Cơ bản
Richard Feynman từ nhỏ đã nhớ đẳng thức sau:
cos 20 ∘ ⋅ cos 40 ∘ ⋅ cos 80 ∘ = 1 8 . {\displaystyle \cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 80^{\circ }={\frac {1}{8}}.}
Tuy nhiên nó là trường hợp riêng của:
∏ j = 0 k − 1 cos ( 2 j x ) = sin ( 2 k x ) 2 k sin ( x ) . {\displaystyle \prod _{j=0}^{k-1}\cos(2^{j}x)={\frac {\sin(2^{k}x)}{2^{k}\sin(x)}}.}
Đẳng thức số sau chưa được tổng quát hóa với biến số:
cos 24 ∘ + cos 48 ∘ + cos 96 ∘ + cos 168 ∘ = 1 2 {\displaystyle \cos 24^{\circ }+\cos 48^{\circ }+\cos 96^{\circ }+\cos 168^{\circ }={\frac {1}{2}}} .
Đẳng thức sau cho thấy đặc điểm của số 21:
cos ( 2 π 21 ) + cos ( 2 ⋅ 2 π 21 ) + cos ( 4 ⋅ 2 π 21 ) {\displaystyle \cos \left({\frac {2\pi }{21}}\right)\,+\,\cos \left(2\cdot {\frac {2\pi }{21}}\right)\,+\,\cos \left(4\cdot {\frac {2\pi }{21}}\right)} + cos ( 5 ⋅ 2 π 21 ) + cos ( 8 ⋅ 2 π 21 ) + cos ( 10 ⋅ 2 π 21 ) = 1 2 . {\displaystyle \,+\,\cos \left(5\cdot {\frac {2\pi }{21}}\right)\,+\,\cos \left(8\cdot {\frac {2\pi }{21}}\right)\,+\,\cos \left(10\cdot {\frac {2\pi }{21}}\right)={\frac {1}{2}}.}
Một cách tính pi có thể dựa vào đẳng thức số sau, do John Machin tìm thấy:
π 4 = 4 arctan 1 5 − arctan 1 239 {\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}
hay dùng công thức Euler:
π 4 = 5 arctan 1 7 + 2 arctan 3 79 . {\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}.}
Một số giá trị lượng giác thông dụng:
sin 0 = sin 0 ∘ = 0 = cos 90 ∘ = cos ( π 2 ) sin ( π 6 ) = sin 30 ∘ = 1 2 = cos 60 ∘ = cos ( π 3 ) sin ( π 4 ) = sin 45 ∘ = 2 2 = cos 45 ∘ = cos ( π 4 ) sin ( π 3 ) = sin 60 ∘ = 3 2 = cos 30 ∘ = cos ( π 6 ) sin ( π 2 ) = sin 90 ∘ = 1 = cos 0 ∘ = cos 0 tan 0 = tan 0 ∘ = 0 = cot 90 ∘ = cot ( π 2 ) tan ( π 6 ) = tan 30 ∘ = 3 3 = cot 60 ∘ = cot ( π 3 ) tan ( π 4 ) = tan 45 ∘ = 1 = cot 45 ∘ = cot ( π 4 ) tan ( π 3 ) = tan 60 ∘ = 3 = cot 30 ∘ = cot ( π 6 ) {\displaystyle {\begin{matrix}\sin 0&=&\sin 0^{\circ }&=&0&=&\cos 90^{\circ }&=&\cos \left({\frac {\pi }{2}}\right)\\\\\sin \left({\frac {\pi }{6}}\right)&=&\sin 30^{\circ }&=&{\frac {1}{2}}&=&\cos 60^{\circ }&=&\cos \left({\frac {\pi }{3}}\right)\\\\\sin \left({\frac {\pi }{4}}\right)&=&\sin 45^{\circ }&=&{\frac {\sqrt {2}}{2}}&=&\cos 45^{\circ }&=&\cos \left({\frac {\pi }{4}}\right)\\\\\sin \left({\frac {\pi }{3}}\right)&=&\sin 60^{\circ }&=&{\frac {\sqrt {3}}{2}}&=&\cos 30^{\circ }&=&\cos \left({\frac {\pi }{6}}\right)\\\\\sin \left({\frac {\pi }{2}}\right)&=&\sin 90^{\circ }&=&1&=&\cos 0^{\circ }&=&\cos 0\\\\\tan 0&=&\tan 0^{\circ }&=&0&=&\cot 90^{\circ }&=&\cot \left({\frac {\pi }{2}}\right)\\\\\tan \left({\frac {\pi }{6}}\right)&=&\tan 30^{\circ }&=&{\frac {\sqrt {3}}{3}}&=&\cot 60^{\circ }&=&\cot \left({\frac {\pi }{3}}\right)\\\\\tan \left({\frac {\pi }{4}}\right)&=&\tan 45^{\circ }&=&1&=&\cot 45^{\circ }&=&\cot \left({\frac {\pi }{4}}\right)\\\\\tan \left({\frac {\pi }{3}}\right)&=&\tan 60^{\circ }&=&{\sqrt {3}}&=&\cot 30^{\circ }&=&\cot \left({\frac {\pi }{6}}\right)\end{matrix}}} sin π 7 = 7 6 − 7 189 ∑ j = 0 ∞ ( 3 j + 1 ) ! 189 j j ! ( 2 j + 2 ) ! {\displaystyle \sin {\frac {\pi }{7}}={\frac {\sqrt {7}}{6}}-{\frac {\sqrt {7}}{189}}\sum _{j=0}^{\infty }{\frac {(3j+1)!}{189^{j}j!\,(2j+2)!}}\!} sin π 18 = 1 6 ∑ j = 0 ∞ ( 3 j ) ! 27 j j ! ( 2 j + 1 ) ! {\displaystyle \sin {\frac {\pi }{18}}={\frac {1}{6}}\sum _{j=0}^{\infty }{\frac {(3j)!}{27^{j}j!\,(2j+1)!}}\!}
Dùng tỷ lệ vàng φ:
cos ( π 5 ) = cos 36 ∘ = 5 + 1 4 = ϕ / 2 {\displaystyle \cos \left({\frac {\pi }{5}}\right)=\cos 36^{\circ }={{\sqrt {5}}+1 \over 4}=\phi /2} sin ( π 10 ) = sin 18 ∘ = 5 − 1 4 = φ − 1 2 = 1 2 φ {\displaystyle \sin \left({\frac {\pi }{10}}\right)=\sin 18^{\circ }={{\sqrt {5}}-1 \over 4}={\varphi -1 \over 2}={1 \over 2\varphi }}
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Nâng cao
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- − sin ( π 7 ) sin 2 ( 2 π 7 ) + sin ( 3 π 7 ) sin 2 ( π 7 ) + sin ( 2 π 7 ) sin − 2 ( 3 π 7 ) = 2 7 {\displaystyle -{\frac {\sin({\frac {\pi }{7}})}{\sin ^{2}({\frac {2\pi }{7}})}}+{\frac {\sin({\frac {3\pi }{7}})}{\sin ^{2}({\frac {\pi }{7}})}}+{\frac {\sin({\frac {2\pi }{7}})}{\sin -^{2}({\frac {3\pi }{7}})}}=2{\sqrt {7}}}
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- sin 2 ( π 7 ) sin 4 ( 2 π 7 ) + sin 2 ( 3 π 7 ) sin 4 ( π 7 ) + sin 2 ( 2 π 7 ) sin − 4 ( 3 π 7 ) = 28 {\displaystyle {\frac {\sin ^{2}({\frac {\pi }{7}})}{\sin ^{4}({\frac {2\pi }{7}})}}+{\frac {\sin ^{2}({\frac {3\pi }{7}})}{\sin ^{4}({\frac {\pi }{7}})}}+{\frac {\sin ^{2}({\frac {2\pi }{7}})}{\sin -^{4}({\frac {3\pi }{7}})}}=28}
