Trigonometrini? funkcij? grafikai: sinusas , kosinusas , tangentas , kotangentas , sekantas , kosekantas 百度 本报讯（沈阳日报、沈报融媒记者吕良德）3月23日，市法律援助中心劳动人事争议仲裁工作站揭牌仪式在市劳动人事争议仲裁院立案大厅举行。
Trigonometrin? funkcija – realaus arba kompleksinio kintamojo elementarioji funkcija: sinusas, kosinusas, tangentas, kotangentas, sekantas, kosekantas.[ 1]
Geometrine prasme trigonometrin?s funkcijos nusako ry?? tarp trikampio kra?tini? ir kamp? .[ 2]
Viena pagrindini? ?i? funkcij? savybi? yra j? periodi?kumas, ta?iau ne kiekviena periodin? funkcija, kurios argumentas yra kampas, yra trigonometrin? funkcija. Pavyzd?iui, 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 ta?ko, atitinkan?io kampo α apskritim?, ordinat? ir abscis?, tai pagal Pitagoro teorem?:
sin
2
α
+
cos
2
α
=
1.
{\displaystyle \sin ^{2}\alpha +\cos ^{2}\alpha =1.\qquad \qquad \,}
Abi ?ios lygties puses padalijus i? 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 lygin? funkcija, nes
cos
(
−
α
)
=
cos
α
.
{\displaystyle \cos(-\alpha )=\cos \alpha .}
Sinusas yra nelygin? funkcija, nes
sin
(
−
α
)
=
−
sin
α
.
{\displaystyle \sin(-\alpha )=-\sin \alpha .}
Tangentas ir kotangentas yra nelygin?s 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 )}.}
? formul?
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)}
?sta?ius
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 formul?je
sin
α
=
cos
(
π
2
−
α
)
(
2
)
{\displaystyle \sin \alpha =\cos({\frac {\pi }{2}}-\alpha )\quad (2)}
?sta?ius
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 ? (1) formul? ?sta?ius
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 i?plaukia i? formul?s (2) ?sta?ius ? j?
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 formul? (3). 
