Sinusi i kosinusi u jediničnoj kružnici Trigonometrijske jednakosti pokazuju poveznice između pojedinih trigonometrijskih funkcija . Ti izrazi su istiniti za svaku odabranu vrijednost određene varijable (kuta ili nekog drugog broja). Kako su trigonometrijske funkcije međusobno povezane pomoću vrijednosti jedne, moguće je izraziti neku drugu funkciju . Jednakosti se koriste za pojednostavljenje izraza koji uključuju trigonometrijske funkcije.
Podrobniji članak o temi:
Kut Nazivi kutova se daju prema slovima grčkog alfabeta kao što su alfa (α ), beta (β ), gama (γ ), delta (δ ) i theta (θ ). Mjerne jedinice za mjerenje kutova su stupnjevi , radijani i gradi :
1 puni krug = 360 stupnjeva = 2
π
{\displaystyle \pi }
Sljedeća tablica prikazuje pretvorbu mjernih jedinica za određene veličine kutova:
Stupnjevi
30°
60°
120°
150°
210°
240°
300°
330°
Radijani
π
6
{\displaystyle {\frac {\pi }{6}}\!}
π
3
{\displaystyle {\frac {\pi }{3}}\!}
2
π
3
{\displaystyle {\frac {2\pi }{3}}\!}
5
π
6
{\displaystyle {\frac {5\pi }{6}}\!}
7
π
6
{\displaystyle {\frac {7\pi }{6}}\!}
4
π
3
{\displaystyle {\frac {4\pi }{3}}\!}
5
π
3
{\displaystyle {\frac {5\pi }{3}}\!}
11
π
6
{\displaystyle {\frac {11\pi }{6}}\!}
Gradi
33⅓
66⅔
133⅓
166⅔
233⅓
266⅔
333⅓
366⅔
Stupnjevi
45°
90°
135°
180°
225°
270°
315°
360°
Radijani
π
4
{\displaystyle {\frac {\pi }{4}}\!}
π
2
{\displaystyle {\frac {\pi }{2}}\!}
3
π
4
{\displaystyle {\frac {3\pi }{4}}\!}
π
{\displaystyle \pi \!}
5
π
4
{\displaystyle {\frac {5\pi }{4}}\!}
3
π
2
{\displaystyle {\frac {3\pi }{2}}\!}
7
π
4
{\displaystyle {\frac {7\pi }{4}}\!}
2
π
{\displaystyle 2\pi \!}
Gradi
50
100
150
200
250
300
350
400
Kutovi se u trigonometriji najčešće izražavaju u radijanima i to bez mjerne jedinice, stupnjevi s oznakom ° se manje koriste, a gradi izrazito rijetko.
Primarne trigonometrijske funkcije su sinus i kosinus kuta. Sinus se označava sa sinθ , a kosinus s cosθ pri čemu je θ naziv kuta.
Tangens (tg, tan) kuta je omjer sinusa i kosinusa:
tg
θ
=
sin
θ
cos
θ
.
{\displaystyle \operatorname {tg} \theta ={\frac {\sin \theta }{\cos \theta }}.}
S druge strane, imamo i recipročne funkcije pri čemu je kosinusu recipročan sekans (sec), sinusu kosekans (csc, cosec), a tangensu kotangens (ctg, cot):
sec
θ
=
1
cos
θ
,
csc
θ
=
1
sin
θ
,
ctg
θ
=
1
tg
θ
=
cos
θ
sin
θ
.
{\displaystyle \sec \theta ={\frac {1}{\cos \theta }},\quad \csc \theta ={\frac {1}{\sin \theta }},\quad \operatorname {ctg} \theta ={\frac {1}{\operatorname {tg} \theta }}={\frac {\cos \theta }{\sin \theta }}.}
Inverzne trigonometrijske funkcije ili arkus funkcije su inverzne funkcije trigonometrijskim funkcijama. Prema tome imamo, arkus sinus (arcsin, asin) je inverzna funkcija sinusnoj funkciji, pri čemu vrijedi da je
sin
(
arcsin
x
)
=
x
{\displaystyle \sin(\arcsin x)=x\!}
i
arcsin
(
sin
θ
)
=
θ
za
−
π
/
2
≤
θ
≤
π
/
2.
{\displaystyle \arcsin(\sin \theta )=\theta \quad {\text{za }}-\pi /2\leq \theta \leq \pi /2.}
U sljedećoj tablici su prikazane i druge komplementarne inverzne funkcije i kratice:
Trigonometrijska funkcija
Sinus
Kosinus
Tangens
Sekans
Kosekans
Kotangens
Kratica
sin
θ
{\displaystyle \operatorname {sin} \theta }
cos
θ
{\displaystyle \operatorname {cos} \theta }
tg
θ
{\displaystyle \operatorname {tg} \theta }
sec
θ
{\displaystyle \operatorname {sec} \theta }
csc
θ
{\displaystyle \operatorname {csc} \theta }
ctg
θ
{\displaystyle \operatorname {ctg} \theta }
Inverzna trigonometrijska funkcija
Arkus sinus
Arkus kosinus
Arkus tangens
Arkus sekans
Arkus kosekans
Arkus kotangens
Kratica
arcsin
θ
{\displaystyle \operatorname {arcsin} \theta }
arccos
θ
{\displaystyle \operatorname {arccos} \theta }
arctg
θ
{\displaystyle \operatorname {arctg} \theta }
arcsec
θ
{\displaystyle \operatorname {arcsec} \theta }
arccsc
θ
{\displaystyle \operatorname {arccsc} \theta }
arcctg
θ
{\displaystyle \operatorname {arcctg} \theta }
Pitagorina trigonometrijska jednakost ili temeljni identitet trigonometrije je jedna od osnovnih trigonometrijskih jednakosti i prikazuje odnos između sinusa i kosinusa:
cos
2
θ
+
sin
2
θ
=
1
{\displaystyle \cos ^{2}\theta +\sin ^{2}\theta =1\!}
gdje cos2 θ znači (cos(θ ))2 i sin2 θ znači (sin(θ ))2 .
Izraz je u biti izvedenica Pitagorinog poučka i proizilazi iz jednakosti
x
2
+
y
2
=
1
{\displaystyle x^{2}+y^{2}=1\!\ }
jediničnu kružnicu . Ova jednadžba može biti riješena za sinus i za kosinus:
sin
θ
=
±
1
−
cos
2
θ
i
cos
θ
=
±
1
−
sin
2
θ
.
{\displaystyle \sin \theta =\pm {\sqrt {1-\cos ^{2}\theta }}\quad {\text{i}}\quad \cos \theta =\pm {\sqrt {1-\sin ^{2}\theta }}.\,}
Podijelivši Pitagorinu jednakost s cos2 θ ili sa sin2 θ dobivamo sljedeće dvije jednakosti:
1
+
tg
2
θ
=
sec
2
θ
i
1
+
ctg
2
θ
=
csc
2
θ
.