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- sin 2 ( π 7 ) sin 4 ( 2 π 7 ) ( 4 sin ( π 7 ) sin ( 2 π 7 ) − 2 sin ( 3 π 7 ) sin − ( π 7 ) ) + sin 2 ( 3 π 7 ) sin 4 ( π 7 ) ( 2 sin ( 2 π 7 ) sin ( 3 π 7 ) + 4 sin ( 3 π 7 ) sin − ( π 7 ) ) − sin 2 ( 2 π 7 ) sin 4 ( 3 π 7 ) ( 2 sin ( π 7 ) sin ( 2 π 7 ) + 4 − sin ( 2 π 7 ) sin ( 3 π 7 ) ) = 280 {\displaystyle {\frac {\sin ^{2}({\frac {\pi }{7}})}{\sin ^{4}({\frac {2\pi }{7}})}}({\frac {4\sin({\frac {\pi }{7}})}{\sin({\frac {2\pi }{7}})}}-{\frac {2\sin({\frac {3\pi }{7}})}{\sin -({\frac {\pi }{7}})}})+{\frac {\sin ^{2}({\frac {3\pi }{7}})}{\sin ^{4}({\frac {\pi }{7}})}}({\frac {2\sin({\frac {2\pi }{7}})}{\sin({\frac {3\pi }{7}})}}+{\frac {4\sin({\frac {3\pi }{7}})}{\sin -({\frac {\pi }{7}})}})-{\frac {\sin ^{2}({\frac {2\pi }{7}})}{\sin ^{4}({\frac {3\pi }{7}})}}({\frac {2\sin({\frac {\pi }{7}})}{\sin({\frac {2\pi }{7}})}}+{\frac {4-\sin({\frac {2\pi }{7}})}{\sin({\frac {3\pi }{7}})}})=280}
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- cos ( π 17 ) = 1 8 ( 2 ( 2 17 ( 17 − 17 ) 2 − 17 − 17 2 − 4 34 + 2 17 + 3 17 + 17 + 34 − 2 17 + 17 + 15 ) ) {\displaystyle \cos({\frac {\pi }{17}})={\frac {1}{8}}{\sqrt {(}}2(2{\sqrt {{\sqrt {\frac {17(17-{\sqrt {17}})}{2}}}-{\sqrt {\frac {17-{\sqrt {17}}}{2}}}-4{\sqrt {34+2{\sqrt {17}}}}+3{\sqrt {17}}+17}}+{\sqrt {34-2{\sqrt {17}}}}+{\sqrt {17}}+15))}
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- tan ( π 120 ) = 8 − 2 ( 2 − 3 ) ( 3 − 5 ) − 2 ( 2 + 3 ) ( 5 + 5 ) 8 + 2 ( 2 − 3 ) ( 3 − 5 ) + 2 − ( 2 + 3 ) ( 5 + 5 ) {\displaystyle \tan({\frac {\pi }{120}})={\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}})}}}}}}
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- cos ( π 240 ) = 1 16 ( 2 − 2 + 2 ( 2 ( 5 + 5 ) + 3 − 15 ) + 2 + 2 + 2 ( 6 ( 5 + 5 ) + 5 − 1 ) ) {\displaystyle \cos({\frac {\pi }{240}})={\frac {1}{16}}({\sqrt {2-{\sqrt {2+{\sqrt {2}}}}}}({\sqrt {2(5+{\sqrt {5}})}}+{\sqrt {3}}-{\sqrt {15}})+{\sqrt {{\sqrt {2+{\sqrt {2}}}}+2}}({\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1))}
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- π 4 = cot − 1 ( 2 ) + cot − 1 ( 3 ) {\displaystyle {\frac {\pi }{4}}=\cot ^{-1}(2)+\cot ^{-1}(3)}
