三角函數精確值

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。 根据尼云定理，有理数度数的角的正弦值，其中的有理数仅有0， ± 1 2 {\displaystyle \pm {\frac {1}{2}}} ，±1。 例如：0°、30°、45° 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)}}} 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。 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 \}}} 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)\,} 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}}} 由于三角函数的特性，大于45°角度的三角函数值，可以经由自0°～45°的角度的三角函数值的相关的计算取得。 sin ⁡ 0 = 0 {\displaystyle \sin 0=0\,} cos ⁡ 0 = 1 {\displaystyle \cos 0=1\,} tan ⁡ 0 = 0 {\displaystyle \tan 0=0\,} 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}}} 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}}} 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}}}} 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 \}}} 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}}}} 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}}}}}}}}}}} 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]\,} 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}}}}}}}}} 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 \}}} 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}}}}}}}} 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}}\,} 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}}}}}}}}} 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}}}}\,} 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}}}}\,} 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}}}}\,} 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}}}}\,} 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} 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]\,} 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}}\,} 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)}}\,} 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)} 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]\,} 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}}} 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\,} 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]\,} 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}}} 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}}}}\,} 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}}\,} 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)\,} 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}}}}\,} 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]\,} 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\,} 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} 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}})}}}}} 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}}}}\,} 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}}}\,} 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\,} 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)}}\,} 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}}\,} 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 )}}}} 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\,} 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}}} 。 可作图多边形 三角數（英语：Trigonometric_number） 十七邊形 埃里克·韦斯坦因. Constructible polygon. MathWorld.  埃里克·韦斯坦因. Trigonometry angles. MathWorld.  π/3 (60°)（页面存档备份，存于互联网档案馆）—π/6 (30°)（页面存档备份，存于互联网档案馆）—π/12 (15°)（页面存档备份，存于互联网档案馆）—π/24 (7.5°)（页面存档备份，存于互联网档案馆） π/4 (45°)（页面存档备份，存于互联网档案馆）—π/8 (22.5°)（页面存档备份，存于互联网档案馆）—π/16 (11.25°)（页面存档备份，存于互联网档案馆）—π/32 (5.625°)（页面存档备份，存于互联网档案馆） π/5 (36°)（页面存档备份，存于互联网档案馆）—π/10 (18°)（页面存档备份，存于互联网档案馆）—π/20 (9°)（页面存档备份，存于互联网档案馆） π/7（页面存档备份，存于互联网档案馆）—π/14 π/9 (20°)（页面存档备份，存于互联网档案馆）—π/18 (10°)（页面存档备份，存于互联网档案馆） π/11（页面存档备份，存于互联网档案馆） π/13（页面存档备份，存于互联网档案馆） π/15 (12°)（页面存档备份，存于互联网档案馆）—π/30 (6°)（页面存档备份，存于互联网档案馆） π/17（页面存档备份，存于互联网档案馆） π/19 π/23（页面存档备份，存于互联网档案馆） 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 由Wolfram Alpha验算：[1] （页面存档备份，存于互联网档案馆） 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