{\displaystyle 1+\operatorname {tg} ^{2}\theta =\sec ^{2}\theta \quad {\text{i}}\quad 1+\operatorname {ctg} ^{2}\theta =\csc ^{2}\theta .\!}
Koristeći navedene jednakosti te omjere koji su korišteni pri definiranju trigonometrijskih funkcija, mogu se izvesti trigonometrijske jednakosti gdje je jedna trigonometrijska funkcija prikazana pomoću druge:
Svaka trigonometrijska funkcija prikazana pomoću druge trigonometrijske funkcije[ 1]
in terms of
sin
θ
{\displaystyle \sin \theta \!}
cos
θ
{\displaystyle \cos \theta \!}
tg
θ
{\displaystyle \operatorname {tg} \theta \!}
csc
θ
{\displaystyle \csc \theta \!}
sec
θ
{\displaystyle \sec \theta \!}
ctg
θ
{\displaystyle \operatorname {ctg} \theta \!}
sin
θ
=
{\displaystyle \sin \theta =\!}
sin
θ
{\displaystyle \sin \theta \ }
±
1
−
cos
2
θ
{\displaystyle \pm {\sqrt {1-\cos ^{2}\theta }}\!}
±
tg
θ
1
+
tg
2
θ
{\displaystyle \pm {\frac {\operatorname {tg} \theta }{\sqrt {1+\operatorname {tg} ^{2}\theta }}}\!}
1
csc
θ
{\displaystyle {\frac {1}{\csc \theta }}\!}
±
sec
2
θ
−
1
sec
θ
{\displaystyle \pm {\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}\!}
±
1
1
+
ctg
2
θ
{\displaystyle \pm {\frac {1}{\sqrt {1+\operatorname {ctg} ^{2}\theta }}}\!}
cos
θ
=
{\displaystyle \cos \theta =\!}
±
1
−
sin
2
θ
{\displaystyle \pm {\sqrt {1-\sin ^{2}\theta }}\!}
cos
θ
{\displaystyle \cos \theta \!}
±
1
1
+
tg
2
θ
{\displaystyle \pm {\frac {1}{\sqrt {1+\operatorname {tg} ^{2}\theta }}}\!}
±
csc
2
θ
−
1
csc
θ
{\displaystyle \pm {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}\!}
1
sec
θ
{\displaystyle {\frac {1}{\sec \theta }}\!}
±
ctg
θ
1
+
ctg
2
θ
{\displaystyle \pm {\frac {\operatorname {ctg} \theta }{\sqrt {1+\operatorname {ctg} ^{2}\theta }}}\!}
tg
θ
=
{\displaystyle \operatorname {tg} \theta =\!}
±
sin
θ
1
−
sin
2
θ
{\displaystyle \pm {\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}\!}
±
1
−
cos
2
θ
cos
θ
{\displaystyle \pm {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}\!}
tg
θ
{\displaystyle \operatorname {tg} \theta \!}
±
1
csc
2
θ
−
1
{\displaystyle \pm {\frac {1}{\sqrt {\csc ^{2}\theta -1}}}\!}
±
sec
2
θ
−
1
{\displaystyle \pm {\sqrt {\sec ^{2}\theta -1}}\!}
1
ctg
θ
{\displaystyle {\frac {1}{\operatorname {ctg} \theta }}\!}
csc
θ
=
{\displaystyle \csc \theta =\!}
1
sin
θ
{\displaystyle {\frac {1}{\sin \theta }}\!}
±
1
1
−
cos
2
θ
{\displaystyle \pm {\frac {1}{\sqrt {1-\cos ^{2}\theta }}}\!}
±
1
+
tg
2
θ
tg
θ
{\displaystyle \pm {\frac {\sqrt {1+\operatorname {tg} ^{2}\theta }}{\operatorname {tg} \theta }}\!}
csc
θ
{\displaystyle \csc \theta \!}
±
sec
θ
sec
2
θ
−
1
{\displaystyle \pm {\frac {\sec \theta }{\sqrt {\sec ^{2}\theta -1}}}\!}
±
1
+
ctg
2
θ
{\displaystyle \pm {\sqrt {1+\operatorname {ctg} ^{2}\theta }}\!}
sec
θ
=
{\displaystyle \sec \theta =\!}
±
1
1
−
sin
2
θ
{\displaystyle \pm {\frac {1}{\sqrt {1-\sin ^{2}\theta }}}\!}
1
cos
θ
{\displaystyle {\frac {1}{\cos \theta }}\!}
±
1
+
tg
2
θ
{\displaystyle \pm {\sqrt {1+\operatorname {tg} ^{2}\theta }}\!}
±
csc
θ
csc
2
θ
−
1
{\displaystyle \pm {\frac {\csc \theta }{\sqrt {\csc ^{2}\theta -1}}}\!}
sec
θ
{\displaystyle \sec \theta \!}
±
1
+
ctg
2
θ
ctg
θ
{\displaystyle \pm {\frac {\sqrt {1+\operatorname {ctg} ^{2}\theta }}{\operatorname {ctg} \theta }}\!}
ctg
θ
=
{\displaystyle \operatorname {ctg} \theta =\!}
±
1
−
sin
2
θ
sin
θ
{\displaystyle \pm {\frac {\sqrt {1-\sin ^{2}\theta }}{\sin \theta }}\!}
±
cos
θ
1
−
cos
2
θ
{\displaystyle \pm {\frac {\cos \theta }{\sqrt {1-\cos ^{2}\theta }}}\!}
1
tg
θ
{\displaystyle {\frac {1}{\operatorname {tg} \theta }}\!}
±
csc
2
θ
−
1
{\displaystyle \pm {\sqrt {\csc ^{2}\theta -1}}\!}
±
1
sec
2
θ
−
1
{\displaystyle \pm {\frac {1}{\sqrt {\sec ^{2}\theta -1}}}\!}
ctg
θ
{\displaystyle \operatorname {ctg} \theta \!}
Sve trigonometrijske funkcije kuta θ mogu biti geometrijski konstruirane s obzirom na jediničnu kružnicu sa središtem u O . Pojedine se višse ne koriste. Pojedine trigonometrijske fukcije više nisu u uporabi. Versinus , koversinus, haversinus i eksekans su se koristile pri navigaciji, a haversinusna formula se koristila za računanje udaljenosti dviju točaka na sferi.
Ime
Kratica
Vrijednost[ 2]
Versinus
versin
(
θ
)
{\displaystyle \operatorname {versin} (\theta )}
vers
(
θ
)
{\displaystyle \operatorname {vers} (\theta )}
ver
(
θ
)
{\displaystyle \operatorname {ver} (\theta )}
1
−
cos
(
θ
)
{\displaystyle 1-\cos(\theta )}
Verkosinus
vercosin
(
θ
)
{\displaystyle \operatorname {vercosin} (\theta )}
1
+
cos
(
θ
)
{\displaystyle 1+\cos(\theta )}
Koversinus
coversin
(
θ
)
{\displaystyle \operatorname {coversin} (\theta )}
cvs
(
θ
)
{\displaystyle \operatorname {cvs} (\theta )}
1
−
sin
(
θ
)
{\displaystyle 1-\sin(\theta )}
Koverkosinus
covercosin
(
θ
)
{\displaystyle \operatorname {covercosin} (\theta )}
1
+
sin
(
θ
)
{\displaystyle 1+\sin(\theta )}
Haversinus
haversin
(
θ
)
{\displaystyle \operatorname {haversin} (\theta )}
1
−
cos
(
θ
)
2
{\displaystyle {\frac {1-\cos(\theta )}{2}}}
Haverkosinus
havercosin
(
θ
)
{\displaystyle \operatorname {havercosin} (\theta )}
1
+
cos
(
θ
)
2
{\displaystyle {\frac {1+\cos(\theta )}{2}}}
Hakoversinus
hacoversin
(
θ
)
{\displaystyle \operatorname {hacoversin} (\theta )}
1
−
sin
(
θ
)
2
{\displaystyle {\frac {1-\sin(\theta )}{2}}}
Hakoverkosinus
hacovercosin
(
θ
)
{\displaystyle \operatorname {hacovercosin} (\theta )}
1
+
sin
(
θ
)
2
{\displaystyle {\frac {1+\sin(\theta )}{2}}}
Eksekans
exsec
(
θ
)
{\displaystyle \operatorname {exsec} (\theta )}
sec
(
θ
)
−
1
{\displaystyle \sec(\theta )-1}
Ekskosekans
excsc
(
θ
)
{\displaystyle \operatorname {excsc} (\theta )}
csc
(
θ
)
−
1
{\displaystyle \csc(\theta )-1}
Tetiva
crd
(
θ
)
{\displaystyle \operatorname {crd} (\theta )}
2
sin
(
θ
2
)
{\displaystyle 2\sin \left({\frac {\theta }{2}}\right)}
Proučavajući jediničnu kružnicu mogu se uvidjeti pojedina svojstva trigonometrijske kružnice kao što su simetrija, razni pomaci i periodičnost funkcija. Formule u sljedeće dvije tablice se često nazivaju formule redukcije.
Kada neku trigonometrijsku funkciju odbijemo za određeni kut (npr. π,π/2) rezultat često bude neka druga trigonometrijska funkcija.
