Trigonometrinė funkcija – realaus arba kompleksinio kintamojo elementarioji funkcija: sinusas, kosinusas, tangentas, kotangentas, sekantas, kosekantas.
Geometrine prasme trigonometrinės funkcijos nusako ryšį tarp trikampio kraštinių ir kampų .
Viena pagrindinių šių funkcijų savybių yra jų periodiškumas, tačiau ne kiekviena periodinė funkcija, kurios argumentas yra kampas, yra trigonometrinė funkcija. Pavyzdžiui, funkcija e sin x + cos x {\displaystyle e^{\sin x}+\cos x}
α {\displaystyle \alpha \,\!} 0° (0 rad ) 30° (π/6) 45° (π/4) 60° (π/3) 90° (π/2) 180° (π) 270° (3π/2) 360° (2π) sin α {\displaystyle \sin \alpha \,\!} 0 {\displaystyle {0}\,\!} 1 2 {\displaystyle {\frac {1}{2}}\,\!} 2 2 {\displaystyle {\frac {\sqrt {2}}{2}}\,\!} 3 2 {\displaystyle {\frac {\sqrt {3}}{2}}\,\!} 1 {\displaystyle {1}\,\!} 0 {\displaystyle {0}\,\!} − 1 {\displaystyle {-1}\,\!} 0 {\displaystyle {0}\,\!} cos α {\displaystyle \cos \alpha \,\!} 1 {\displaystyle {1}\,\!} 3 2 {\displaystyle {\frac {\sqrt {3}}{2}}\,\!} 2 2 {\displaystyle {\frac {\sqrt {2}}{2}}\,\!} 1 2 {\displaystyle {\frac {1}{2}}\,\!} 0 {\displaystyle {0}\,\!} − 1 {\displaystyle {-1}\,\!} 0 {\displaystyle {0}\,\!} 1 {\displaystyle {1}\,\!} t g α {\displaystyle \mathop {\mathrm {tg} } \,\alpha \,\!} 0 {\displaystyle {0}\,\!} 1 3 {\displaystyle {\frac {1}{\sqrt {3}}}\,\!} 1 {\displaystyle {1}\,\!} 3 {\displaystyle {\sqrt {3}}\,\!} ∞ {\displaystyle \infty } 0 {\displaystyle {0}\,\!} ∞ {\displaystyle \infty } 0 {\displaystyle {0}\,\!} c t g α {\displaystyle \mathop {\mathrm {ctg} } \,\alpha \,\!} ∞ {\displaystyle \infty } 3 {\displaystyle {\sqrt {3}}\,\!} 1 {\displaystyle {1}\,\!} 1 3 {\displaystyle {\frac {1}{\sqrt {3}}}\,\!} 0 {\displaystyle {0}\,\!} ∞ {\displaystyle \infty } 0 {\displaystyle {0}\,\!} ∞ {\displaystyle \infty } sec α {\displaystyle \sec \alpha \,\!} 1 {\displaystyle {1}\,\!} 2 3 {\displaystyle {\frac {2}{\sqrt {3}}}\,\!} 2 {\displaystyle {\sqrt {2}}\,\!} 2 {\displaystyle {2}\,\!} ∞ {\displaystyle \infty } − 1 {\displaystyle {-1}\,\!} ∞ {\displaystyle \infty } 1 {\displaystyle {1}\,\!} cosec α {\displaystyle \operatorname {cosec} \,\alpha \,\!} ∞ {\displaystyle \infty } 2 {\displaystyle {2}\,\!} 2 {\displaystyle {\sqrt {2}}\,\!} 2 3 {\displaystyle {\frac {2}{\sqrt {3}}}\,\!} 1 {\displaystyle {1}\,\!} ∞ {\displaystyle \infty } − 1 {\displaystyle {-1}\,\!} ∞ {\displaystyle \infty }
α {\displaystyle \alpha \,} π 12 = 15 ∘ {\displaystyle {\frac {\pi }{12}}=15^{\circ }} π 10 = 18 ∘ {\displaystyle {\frac {\pi }{10}}=18^{\circ }} π 8 = 22 , 5 ∘ {\displaystyle {\frac {\pi }{8}}=22,5^{\circ }} π 5 = 36 ∘ {\displaystyle {\frac {\pi }{5}}=36^{\circ }} 3 π 10 = 54 ∘ {\displaystyle {\frac {3\,\pi }{10}}=54^{\circ }} 3 π 8 = 67 , 5 ∘ {\displaystyle {\frac {3\,\pi }{8}}=67,5^{\circ }} 2 π 5 = 72 ∘ {\displaystyle {\frac {2\,\pi }{5}}=72^{\circ }} sin α {\displaystyle \sin \alpha \,} 3 − 1 2 2 {\displaystyle {\frac {{\sqrt {3}}-1}{2\,{\sqrt {2}}}}} 5 − 1 4 {\displaystyle {\frac {{\sqrt {5}}-1}{4}}} 2 − 2 2 {\displaystyle {\frac {\sqrt {2-{\sqrt {2}}}}{2}}} 5 − 5 2 2 {\displaystyle {\frac {\sqrt {5-{\sqrt {5}}}}{2\,{\sqrt {2}}}}} 5 + 1 4 {\displaystyle {\frac {{\sqrt {5}}+1}{4}}} 2 + 2 2 {\displaystyle {\frac {\sqrt {2+{\sqrt {2}}}}{2}}} 5 + 5 2 2 {\displaystyle {\frac {\sqrt {5+{\sqrt {5}}}}{2\,{\sqrt {2}}}}} cos α {\displaystyle \cos \alpha \,} 3 + 1 2 2 {\displaystyle {\frac {{\sqrt {3}}+1}{2\,{\sqrt {2}}}}} 5 + 5 2 2 {\displaystyle {\frac {\sqrt {5+{\sqrt {5}}}}{2\,{\sqrt {2}}}}} 2 + 2 2 {\displaystyle {\frac {\sqrt {2+{\sqrt {2}}}}{2}}} 5 + 1 4 {\displaystyle {\frac {{\sqrt {5}}+1}{4}}} 5 − 5 2 2 {\displaystyle {\frac {\sqrt {5-{\sqrt {5}}}}{2\,{\sqrt {2}}}}} 2 − 2 2 {\displaystyle {\frac {\sqrt {2-{\sqrt {2}}}}{2}}} 5 − 1 4 {\displaystyle {\frac {{\sqrt {5}}-1}{4}}} tg α {\displaystyle \operatorname {tg} \,\alpha } 2 − 3 {\displaystyle 2-{\sqrt {3}}} 1 − 2 5 {\displaystyle {\sqrt {1-{\frac {2}{\sqrt {5}}}}}} 2 − 1 2 + 1 {\displaystyle {\sqrt {\frac {{\sqrt {2}}-1}{{\sqrt {2}}+1}}}} 5 − 2 5 {\displaystyle {\sqrt {5-2\,{\sqrt {5}}}}} 1 + 2 5 {\displaystyle {\sqrt {1+{\frac {2}{\sqrt {5}}}}}} 2 + 1 2 − 1 {\displaystyle {\sqrt {\frac {{\sqrt {2}}+1}{{\sqrt {2}}-1}}}} 5 + 2 5 {\displaystyle {\sqrt {5+2\,{\sqrt {5}}}}} ctg α {\displaystyle \operatorname {ctg} \,\alpha } 2 + 3 {\displaystyle 2+{\sqrt {3}}} 5 + 2 5 {\displaystyle {\sqrt {5+2\,{\sqrt {5}}}}} 2 + 1 2 − 1 {\displaystyle {\sqrt {\frac {{\sqrt {2}}+1}{{\sqrt {2}}-1}}}} 1 + 2 5 {\displaystyle {\sqrt {1+{\frac {2}{\sqrt {5}}}}}} 5 − 2 5 {\displaystyle {\sqrt {5-2\,{\sqrt {5}}}}} 2 − 1 2 + 1 {\displaystyle {\sqrt {\frac {{\sqrt {2}}-1}{{\sqrt {2}}+1}}}} 1 − 2 5 {\displaystyle {\sqrt {1-{\frac {2}{\sqrt {5}}}}}}
tg π 120 = tg 1 , 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 {tg} 1,5^{\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 = 1 16 ( 2 − k ( 2 ( 5 + 5 ) + 3 − 15 ) + 2 + k ( 6 ( 5 + 5 ) + 5 − 1 ) ) {\displaystyle \cos {\frac {\pi }{240}}={\frac {1}{16}}\left({\sqrt {2-k}}\left({\sqrt {2(5+{\sqrt {5}})}}+{\sqrt {3}}-{\sqrt {15}}\right)+{\sqrt {2+k}}\left({\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right)\right)} k = 2 + 2 {\displaystyle k={\sqrt {2+{\sqrt {2}}}}}
cos π 17 = 1 8 2 ( 2 17 k 2 − k 2 − 4 2 ( 17 + 17 ) + 3 17 + 17 + 2 k + 17 + 15 ) {\displaystyle \cos {\frac {\pi }{17}}={\frac {1}{8}}{\sqrt {2\left(2{\sqrt {{\sqrt {\frac {17k}{2}}}-{\sqrt {\frac {k}{2}}}-4{\sqrt {2(17+{\sqrt {17}})}}+3{\sqrt {17}}+17}}+{\sqrt {2k}}+{\sqrt {17}}+15\right)}}} k = 17 − 17 {\displaystyle k=17-{\sqrt {17}}}
