\frac {2\pi }{15}}=\sin 24^{\circ }={\tfrac {1}{8}}\left[{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/748d70aaab6ed6e9d613d6ef799e612174f96970)
![{\displaystyle \tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\tfrac {1}{2}}\left[{\sqrt {2(25+11{\sqrt {5}})}}-{\sqrt {3}}(3+{\sqrt {5}})\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/751f49b7e0a84e36750d6df47e69f7e5ca2f4ba2)
![{\displaystyle \cos {\frac {\pi }{7}}=\cos {\frac {180}{7}}^{\circ }=\cos 25{\frac {5}{7}}^{\circ }={\frac {1}{6}}+{\frac {1-{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}+{\frac {1+{\sqrt {3}}i}{24}}{\sqrt[{3}]{28-84{\sqrt {3}}i}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/9e6f0da045735bbe37ae69faa6cb417ce734400d)
![{\displaystyle \sin {\frac {3\pi }{20}}=\sin 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}\;({\sqrt {5}}-1)\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/56f979b3d0d06f3124dd266ee7860dd955e654c1)
![{\displaystyle \cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\tfrac {1}{8}}\left[2{\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}\;\left({\sqrt {5}}-1\right)\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/96ed3130d2995af89ffd7fed92ca75b79e11b580)
![{\displaystyle \sin {\frac {\pi }{5}}=\sin 36^{\circ }={\tfrac {1}{4}}\left[{\sqrt {2\left(5-{\sqrt {5}}\right)}}\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/a5a125bab3d72d8e6ca59b000aa0de58afe4e645)
![{\displaystyle \tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\tfrac {1}{4}}\left[\left(2-{\sqrt {3}}\right)\left(3-{\sqrt {5}}\right)-2\right]\left[2-{\sqrt {2\left(5+{\sqrt {5}}\right)}}\right]\,}](http://wikimedia.org/api/rest_v1/media/math/render/svg/3f5d40feb83e05d4be7784892beb0641f8740e59)
![{\displaystyle {\frac {1}{2}}{\sqrt {{\frac {1}{3}}\left(7-\omega ^{2}{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}-\omega {\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/8cc7dc542d7b26bf260027485f6a30538ffeffed)
![{\displaystyle {\frac {1}{6}}\left(-1+{\sqrt[{3}]{\frac {7+21{\sqrt {-3}}}{2}}}+{\sqrt[{3}]{\frac {7-21{\sqrt {-3}}}{2}}}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/b9770256721ca5edf98ee18d025cdcea444febf4)
![{\displaystyle {\frac {1}{4}}\left({\sqrt[{3}]{4(-1+{\sqrt {-3}})}}+{\sqrt[{3}]{4(-1-{\sqrt {-3}})}}\right)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/559fcd1611f8dce6e41277a61ed398ae11b67dad)
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