Odbitak za
θ
=
0
{\displaystyle \theta =0}
[ 3]
Odbitak za
θ
=
π
/
2
{\displaystyle \theta =\pi /2}
[ 4]
Odbitak za
θ
=
π
{\displaystyle \theta =\pi }
sin
(
−
θ
)
=
−
sin
θ
cos
(
−
θ
)
=
+
cos
θ
tg
(
−
θ
)
=
−
tg
θ
csc
(
−
θ
)
=
−
csc
θ
sec
(
−
θ
)
=
+
sec
θ
ctg
(
−
θ
)
=
−
ctg
θ
{\displaystyle {\begin{aligned}\sin(-\theta )&=-\sin \theta \\\cos(-\theta )&=+\cos \theta \\\operatorname {tg} (-\theta )&=-\operatorname {tg} \theta \\\csc(-\theta )&=-\csc \theta \\\sec(-\theta )&=+\sec \theta \\\operatorname {ctg} (-\theta )&=-\operatorname {ctg} \theta \end{aligned}}}
sin
(
π
2
−
θ
)
=
+
cos
θ
cos
(
π
2
−
θ
)
=
+
sin
θ
tg
(
π
2
−
θ
)
=
+
ctg
θ
csc
(
π
2
−
θ
)
=
+
sec
θ
sec
(
π
2
−
θ
)
=
+
csc
θ
ctg
(
π
2
−
θ
)
=
+
tg
θ
{\displaystyle {\begin{aligned}\sin({\tfrac {\pi }{2}}-\theta )&=+\cos \theta \\\cos({\tfrac {\pi }{2}}-\theta )&=+\sin \theta \\\operatorname {tg} ({\tfrac {\pi }{2}}-\theta )&=+\operatorname {ctg} \theta \\\csc({\tfrac {\pi }{2}}-\theta )&=+\sec \theta \\\sec({\tfrac {\pi }{2}}-\theta )&=+\csc \theta \\\operatorname {ctg} ({\tfrac {\pi }{2}}-\theta )&=+\operatorname {tg} \theta \end{aligned}}}
sin
(
π
−
θ
)
=
+
sin
θ
cos
(
π
−
θ
)
=
−
cos
θ
tg
(
π
−
θ
)
=
−
tg
θ
csc
(
π
−
θ
)
=
+
csc
θ
sec
(
π
−
θ
)
=
−
sec
θ
ctg
(
π
−
θ
)
=
−
ctg
θ
{\displaystyle {\begin{aligned}\sin(\pi -\theta )&=+\sin \theta \\\cos(\pi -\theta )&=-\cos \theta \\\operatorname {tg} (\pi -\theta )&=-\operatorname {tg} \theta \\\csc(\pi -\theta )&=+\csc \theta \\\sec(\pi -\theta )&=-\sec \theta \\\operatorname {ctg} (\pi -\theta )&=-\operatorname {ctg} \theta \\\end{aligned}}}
Pomicanjem funkcije za određeni kut također se kao rezultat dobije neka druga trigonometrijska funkcija koja rezultat prikaže jednostavnije. To možemo vidjeti u primjerima pomaka za
π/2, π i 2π radijana. S obzirom na to da su trigonomterijeske funkcije periodične, ovisno o funkciji za π (tangens i kotangens funkcija) ili 2π (sinus i kosinus funkcija), tada nova funkcija poprima istu vrijednost.
Pomak za π/2
Pomak za π [ 5]
Pomak za 2π [ 6]
sin
(
θ
+
π
2
)
=
+
cos
θ
cos
(
θ
+
π
2
)
=
−
sin
θ
tg
(
θ
+
π
2
)
=
−
ctg
θ
csc
(
θ
+
π
2
)
=
+
sec
θ
sec
(
θ
+
π
2
)
=
−
csc
θ
ctg
(
θ
+
π
2
)
=
−
tg
θ
{\displaystyle {\begin{aligned}\sin(\theta +{\tfrac {\pi }{2}})&=+\cos \theta \\\cos(\theta +{\tfrac {\pi }{2}})&=-\sin \theta \\\operatorname {tg} (\theta +{\tfrac {\pi }{2}})&=-\operatorname {ctg} \theta \\\csc(\theta +{\tfrac {\pi }{2}})&=+\sec \theta \\\sec(\theta +{\tfrac {\pi }{2}})&=-\csc \theta \\\operatorname {ctg} (\theta +{\tfrac {\pi }{2}})&=-\operatorname {tg} \theta \end{aligned}}}
sin
(
θ
+
π
)
=
−
sin
θ
cos
(
θ
+
π
)
=
−
cos
θ
tg
(
θ
+
π
)
=
+
tg
θ
csc
(
θ
+
π
)
=
−
csc
θ
sec
(
θ
+
π
)
=
−
sec
θ
ctg
(
θ
+
π
)
=
+
ctg
θ
{\displaystyle {\begin{aligned}\sin(\theta +\pi )&=-\sin \theta \\\cos(\theta +\pi )&=-\cos \theta \\\operatorname {tg} (\theta +\pi )&=+\operatorname {tg} \theta \\\csc(\theta +\pi )&=-\csc \theta \\\sec(\theta +\pi )&=-\sec \theta \\\operatorname {ctg} (\theta +\pi )&=+\operatorname {ctg} \theta \\\end{aligned}}}
sin
(
θ
+
2
π
)
=
+
sin
θ
cos
(
θ
+
2
π
)
=
+
cos
θ
tg
(
θ
+
2
π
)
=
+
tg
θ
csc
(
θ
+
2
π
)
=
+
csc
θ
sec
(
θ
+
2
π
)
=
+
sec
θ
ctg
(
θ
+
2
π
)
=
+
ctg
θ
{\displaystyle {\begin{aligned}\sin(\theta +2\pi )&=+\sin \theta \\\cos(\theta +2\pi )&=+\cos \theta \\\operatorname {tg} (\theta +2\pi )&=+\operatorname {tg} \theta \\\csc(\theta +2\pi )&=+\csc \theta \\\sec(\theta +2\pi )&=+\sec \theta \\\operatorname {ctg} (\theta +2\pi )&=+\operatorname {ctg} \theta \end{aligned}}}
Ove trigonometrijske jednakosti se nazivaju adicijske formule. Otkrio ih je prezijski matematičar Abū al-Wafā' Būzjānī u 10. stoljeću. Eulerova formula može pomoći pri dokazivanju ovih jednakosti.
Sinus
sin
(
α
±
β
)
=
sin
α
cos
β
±
cos
α
sin
β
{\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \!}
[ 7]
Kosinus
cos
(
α
±
β
)
=
cos
α
cos
β
∓
sin
α
sin
β
{\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \,}
[ 8]
Tangens
tg
(
α
±
β
)
=
tg
α
±
tg
β
1
∓
tg
α
tg
β
{\displaystyle \operatorname {tg} (\alpha \pm \beta )={\frac {\operatorname {tg} \alpha \pm \operatorname {tg} \beta }{1\mp \operatorname {tg} \alpha \operatorname {tg} \beta }}}
[ 9]
Arkus sinus
arcsin
α
±
arcsin
β
=
arcsin
(
α
1
−
β
2
±
β
1
−
α
2
)
{\displaystyle \arcsin \alpha \pm \arcsin \beta =\arcsin(\alpha {\sqrt {1-\beta ^{2}}}\pm \beta {\sqrt {1-\alpha ^{2}}})}
[ 10]
Arkus kosinus
arccos
α
±
arccos
β
=
arccos
(
α
β
∓
(
1
−
α
2
)
(
1
−
β
2
)
)
{\displaystyle \arccos \alpha \pm \arccos \beta =\arccos(\alpha \beta \mp {\sqrt {(1-\alpha ^{2})(1-\beta ^{2})}})}
[ 11]
Arkus tangens
arctg
α
±
arctg
β
=
arctg
(
α
±
β
1
∓
α
β
)
{\displaystyle \operatorname {arctg} \alpha \pm \operatorname {arctg} \beta =\operatorname {arctg} \left({\frac {\alpha \pm \beta }{1\mp \alpha \beta }}\right)}
[ 12]
Trigonometrijske formule zbroja i razlike za sinus i kosinus mogu biti zapisani u obliku matrice .