u {\displaystyle u} π 2 + α {\displaystyle {\frac {\pi }{2}}+\alpha } π + α {\displaystyle \pi +\alpha } 3 π 2 + α {\displaystyle {\frac {3\pi }{2}}+\alpha } − α {\displaystyle -\alpha } π 2 − α {\displaystyle {\frac {\pi }{2}}-\alpha } π − α {\displaystyle \pi -\alpha } 3 π 2 − α {\displaystyle {\frac {3\pi }{2}}-\alpha } sin u {\displaystyle \sin u\,} cos α {\displaystyle \cos \alpha } − sin α {\displaystyle -\sin \alpha } − cos α {\displaystyle -\cos \alpha } − sin α {\displaystyle -\sin \alpha } cos α {\displaystyle \cos \alpha } sin α {\displaystyle \sin \alpha } − cos α {\displaystyle -\cos \alpha } cos u {\displaystyle \cos u\,} − sin α {\displaystyle -\sin \alpha } − cos α {\displaystyle -\cos \alpha } sin α {\displaystyle \sin \alpha } cos α {\displaystyle \cos \alpha } sin α {\displaystyle \sin \alpha } − cos α {\displaystyle -\cos \alpha } − sin α {\displaystyle -\sin \alpha } tg u {\displaystyle \operatorname {tg} u} − ctg α {\displaystyle -\operatorname {ctg} \alpha } tg α {\displaystyle \operatorname {tg} \alpha } − ctg α {\displaystyle -\operatorname {ctg} \alpha } − tg α {\displaystyle -\operatorname {tg} \alpha } ctg α {\displaystyle \operatorname {ctg} \alpha } − tg α {\displaystyle -\operatorname {tg} \alpha } ctg α {\displaystyle \operatorname {ctg} \alpha } ctg u {\displaystyle \operatorname {ctg} u} − tg α {\displaystyle -\operatorname {tg} \alpha } ctg α {\displaystyle \operatorname {ctg} \alpha } − tg α {\displaystyle -\operatorname {tg} \alpha } − ctg α {\displaystyle -\operatorname {ctg} \alpha } tg α {\displaystyle \operatorname {tg} \alpha } − ctg α {\displaystyle -\operatorname {ctg} \alpha } tg α {\displaystyle \operatorname {tg} \alpha }
Kadangi sinusas ir kosinusas yra atitinkamai taško, atitinkančio kampo α apskritimą, ordinatė ir abscisė, tai pagal Pitagoro teoremą:
sin 2 α + cos 2 α = 1. {\displaystyle \sin ^{2}\alpha +\cos ^{2}\alpha =1.\qquad \qquad \,} Abi šios lygties puses padalijus iš sinuso kvadrato arba kosinuso kvadrato, gaunama:
1 + t g 2 α = 1 cos 2 α , {\displaystyle 1+\mathop {\mathrm {tg} } \,^{2}\alpha ={\frac {1}{\cos ^{2}\alpha }},\qquad \qquad \,} 1 + c t g 2 α = 1 sin 2 α . {\displaystyle 1+\mathop {\mathrm {ctg} } \,^{2}\alpha ={\frac {1}{\sin ^{2}\alpha }}.\qquad \qquad \,}
Funkcijos y = sin α {\displaystyle y=\sin \alpha } y = cos α {\displaystyle y=\cos \alpha } y = sec α {\displaystyle y=\sec \alpha } y = csc α {\displaystyle y=\csc \alpha } 2 π {\displaystyle 2\pi } y = tg α {\displaystyle y=\operatorname {tg} \alpha } y = ctg α {\displaystyle y=\operatorname {ctg} \alpha } π {\displaystyle \pi }
Kosinusas yra lyginė funkcija, nes
cos ( − α ) = cos α . {\displaystyle \cos(-\alpha )=\cos \alpha .} Sinusas yra nelyginė funkcija, nes
sin ( − α ) = − sin α . {\displaystyle \sin(-\alpha )=-\sin \alpha .} Tangentas ir kotangentas yra nelyginės funkcijos, t. y.