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- π 4 = cot − 1 ( 2 ) + cot − 1 ( 5 ) + cot − 1 ( 8 ) {\displaystyle {\frac {\pi }{4}}=\cot ^{-1}(2)+\cot ^{-1}(5)+\cot ^{-1}(8)}
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- π 4 = 2 cot − 1 ( 3 ) + cot − 1 ( 7 ) {\displaystyle {\frac {\pi }{4}}=2\cot ^{-1}(3)+\cot ^{-1}(7)}
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- π 4 = 3 cot − 1 ( 4 ) + cot − 1 ( 99 5 ) {\displaystyle {\frac {\pi }{4}}=3\cot ^{-1}(4)+\cot ^{-1}({\frac {99}{5}})}
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- π 4 = 4 cot − 1 ( 5 ) − cot − 1 ( 239 ) {\displaystyle {\frac {\pi }{4}}=4\cot ^{-1}(5)-\cot ^{-1}(239)}
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- π 4 = 4 cot − 1 ( 5 ) − cot − 1 ( 70 ) + cot − 1 ( 99 ) π 4 = 5 cot − 1 ( 6 ) − cot − 1 ( 503 16 ) − cot − 1 ( 117 ) {\displaystyle {\frac {\pi }{4}}=4\cot ^{-1}(5)-\cot ^{-1}(70)+\cot ^{-1}(99){\frac {\pi }{4}}=5\cot ^{-1}(6)-\cot ^{-1}({\frac {503}{16}})-\cot ^{-1}(117)}
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- π 4 = 5 cot − 1 ( 7 ) + 2 cot − 1 ( 79 3 ) π 4 = 6 cot − 1 ( 8 ) + cot − 1 ( 99 5 ) − 3 cot − 1 ( 268 ) {\displaystyle {\frac {\pi }{4}}=5\cot ^{-1}(7)+2\cot ^{-1}({\frac {79}{3}}){\frac {\pi }{4}}=6\cot ^{-1}(8)+\cot ^{-1}({\frac {99}{5}})-3\cot ^{-1}(268)}
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- π 4 = 8 cot − 1 ( 10 ) − cot − 1 ( 239 ) − 4 cot − 1 ( 515 ) π 4 = 8 cot − 1 ( 10 ) − 2 cot − 1 ( 452761 2543 ) − cot − 1 ( 1393 ) {\displaystyle {\frac {\pi }{4}}=8\cot ^{-1}(10)-\cot ^{-1}(239)-4\cot ^{-1}(515){\frac {\pi }{4}}=8\cot ^{-1}(10)-2\cot ^{-1}({\frac {452761}{2543}})-\cot ^{-1}(1393)}
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- π 4 = 8 cot − 1 ( 10 ) − cot − 1 ( 100 ) − cot − 1 ( 515 ) − cot − 1 ( 371498882 3583 ) π 4 = 12 cot − 1 ( 18 ) + 3 cot − 1 ( 70 ) + 5 cot − 1 ( 99 ) + 8 cot − 1 ( 307 ) {\displaystyle {\frac {\pi }{4}}=8\cot ^{-1}(10)-\cot ^{-1}(100)-\cot ^{-1}(515)-\cot ^{-1}({\frac {371498882}{3583}}){\frac {\pi }{4}}=12\cot ^{-1}(18)+3\cot ^{-1}(70)+5\cot ^{-1}(99)+8\cot ^{-1}(307)}
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- π 4 = 12 cot − 1 ( 18 ) + 8 cot − 1 ( 99 ) + 3 cot − 1 ( 239 ) + 8 cot − 1 ( 307 ) {\displaystyle {\frac {\pi }{4}}=12\cot ^{-1}(18)+8\cot ^{-1}(99)+3\cot ^{-1}(239)+8\cot ^{-1}(307)}