(
cos
ϕ
−
sin
ϕ
sin
ϕ
cos
ϕ
)
(
cos
θ
−
sin
θ
sin
θ
cos
θ
)
=
(
cos
ϕ
cos
θ
−
sin
ϕ
sin
θ
−
cos
ϕ
sin
θ
−
sin
ϕ
cos
θ
sin
ϕ
cos
θ
+
cos
ϕ
sin
θ
−
sin
ϕ
sin
θ
+
cos
ϕ
cos
θ
)
=
(
cos
(
θ
+
ϕ
)
−
sin
(
θ
+
ϕ
)
sin
(
θ
+
ϕ
)
cos
(
θ
+
ϕ
)
)
{\displaystyle {\begin{aligned}&{}\quad \left({\begin{array}{rr}\cos \phi &-\sin \phi \\\sin \phi &\cos \phi \end{array}}\right)\left({\begin{array}{rr}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{array}}\right)\\[12pt]&=\left({\begin{array}{rr}\cos \phi \cos \theta -\sin \phi \sin \theta &-\cos \phi \sin \theta -\sin \phi \cos \theta \\\sin \phi \cos \theta +\cos \phi \sin \theta &-\sin \phi \sin \theta +\cos \phi \cos \theta \end{array}}\right)\\[12pt]&=\left({\begin{array}{rr}\cos(\theta +\phi )&-\sin(\theta +\phi )\\\sin(\theta +\phi )&\cos(\theta +\phi )\end{array}}\right)\end{aligned}}}
sin
(
∑
i
=
1
∞
θ
i
)
=
∑
odd
k
≥
1
(
−
1
)
(
k
−
1
)
/
2
∑
A
⊆
{
1
,
2
,
3
,
…
}
|
A
|
=
k
(
∏
i
∈
A
sin
θ
i
∏
i
∉
A
cos
θ
i
)
{\displaystyle \sin \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{{\text{odd}}\ k\geq 1}(-1)^{(k-1)/2}\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}
cos
(
∑
i
=
1
∞
θ
i
)
=
∑
even
k
≥
0
(
−
1
)
k
/
2
∑
A
⊆
{
1
,
2
,
3
,
…
}
|
A
|
=
k
(
∏
i
∈
A
sin
θ
i
∏
i
∉
A
cos
θ
i
)
{\displaystyle \cos \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{{\text{even}}\ k\geq 0}~(-1)^{k/2}~~\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}
Neka je
e
k
{\displaystyle e_{k}\,}
k ∈ {0, ..., n }) k -ti stupanj osnovnog simetričnog polinoma pri čemu je
x
i
=
tg
θ
i
{\displaystyle x_{i}=\operatorname {tg} \theta _{i}\,}
za i ∈ {0, ..., n } pa slijedi
e
0
=
1
e
1
=
∑
1
≤
i
≤
n
x
i
=
∑
1
≤
i
≤
n
tg
θ
i
e
2
=
∑
1
≤
i
<
j
≤
n
x
i
x
j
=
∑
1
≤
i
<
j
≤
n
tg
θ
i
tg
θ
j
e
3
=
∑
1
≤
i
<
j
<
k
≤
n
x
i
x
j
x
k
=
∑
1
≤
i
<
j
<
k
≤
n
tg
θ
i
tg
θ
j
tg
θ
k
⋮
⋮
{\displaystyle {\begin{aligned}e_{0}&=1\\[6pt]e_{1}&=\sum _{1\leq i\leq n}x_{i}&&=\sum _{1\leq i\leq n}\operatorname {tg} \theta _{i}\\[6pt]e_{2}&=\sum _{1\leq i<j\leq n}x_{i}x_{j}&&=\sum _{1\leq i<j\leq n}\operatorname {tg} \theta _{i}\operatorname {tg} \theta _{j}\\[6pt]e_{3}&=\sum _{1\leq i<j<k\leq n}x_{i}x_{j}x_{k}&&=\sum _{1\leq i<j<k\leq n}\operatorname {tg} \theta _{i}\operatorname {tg} \theta _{j}\operatorname {tg} \theta _{k}\\&{}\ \ \vdots &&{}\ \ \vdots \end{aligned}}}
Tada vrijedi da je
tg
(
θ
1
+
⋯
+
θ
n
)
=
e
1
−
e
3
+
e
5
−
⋯
e
0
−
e
2
+
e
4
−
⋯
,
{\displaystyle \operatorname {tg} (\theta _{1}+\cdots +\theta _{n})={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }},\!}
u ovisnosti o broju n .
Na primjer:
tg
(
θ
1
+
θ
2
)
=
e
1
e
0
−
e
2
=
x
1
+
x
2
1
−
x
1
x
2
=
tg
θ
1
+
tg
θ
2
1
−
tg
θ
1
tg
θ
2
,
tg
(
θ
1
+
θ
2
+
θ
3
)
=
e
1
−
e
3
e
0
−
e
2
=
(
x
1
+
x
2
+
x
3
)
−
(
x
1
x
2
x
3
)
1
−
(
x
1
x
2
+
x
1
x
3
+
x
2
x
3
)
,
tg
(
θ
1
+
θ
2
+
θ
3
+
θ
4
)
=
e
1
−
e
3
e
0
−
e
2
+
e
4
=
(
x
1
+
x
2
+
x
3
+
x
4
)
−
(
x
1
x
2
x
3
+
x
1
x
2
x
4
+
x
1
x
3
x
4
+
x
2
x
3
x
4
)
1
−
(
x
1
x
2
+
x
1
x
3
+
x
1
x
4
+
x
2
x
3
+
x
2
x
4
+
x
3
x
4
)
+
(
x
1
x
2
x
3
x
4
)
,
{\displaystyle {\begin{aligned}\operatorname {tg} (\theta _{1}+\theta _{2})&={\frac {e_{1}}{e_{0}-e_{2}}}={\frac {x_{1}+x_{2}}{1\ -\ x_{1}x_{2}}}={\frac {\operatorname {tg} \theta _{1}+\operatorname {tg} \theta _{2}}{1\ -\ \operatorname {tg} \theta _{1}\operatorname {tg} \theta _{2}}},\\\\\operatorname {tg} (\theta _{1}+\theta _{2}+\theta _{3})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\\\\operatorname {tg} (\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\\\&={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}
i tako dalje. Navedena jednakost se može dokazati matematičkom indukcijom .[ 13]
sec
(
θ
1
+
⋯
+
θ
n
)
=
sec
θ
1
⋯
sec
θ
n
e
0
−
e
2
+
e
4
−
⋯
csc
(
θ
1
+
⋯
+
θ
n
)
=
sec
θ
1
⋯
sec
θ
n
e
1
−
e
3
+
e
5
−
⋯
{\displaystyle {\begin{aligned}\sec(\theta _{1}+\cdots +\theta _{n})&={\frac {\sec \theta _{1}\cdots \sec \theta _{n}}{e_{0}-e_{2}+e_{4}-\cdots }}\\[8pt]\csc(\theta _{1}+\cdots +\theta _{n})&={\frac {\sec \theta _{1}\cdots \sec \theta _{n}}{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}
gdje je
e
k
{\displaystyle e_{k}\,}
k -ti stupanj osnovnog simetričnog polinoma za n varijabla x i θ i i = 1, ..., n , a broj veličina u nazivniku ovisi o n .
Na primjer,
sec
(
α
+
β
+
γ
)
=
sec
α
sec
β
sec
γ
1
−
tg
α
tg
β
−
tg
α
tg
γ
−
tg
β
tg
γ
csc
(
α
+
β
+
γ
)
=
sec
α
sec
β
sec
γ
tg
α
+
tg
β
+
tg
γ
−
tg
α
tg
β
tg
γ
{\displaystyle {\begin{aligned}\sec(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{1-\operatorname {tg} \alpha \operatorname {tg} \beta -\operatorname {tg} \alpha \operatorname {tg} \gamma -\operatorname {tg} \beta \operatorname {tg} \gamma }}\\[8pt]\csc(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{\operatorname {tg} \alpha +\operatorname {tg} \beta +\operatorname {tg} \gamma -\operatorname {tg} \alpha \operatorname {tg} \beta \operatorname {tg} \gamma }}\end{aligned}}}
Tn je n -ti Čebiševljev polinom
cos
n
θ
=
T
n
(
cos
θ
)
{\displaystyle \cos n\theta =T_{n}(\cos \theta )\,}
S n n -ti polinom širine
sin
2
n
θ
=
S
n
(
sin
2
θ
)
{\displaystyle \sin ^{2}n\theta =S_{n}(\sin ^{2}\theta )\,}
De Moivreova formula ,
i
{\displaystyle i}
imaginarna jedinica
cos
n
θ
+
i
sin
n
θ
=
(
cos
(
θ
)
+
i
sin
(
θ
)
)
n
{\displaystyle \cos n\theta +i\sin n\theta =(\cos(\theta )+i\sin(\theta ))^{n}\,}
[ 14]
[ uredi | uredi kôd ]
sin
n
θ
=
∑
k
=
0
n
(
n
k
)
cos
k
θ
sin
n
−
k
θ
sin
(
1
2
(
n
−
k
)
π
)
{\displaystyle \sin n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\sin \left({\frac {1}{2}}(n-k)\pi \right)}
cos
n
θ
=
∑
k
=
0
n
(
n
k
)
cos
k
θ
sin
n
−
k
θ
cos
(
1
2
(
n
−
k
)
π
)
{\displaystyle \cos n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\cos \left({\frac {1}{2}}(n-k)\pi \right)}
tg
(
n
+
1
)
θ
=
tg
n
θ
+
tg
θ
1
−
tg
n
θ
tg
θ
.
{\displaystyle \operatorname {tg} \,(n{+}1)\theta ={\frac {\operatorname {tg} n\theta +\operatorname {tg} \theta }{1-\operatorname {tg} n\theta \,\operatorname {tg} \theta }}.}
ctg
(
n
+
1
)
θ
=
ctg
n
θ
ctg
θ
−
1
ctg
n
θ
+
ctg
θ
.