tg ( − α ) = − tg α ; {\displaystyle {\text{tg}}(-\alpha )=-{\text{tg}}\;\alpha ;} ctg ( − α ) = − ctg α . {\displaystyle {\text{ctg}}(-\alpha )=-{\text{ctg}}\;\alpha .}
cos x = sin ( x + π 2 ) . {\displaystyle \cos x=\sin {\Big (}x+{\frac {\pi }{2}}{\Big )}.} Į formulę cos ( α − β ) = cos α cos β + sin α sin β ( 1 ) {\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta \quad (1)} įstačius π 2 {\displaystyle {\frac {\pi }{2}}} α {\displaystyle \alpha } α {\displaystyle \alpha } β {\displaystyle \beta } cos ( π 2 − α ) = cos π 2 cos α + sin π 2 sin α = sin α . {\displaystyle \cos({\frac {\pi }{2}}-\alpha )=\cos {\frac {\pi }{2}}\cos \alpha +\sin {\frac {\pi }{2}}\sin \alpha =\sin \alpha .} Gautoje formulėje sin α = cos ( π 2 − α ) ( 2 ) {\displaystyle \sin \alpha =\cos({\frac {\pi }{2}}-\alpha )\quad (2)} įstačius α + β {\displaystyle \alpha +\beta } α , {\displaystyle \alpha ,} sin ( α + β ) = cos ( π 2 − α − β ) = cos ( ( π 2 − α ) − β ) . {\displaystyle \sin(\alpha +\beta )=\cos({\frac {\pi }{2}}-\alpha -\beta )=\cos(({\frac {\pi }{2}}-\alpha )-\beta ).} Toliau į (1) formulę įstačius π 2 − α {\displaystyle {\frac {\pi }{2}}-\alpha } α , {\displaystyle \alpha ,} sin ( α + β ) = cos ( ( π 2 − α ) − β ) = {\displaystyle \sin(\alpha +\beta )=\cos(({\frac {\pi }{2}}-\alpha )-\beta )=} = cos ( π 2 − α ) cos β + sin ( π 2 − α ) sin β = {\displaystyle =\cos({\frac {\pi }{2}}-\alpha )\cos \beta +\sin({\frac {\pi }{2}}-\alpha )\sin \beta =} = sin α cos β + cos α sin β . {\displaystyle =\sin \alpha \cos \beta +\cos \alpha \sin \beta .} Pasinaudojome formule sin ( π 2 − α ) = cos α , ( 3 ) {\displaystyle \sin({\frac {\pi }{2}}-\alpha )=\cos \alpha ,\quad (3)} kuri išplaukia iš formulės (2) įstačius į ją π 2 − α {\displaystyle {\frac {\pi }{2}}-\alpha } α ; {\displaystyle \alpha ;} [sin α = cos ( π 2 − α ) ( 2 ) {\displaystyle \sin \alpha =\cos({\frac {\pi }{2}}-\alpha )\quad (2)} sin ( π 2 − α ) = cos ( π 2 − ( π 2 − α ) ) = cos α . {\displaystyle \sin({\frac {\pi }{2}}-\alpha )=\cos({\frac {\pi }{2}}-({\frac {\pi }{2}}-\alpha ))=\cos \alpha .} Taigi, gavome formulę (3).