{\displaystyle \operatorname {ctg} \,(n{+}1)\theta ={\frac {\operatorname {ctg} n\theta \,\operatorname {ctg} \theta -1}{\operatorname {ctg} n\theta +\operatorname {ctg} \theta }}.}
Čebiševljeva metoda je rekurzivni algoritam za nalaženje formula n -tih višestrukih kutova poznavajući (n − 1)-te i (n − 2)-te formule.[ 17]
cos
n
x
=
2
⋅
cos
x
⋅
cos
(
n
−
1
)
x
−
cos
(
n
−
2
)
x
{\displaystyle \cos nx=2\cdot \cos x\cdot \cos(n-1)x-\cos(n-2)x\,}
sin
n
x
=
2
⋅
cos
x
⋅
sin
(
n
−
1
)
x
−
sin
(
n
−
2
)
x
{\displaystyle \sin nx=2\cdot \cos x\cdot \sin(n-1)x-\sin(n-2)x\,}
tg
n
x
=
H
+
K
tg
x
K
−
H
tg
x
{\displaystyle \operatorname {tg} nx={\frac {H+K\operatorname {tg} x}{K-H\operatorname {tg} x}}\,}
gdje je H /K = tan(n − 1)x .
tg
(
α
+
β
2
)
=
sin
α
+
sin
β
cos
α
+
cos
β
=
−
cos
α
−
cos
β
sin
α
−
sin
β
{\displaystyle \operatorname {tg} \left({\frac {\alpha +\beta }{2}}\right)={\frac {\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }}=-\,{\frac {\cos \alpha -\cos \beta }{\sin \alpha -\sin \beta }}}
Ako su α ili β jednaki 0 tada dobivamo formulu za tangens polovičnog kuta.
cos
(
θ
2
)
⋅
cos
(
θ
4
)
⋅
cos
(
θ
8
)
⋯
=
∏
n
=
1
∞
cos
(
θ
2
n
)
=
sin
(
θ
)
θ
=
sinc
θ
.
{\displaystyle \cos \left({\theta \over 2}\right)\cdot \cos \left({\theta \over 4}\right)\cdot \cos \left({\theta \over 8}\right)\cdots =\prod _{n=1}^{\infty }\cos \left({\theta \over 2^{n}}\right)={\sin(\theta ) \over \theta }=\operatorname {sinc} \,\theta .}
Sinus
Kosinus
Druge
sin
2
θ
=
1
−
cos
2
θ
2
{\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}\!}
cos
2
θ
=
1
+
cos
2
θ
2
{\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}\!}
sin
2
θ
cos
2
θ
=
1
−
cos
4
θ
8
{\displaystyle \sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos 4\theta }{8}}\!}
sin
3
θ
=
3
sin
θ
−
sin
3
θ
4
{\displaystyle \sin ^{3}\theta ={\frac {3\sin \theta -\sin 3\theta }{4}}\!}
cos
3
θ
=
3
cos
θ
+
cos
3
θ
4
{\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos 3\theta }{4}}\!}
sin
3
θ
cos
3
θ
=
3
sin
2
θ
−
sin
6
θ
32
{\displaystyle \sin ^{3}\theta \cos ^{3}\theta ={\frac {3\sin 2\theta -\sin 6\theta }{32}}\!}
sin
4
θ
=
3
−
4
cos
2
θ
+
cos
4
θ
8
{\displaystyle \sin ^{4}\theta ={\frac {3-4\cos 2\theta +\cos 4\theta }{8}}\!}
cos
4
θ
=
3
+
4
cos
2
θ
+
cos
4
θ
8
{\displaystyle \cos ^{4}\theta ={\frac {3+4\cos 2\theta +\cos 4\theta }{8}}\!}
sin
4
θ
cos
4
θ
=
3
−
4
cos
4
θ
+
cos
8
θ
128
{\displaystyle \sin ^{4}\theta \cos ^{4}\theta ={\frac {3-4\cos 4\theta +\cos 8\theta }{128}}\!}
sin
5
θ
=
10
sin
θ
−
5
sin
3
θ
+
sin
5
θ
16
{\displaystyle \sin ^{5}\theta ={\frac {10\sin \theta -5\sin 3\theta +\sin 5\theta }{16}}\!}
cos
5
θ
=
10
cos
θ
+
5
cos
3
θ
+
cos
5
θ
16
{\displaystyle \cos ^{5}\theta ={\frac {10\cos \theta +5\cos 3\theta +\cos 5\theta }{16}}\!}
sin
5
θ
cos
5
θ
=
10
sin
2
θ
−
5
sin
6
θ
+
sin
10
θ
512
{\displaystyle \sin ^{5}\theta \cos ^{5}\theta ={\frac {10\sin 2\theta -5\sin 6\theta +\sin 10\theta }{512}}\!}
Za izvode potencija sinus i kosinusa kuta se koriste De Moivreova formula , Eulerov poučak i binomni poučak .
Kosinus
Sinus
ako je
n
neparan
{\displaystyle {\text{ako je }}n{\text{ neparan}}}
cos
n
θ
=
2
2
n
∑
k
=
0
n
−
1
2
(
n
k
)
cos
(
(
n
−
2
k
)
θ
)
{\displaystyle \cos ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}{\binom {n}{k}}\cos {((n-2k)\theta )}}
sin
n
θ
=
2
2
n
∑
k
=
0
n
−
1
2
(
−
1
)
(
n
−
1
2
−
k
)
(
n
k
)
sin
(
(
n
−
2
k
)
θ
)
{\displaystyle \sin ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}(-1)^{({\frac {n-1}{2}}-k)}{\binom {n}{k}}\sin {((n-2k)\theta )}}
ako je
n
paran
{\displaystyle {\text{ako je }}n{\text{ paran}}}
cos
n
θ
=
1
2
n
(
n
n
2
)
+
2
2
n
∑
k
=
0
n
2
−
1
(
n
k
)
cos
(
(
n
−
2
k
)
θ
)
{\displaystyle \cos ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}{\binom {n}{k}}\cos {((n-2k)\theta )}}
sin
n
θ
=
1
2
n
(
n
n
2
)
+
2
2
n
∑
k
=
0
n
2
−
1
(
−
1
)
(
n
2
−
k
)
(
n
k
)
cos
(
(
n
−
2
k
)
θ
)
{\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{({\frac {n}{2}}-k)}{\binom {n}{k}}\cos {((n-2k)\theta )}}
Umnožak u zbroj[ 18]
cos
θ
cos
φ
=
cos
(
θ
−
φ
)
+
cos
(
θ
+
φ
)
2
{\displaystyle \cos \theta \cos \varphi ={\cos(\theta -\varphi )+\cos(\theta +\varphi ) \over 2}}
sin
θ
sin
φ
=
cos
(
θ
−
φ
)
−
cos
(
θ
+
φ
)
2
{\displaystyle \sin \theta \sin \varphi ={\cos(\theta -\varphi )-\cos(\theta +\varphi ) \over 2}}
sin
θ
cos
φ
=
sin
(
θ
+
φ
)
+
sin
(
θ
−
φ
)
2
{\displaystyle \sin \theta \cos \varphi ={\sin(\theta +\varphi )+\sin(\theta -\varphi ) \over 2}}
cos
θ
sin
φ
=
sin
(
θ
+
φ
)
−
sin
(
θ
−
φ
)
2
{\displaystyle \cos \theta \sin \varphi ={\sin(\theta +\varphi )-\sin(\theta -\varphi ) \over 2}}
Zbroj u umnožak[ 19]
sin
θ
±
sin
φ
=
2
sin
(
θ
±
φ
2
)
cos
(
θ
∓
φ
2
)
{\displaystyle \sin \theta \pm \sin \varphi =2\sin \left({\frac {\theta \pm \varphi }{2}}\right)\cos \left({\frac {\theta \mp \varphi }{2}}\right)}
cos
θ
+
cos
φ
=
2
cos
(
θ
+
φ
2
)
cos
(
θ
−
φ
2
)
{\displaystyle \cos \theta +\cos \varphi =2\cos \left({\frac {\theta +\varphi }{2}}\right)\cos \left({\frac {\theta -\varphi }{2}}\right)}
cos
θ
−
cos
φ
=
−
2
sin
(
θ
+
φ
2
)
sin
(
θ
−
φ
2
)
{\displaystyle \cos \theta -\cos \varphi =-2\sin \left({\theta +\varphi \over 2}\right)\sin \left({\theta -\varphi \over 2}\right)}
Ako su x , y i z bilo kojeg trokuta, tada vrijedi
ako je zbroj
x
+
y
+
z
=
π
=
polukrug,
{\displaystyle {\text{ako je zbroj }}x+y+z=\pi ={\text{polukrug,}}\,}
onda je
tg
(
x
)
+
tg
(
y
)
+
tg
(
z
)
=
tg
(
x
)
tg
(
y
)
tg
(
z
)
.
{\displaystyle {\text{onda je }}\operatorname {tg} (x)+\operatorname {tg} (y)+\operatorname {tg} (z)=\operatorname {tg} (x)\operatorname {tg} (y)\operatorname {tg} (z).\,}
odnosno
ako je zbroj
x
+
y
+
z
=
π
=
polukrug,
{\displaystyle {\text{ako je zbroj }}x+y+z=\pi ={\text{polukrug,}}\,}
onda je
sin
(
2
x
)
+
sin
(
2
y
)
+
sin
(
2
z
)
=
4
sin
(
x
)
sin
(
y
)
sin
(
z
)
.
{\displaystyle {\text{onda je }}\sin(2x)+\sin(2y)+\sin(2z)=4\sin(x)\sin(y)\sin(z).\,}
Charles Hermite je pokazao da vrijedi određena jednakost[ 20] a 1 , ..., a n kompleksni brojevi . Neka je
A
n
,
k
=
∏
1
≤
j
≤
n
j
≠
k
ctg
(
a
k
−
a
j
)
{\displaystyle A_{n,k}=\prod _{\begin{smallmatrix}1\leq j\leq n\\j\neq k\end{smallmatrix}}\operatorname {ctg} (a_{k}-a_{j})}
te u slučaju kada je A 1,1 , dobiva se prazan produkt , koji je jednak 1. Općenito se dobiva sljedeća vrijednost:
ctg
(
z
−
a
1
)
⋯
ctg
(
z
−
a
n
)
=
cos
n
π
2
+
∑
k
=
1
n
A
n
,
k
ctg
(
z
−
a
k
)
.
{\displaystyle \operatorname {ctg} (z-a_{1})\cdots \operatorname {ctg} (z-a_{n})=\cos {\frac {n\pi }{2}}+\sum _{k=1}^{n}A_{n,k}\operatorname {ctg} (z-a_{k}).}
U najjednostavnijem slučaju za n = 2 vrijedi:
ctg
(
z
−
a
1
)
ctg
(
z
−
a
2
)
=
−
1
+
ctg
(
a
1
−
a
2
)
ctg
(
z
−
a
1
)
+
ctg
(
a
2
−
a
1
)
ctg
(
z
−
a
2
)
.
{\displaystyle \operatorname {ctg} (z-a_{1})\operatorname {ctg} (z-a_{2})=-1+\operatorname {ctg} (a_{1}-a_{2})\operatorname {ctg} (z-a_{1})+\operatorname {ctg} (a_{2}-a_{1})\operatorname {ctg} (z-a_{2}).}
Ove jednakosti predstavljaju trigonometrijski oblik ptolomejevog teorema .
Ako su
w
+
x
+
y
+
z
=
π
=
polukrug,
{\displaystyle {\text{Ako su }}w+x+y+z=\pi ={\text{polukrug,}}\,}
tada vrijedi
sin
(
w
+
x
)
sin
(
x
+
y
)
=
sin
(
x
+
y
)
sin
(
y
+
z
)
=
sin
(
y
+
z
)
sin
(
z
+
w
)
=
sin
(
z
+
w
)
sin
(
w
+
x
)
=
sin
(
w
)
sin
(
y
)
+
sin
(
x
)
sin
(
z
)
.
{\displaystyle {\begin{aligned}{\text{tada vrijedi }}&\sin(w+x)\sin(x+y)\\&{}=\sin(x+y)\sin(y+z)\\&{}=\sin(y+z)\sin(z+w)\\&{}=\sin(z+w)\sin(w+x)=\sin(w)\sin(y)+\sin(x)\sin(z).\end{aligned}}}
Bilo koja linearna kombinacija sinusnih valova istih perioda ili frekvencija s različitim faznim pomacima je također sinusni val s istom periodom ili frekvencijom s različitim faznim pomakom. Kod nenulte linearne kombinacije sinusnog i kosinusnog vala
,[ 21]
a
sin
x
+
b
cos
x
=
a
2
+
b
2
⋅
sin
(
x
+
φ
)
{\displaystyle a\sin x+b\cos x={\sqrt {a^{2}+b^{2}}}\cdot \sin(x+\varphi )\,}
gdje je
φ
=
{
arcsin
(
b
a
2
+
b
2
)
ako je
a
≥
0
,
π
−
arcsin
(
b
a
2
+
b
2
)
ako je
a
<
0
,
{\displaystyle \varphi ={\begin{cases}\arcsin \left({\frac {b}{\sqrt {a^{2}+b^{2}}}}\right)&{\text{ako je }}a\geq 0,\\\pi -\arcsin \left({\frac {b}{\sqrt {a^{2}+b^{2}}}}\right)&{\text{ako je }}a<0,\end{cases}}}
što je ekvivalentno s
φ
=
arctg
(
b
a
)
+
{
0
ako je
a
≥
0
,
π
ako je
a
<
0
,
{\displaystyle \varphi =\operatorname {arctg} \left({\frac {b}{a}}\right)+{\begin{cases}0&{\text{ako je }}a\geq 0,\\\pi &{\text{ako je }}a<0,\end{cases}}}
ili čak s
φ
=
sgn
(
b
)
cos
−
1
(
a
a
2
+
b
2
)
{\displaystyle \varphi ={\text{sgn}}(b)\cos ^{-1}\left({\tfrac {a}{\sqrt {a^{2}+b^{2}}}}\right)}
Općenito za proizvoljan fazni pomak vrijedi
a
sin
x
+
b
sin
(
x
+
α
)
=
c
sin
(
x
+
β
)
{\displaystyle a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,}
gdje je
c
=
a
2
+
b
2
+
2
a
b
cos
α
,
{\displaystyle c={\sqrt {a^{2}+b^{2}+2ab\cos \alpha }},\,}
i
β
=
arctg
(
b
sin
α
a
+
b
cos
α
)
+
{
0
ako je
a
+
b
cos
α
≥
0
,
π
ako je
a
+
b
cos
α
<
0.
{\displaystyle \beta =\operatorname {arctg} \left({\frac {b\sin \alpha }{a+b\cos \alpha }}\right)+{\begin{cases}0&{\text{ako je }}a+b\cos \alpha \geq 0,\\\pi &{\text{ako je }}a+b\cos \alpha <0.\end{cases}}}
Ove jednakosti su ime dobili po Josephu Louisu Lagrangeu .[ 22] [ 23]
∑
n
=
1
N
sin
n
θ
=
1
2
ctg
θ
−
cos
(
N
+
1
2
)
θ
2
sin
1
2
θ
∑
n
=
1
N
cos
n
θ
=
−
1
2
+
sin
(
N
+
1
2
)
θ
2
sin
1
2
θ
{\displaystyle {\begin{aligned}\sum _{n=1}^{N}\sin n\theta &={\frac {1}{2}}\operatorname {ctg} \theta -{\frac {\cos(N+{\frac {1}{2}})\theta }{2\sin {\frac {1}{2}}\theta }}\\\sum _{n=1}^{N}\cos n\theta &=-{\frac {1}{2}}+{\frac {\sin(N+{\frac {1}{2}})\theta }{2\sin {\frac {1}{2}}\theta }}\end{aligned}}}
S njima je povezana funkcija koja se naziva Dirichletova jezgra .
1
+
2
cos
(
x
)
+
2
cos
(
2
x
)
+
2
cos
(
3
x
)
+
⋯
+
2
cos
(
n
x
)
=
sin
(
(
n
+
1
2
)
x
)
sin
(
x
/
2
)
.
{\displaystyle 1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots +2\cos(nx)={\frac {\sin \left(\left(n+{\frac {1}{2}}\right)x\right)}{\sin(x/2)}}.}
Zbroj sinusa i kosinusa s varijablama u aritmetičkom nizu
[ 24]
sin
φ
+
sin
(
φ
+
α
)
+
sin
(
φ
+
2
α
)
+
⋯
⋯
+
sin
(
φ
+
n
α
)
=
sin
(
(
n
+
1
)
α
2
)
⋅
sin
(
φ
+
n
α
2
)
sin
α
2
.
cos
φ
+
cos
(
φ
+
α
)
+
cos
(
φ
+
2
α
)
+
⋯
⋯
+
cos
(
φ
+
n
α
)
=
sin
(
(
n
+
1
)
α
2
)
⋅
cos
(
φ
+
n
α
2
)
sin
α
2
.
{\displaystyle {\begin{aligned}&\sin {\varphi }+\sin {(\varphi +\alpha )}+\sin {(\varphi +2\alpha )}+\cdots {}\\[8pt]&{}\qquad \qquad \cdots +\sin {(\varphi +n\alpha )}={\frac {\sin {\left({\frac {(n+1)\alpha }{2}}\right)}\cdot \sin {(\varphi +{\frac {n\alpha }{2}})}}{\sin {\frac {\alpha }{2}}}}.\\[10pt]&\cos {\varphi }+\cos {(\varphi +\alpha )}+\cos {(\varphi +2\alpha )}+\cdots {}\\[8pt]&{}\qquad \qquad \cdots +\cos {(\varphi +n\alpha )}={\frac {\sin {\left({\frac {(n+1)\alpha }{2}}\right)}\cdot \cos {(\varphi +{\frac {n\alpha }{2}})}}{\sin {\frac {\alpha }{2}}}}.\end{aligned}}}
Za bilo koji a i b vrijedi:
a
cos
(
x
)
+
b
sin
(
x
)
=
a
2
+
b
2
cos
(
x
−
atan2
(
b
,
a
)
)
{\displaystyle a\cos(x)+b\sin(x)={\sqrt {a^{2}+b^{2}}}\cos(x-\operatorname {atan2} \,(b,a))\;}
gdje je atan2(y , x ) poopćenje funkcije arctan(y /x ) koja pokriva cijeli kružni opseg.
Koristeći Gudermannovu funkciju koja povezuje cirkularne i hiperbolne trigonometrijske funkcije bez korištenja kompleksnih brojeva može se iskoristiti sljedeći izraz:
tg
(
x
)
+
sec
(
x
)
=
tg
(
x
2
+
π
4
)
.
{\displaystyle \operatorname {tg} (x)+\sec(x)=\operatorname {tg} \left({x \over 2}+{\pi \over 4}\right).}
Ako su x , y i z ako su kutovi bilo kojeg trokuta odnosno x + y + z = π, tada je
ctg
(
x
)
ctg
(
y
)
+
ctg
(
y
)
ctg
(
z
)
+
ctg
(
z
)
ctg
(
x
)
=
1.
{\displaystyle \operatorname {ctg} (x)\operatorname {ctg} (y)+\operatorname {ctg} (y)\operatorname {ctg} (z)+\operatorname {ctg} (z)\operatorname {ctg} (x)=1.\,}
Ako je ƒ (x ) dan linearnom frakcionalnom transformacijom
f
(
x
)
=
(
cos
α
)
x
−
sin
α
(
sin
α
)
x
+
cos
α
,
{\displaystyle f(x)={\frac {(\cos \alpha )x-\sin \alpha }{(\sin \alpha )x+\cos \alpha }},}
i slično tome
g
(
x
)
=
(
cos
β
)
x
−
sin
β
(
cos
β
)
x
+
sin
β
,
{\displaystyle g(x)={\frac {(\cos \beta )x-\sin \beta }{(\cos \beta )x+\sin \beta }},}
tada vrijedi
f
(
g
(
x
)
)
=
g
(
f
(
x
)
)
=
(
cos
(
α
+
β
)
)
x
−
sin
(
α
+
β
)
(
sin
(
α
+
β
)
)
x
+
cos
(
α
+
β
)
.
{\displaystyle f(g(x))=g(f(x))={\frac {(\cos(\alpha +\beta ))x-\sin(\alpha +\beta )}{(\sin(\alpha +\beta ))x+\cos(\alpha +\beta )}}.}
Kraće rečeno, ako je za sve α funkcija ƒ α ƒ tada vrijedi da je
f
α
∘
f
β
=
f
α
+
β
.
{\displaystyle f_{\alpha }\circ f_{\beta }=f_{\alpha +\beta }.\,}
arcsin
(
x
)
+
arccos
(
x
)
=
π
/
2
{\displaystyle \arcsin(x)+\arccos(x)=\pi /2\;}
arctg
(
x
)
+
arcctg
(
x
)
=
π
/
2.
{\displaystyle \operatorname {arctg} (x)+\operatorname {arcctg} (x)=\pi /2.\;}
arctg
(
x
)
+
arctg
(
1
/
x
)
=
{
π
/
2
,
ako je
x
>
0
−
π
/
2
,
ako je
x
<
0
{\displaystyle \operatorname {arctg} (x)+\operatorname {arctg} (1/x)=\left\{{\begin{matrix}\pi /2,&{\mbox{ako je }}x>0\\-\pi /2,&{\mbox{ako je }}x<0\end{matrix}}\right.}
Kompozicija trigonometrijskih i inverznih trigonometrijskih funkcija [ uredi | uredi kôd ]
sin
[
arccos
(
x
)
]
=
1
−
x
2
{\displaystyle \sin[\arccos(x)]={\sqrt {1-x^{2}}}\,}
tg
[
arcsin
(
x
)
]
=
x
1
−
x
2
{\displaystyle \operatorname {tg} [\arcsin(x)]={\frac {x}{\sqrt {1-x^{2}}}}}
sin
[
arctg
(
x
)
]
=
x
1
+
x
2
{\displaystyle \sin[\operatorname {arctg} (x)]={\frac {x}{\sqrt {1+x^{2}}}}}
tg
[
arccos
(
x
)
]
=
1
−
x
2
x
{\displaystyle \operatorname {tg} [\arccos(x)]={\frac {\sqrt {1-x^{2}}}{x}}}
cos
[
arctg
(
x
)
]
=
1
1
+
x
2
{\displaystyle \cos[\operatorname {arctg} (x)]={\frac {1}{\sqrt {1+x^{2}}}}}
ctg
[
arcsin
(
x
)
]
=
1
−
x
2
x
{\displaystyle \operatorname {ctg} [\arcsin(x)]={\frac {\sqrt {1-x^{2}}}{x}}}
cos
[
arcsin
(
x
)
]
=
1
−
x
2
{\displaystyle \cos[\arcsin(x)]={\sqrt {1-x^{2}}}\,}
ctg
[
arccos
(
x
)
]
=
x
1
−
x
2
{\displaystyle \operatorname {ctg} [\arccos(x)]={\frac {x}{\sqrt {1-x^{2}}}}}
Povezanost s kompleksnom eksponencijalnom funkcijom [ uredi | uredi kôd ]
e
i
x
=
cos
(
x
)
+
i
sin
(
x
)
{\displaystyle e^{ix}=\cos(x)+i\sin(x)\,}
[ 25] Eulerova formula ,
e
−
i
x
=
cos
(
−
x
)
+
i
sin
(
−
x
)
=
cos
(
x
)
−
i
sin
(
x
)
{\displaystyle e^{-ix}=\cos(-x)+i\sin(-x)=\cos(x)-i\sin(x)\,}
e
i
π
=
−
1
{\displaystyle e^{i\pi }=-1}
Eulerov identitet ,
cos
(
x
)
=
e
i
x
+
e
−
i
x
2
{\displaystyle \cos(x)={\frac {e^{ix}+e^{-ix}}{2}}\;}
[ 26]
sin
(
x
)
=
e
i
x
−
e
−
i
x
2
i
{\displaystyle \sin(x)={\frac {e^{ix}-e^{-ix}}{2i}}\;}
[ 27] odnosno
tg
(
x
)
=
e
i
x
−
e
−
i
x
i
(
e
i
x
+
e
−
i
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle \operatorname {tg} (x)={\frac {e^{ix}-e^{-ix}}{i({e^{ix}+e^{-ix}})}}\;={\frac {\sin(x)}{\cos(x)}}}
gdje je
i
2
=
−
1
{\displaystyle i^{2}=-1}
Pri rješavanju specijalnih funkcija , različite koristimo formule koje povezuju beskonačni produkt i trigonometrijske funkcije:[ 28] [ 29]
sin
x
=
x
∏
n
=
1
∞
(
1
−
x
2
π
2
n
2
)
{\displaystyle \sin x=x\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)}
sinh
x
=
x
∏
n
=
1
∞
(
1
+
x
2
π
2
n
2
)
{\displaystyle \sinh x=x\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)}
sin
x
x
=
∏
n
=
1
∞
cos
(
x
2
n
)
{\displaystyle {\frac {\sin x}{x}}=\prod _{n=1}^{\infty }\cos \left({\frac {x}{2^{n}}}\right)}
cos
x
=
∏
n
=
1
∞
(
1
−
x
2
π
2
(
n
−
1
2
)
2
)
{\displaystyle \cos x=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}(n-{\frac {1}{2}})^{2}}}\right)}
cosh
x
=
∏
n
=
1
∞
(
1
+
x
2
π
2
(
n
−
1
2
)
2
)
{\displaystyle \cosh x=\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}(n-{\frac {1}{2}})^{2}}}\right)}
|
sin
x
|
=
1
2
∏
n
=
0
∞
|
tg
(
2
n
x
)
|
2
n
+
1
{\displaystyle |\sin x|={\frac {1}{2}}\prod _{n=0}^{\infty }{\sqrt[{2^{n+1}}]{\left|\operatorname {tg} \left(2^{n}x\right)\right|}}}
Jednakost bez varijabli
cos
20
∘
⋅
cos
40
∘
⋅
cos
80
∘
=
1
8
{\displaystyle \cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 80^{\circ }={\frac {1}{8}}}
je poseban slučaj jednakosti s jednom varijablom:
∏
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)}}.}
Nadalje, također vrijedi da je
sin
20
∘
⋅
sin
40
∘
⋅
sin
80
∘
=
3
8
.
{\displaystyle \sin 20^{\circ }\cdot \sin 40^{\circ }\cdot \sin 80^{\circ }={\frac {\sqrt {3}}{8}}.}
cos
π
7
cos
2
π
7
cos
3
π
7
=
1
8
,
{\displaystyle \cos {\frac {\pi }{7}}\cos {\frac {2\pi }{7}}\cos {\frac {3\pi }{7}}={\frac {1}{8}},}
tg
50
∘
⋅
tg
60
∘
⋅
tg
70
∘
=
tg
80
∘
.
{\displaystyle \operatorname {tg} 50^{\circ }\cdot \operatorname {tg} 60^{\circ }\cdot \operatorname {tg} 70^{\circ }=\operatorname {tg} 80^{\circ }.}
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}}.}
cos
(
2
π
21
)
+
cos
(
2
⋅
2
π
21
)
+
cos
(
4
⋅
2
π
21
)
+
cos
(
5
⋅
2
π
21
)
+
cos
(
8
⋅
2
π
21
)
+
cos
(
10
⋅
2
π
21
)
=
1
2
.
{\displaystyle {\begin{aligned}&\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)\\[10pt]&{}\qquad {}+\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}}.\end{aligned}}}
Mnogo jednakosti ima osnovu u izrazima kao što su[ 30]
∏
k
=
1
n
−
1
sin
(
k
π
n
)
=
n
2
n
−
1
{\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{n}}\right)={\frac {n}{2^{n-1}}}}
i
∏
k
=
1
n
−
1
cos
(
k
π
n
)
=
sin
(
π
n
/
2
)
2
n
−
1
{\displaystyle \prod _{k=1}^{n-1}\cos \left({\frac {k\pi }{n}}\right)={\frac {\sin(\pi n/2)}{2^{n-1}}}}
Njihovom kombinacijom dobivamo:
∏
k
=
1
n
−
1
tg
(
k
π
n
)
=
n
sin
(
π
n
/
2
)
{\displaystyle \prod _{k=1}^{n-1}\operatorname {tg} \left({\frac {k\pi }{n}}\right)={\frac {n}{\sin(\pi n/2)}}}
Ako je n neparan broj(n = 2m + 1) korištenjem simetrije dobivamo
∏
k
=
1
m
tg
(
k
π
2
m
+
1
)
=
2
m
+
1
{\displaystyle \prod _{k=1}^{m}\operatorname {tg} \left({\frac {k\pi }{2m+1}}\right)={\sqrt {2m+1}}}
π
4
=
4
arctg
1
5
−
arctg
1
239
{\displaystyle {\frac {\pi }{4}}=4\operatorname {arctg} {\frac {1}{5}}-\operatorname {arctg} {\frac {1}{239}}}
π
4
=
5
arctg
1
7
+
2
arctg
3
79
.
{\displaystyle {\frac {\pi }{4}}=5\operatorname {arctg} {\frac {1}{7}}+2\operatorname {arctg} {\frac {3}{79}}.}
[ uredi | uredi kôd ]
sin
0
=
sin
0
∘
=
0
/
2
=
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
∘
=
4
/
2
=
cos
0
∘
=
cos
0
{\displaystyle {\begin{matrix}\sin 0&=&\sin 0^{\circ }&=&{\sqrt {0}}/2&=&\cos 90^{\circ }&=&\cos \left({\frac {\pi }{2}}\right)\\\\\sin \left({\frac {\pi }{6}}\right)&=&\sin 30^{\circ }&=&{\sqrt {1}}/2&=&\cos 60^{\circ }&=&\cos \left({\frac {\pi }{3}}\right)\\\\\sin \left({\frac {\pi }{4}}\right)&=&\sin 45^{\circ }&=&{\sqrt {2}}/2&=&\cos 45^{\circ }&=&\cos \left({\frac {\pi }{4}}\right)\\\\\sin \left({\frac {\pi }{3}}\right)&=&\sin 60^{\circ }&=&{\sqrt {3}}/2&=&\cos 30^{\circ }&=&\cos \left({\frac {\pi }{6}}\right)\\\\\sin \left({\frac {\pi }{2}}\right)&=&\sin 90^{\circ }&=&{\sqrt {4}}/2&=&\cos 0^{\circ }&=&\cos 0\end{matrix}}}
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}={\frac {\varphi }{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 }}
sin
2
(
18
∘
)
+
sin
2
(
30
∘
)
=
sin
2
(
36
∘
)
.
{\displaystyle \sin ^{2}(18^{\circ })+\sin ^{2}(30^{\circ })=\sin ^{2}(36^{\circ }).\,}
Koristeći infinitezimalni račun , kutovi pri računanju moraju biti u radijanima. Derivacije trigonometrijskih funkcija mogu se odrediti pomoću dva limesa :
lim
x
→
0
sin
x
x
=
1
,
{\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1,}
lim
x
→
0
1
−
cos
x
x
=
0
,
{\displaystyle \lim _{x\rightarrow 0}{\frac {1-\cos x}{x}}=0,}
Deriviranjem trigonometrijskih funkcija dobivaju se sljedeće jednakosti i pravila:[ 31] [ 32] [ 33]
d
d
x
sin
x
=
cos
x
,
d
d
x
arcsin
x
=
1
1
−
x
2
d
d
x
cos
x
=
−
sin
x
,
d
d
x
arccos
x
=
−
1
1
−
x
2
d
d
x
tg
x
=
sec
2
x
,
d
d
x
arctg
x
=
1
1
+
x
2
d
d
x
ctg
x
=
−
csc
2
x
,
d
d
x
arcctg
x
=
−
1
1
+
x
2
d
d
x
sec
x
=
tg
x
sec
x
,
d
d
x
arcsec
x
=
1
|
x
|
x
2
−
1
d
d
x
csc
x
=
−
csc
x
ctg
x
,
d
d
x
arccsc
x
=
−
1
|
x
|
x
2
−
1
{\displaystyle {\begin{aligned}{d \over dx}\sin x&=\cos x,&{d \over dx}\arcsin x&={1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\cos x&=-\sin x,&{d \over dx}\arccos x&={-1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\operatorname {tg} x&=\sec ^{2}x,&{d \over dx}\operatorname {arctg} x&={1 \over 1+x^{2}}\\\\{d \over dx}\operatorname {ctg} x&=-\csc ^{2}x,&{d \over dx}\operatorname {arcctg} x&={-1 \over 1+x^{2}}\\\\{d \over dx}\sec x&=\operatorname {tg} x\sec x,&{d \over dx}\operatorname {arcsec} x&={1 \over |x|{\sqrt {x^{2}-1}}}\\\\{d \over dx}\csc x&=-\csc x\operatorname {ctg} x,&{d \over dx}\operatorname {arccsc} x&={-1 \over |x|{\sqrt {x^{2}-1}}}\end{aligned}}}
∫
d
u
a
2
−
u
2
=
sin
−
1
(
u
a
)
+
C
{\displaystyle \int {\frac {du}{\sqrt {a^{2}-u^{2}}}}=\sin ^{-1}\left({\frac {u}{a}}\right)+C}
∫
d
u
a
2
+
u
2
=
1
a
tg
−
1
(
u
a
)
+
C
{\displaystyle \int {\frac {du}{a^{2}+u^{2}}}={\frac {1}{a}}\operatorname {tg} ^{-1}\left({\frac {u}{a}}\right)+C}
∫
d
u
u
u
2
−
a
2
=
1
a
sec
−
1
|
u
a
|
+
C
{\displaystyle \int {\frac {du}{u{\sqrt {u^{2}-a^{2}}}}}={\frac {1}{a}}\sec ^{-1}\left|{\frac {u}{a}}\right|+C}
Eksponencijalne definicije trigonometrijskih funkcija [ uredi | uredi kôd ]
Funkcija
Inverzna funkcija[ 34]
sin
θ
=
e
i
θ
−
e
−
i
θ
2
i
{\displaystyle \sin \theta ={\frac {e^{i\theta }-e^{-i\theta }}{2i}}\,}
arcsin
x
=
−
i
ln
(
i
x
+
1
−
x
2
)
{\displaystyle \arcsin x=-i\ln \left(ix+{\sqrt {1-x^{2}}}\right)\,}
cos
θ
=
e
i
θ
+
e
−
i
θ
2
{\displaystyle \cos \theta ={\frac {e^{i\theta }+e^{-i\theta }}{2}}\,}
arccos
x
=
−
i
ln
(
x
+
x
2
−
1
)
{\displaystyle \arccos x=-i\ln \left(x+{\sqrt {x^{2}-1}}\right)\,}
tg
θ
=
e
i
θ
−
e
−
i
θ
i
(
e
i
θ
+
e
−
i
θ
)
{\displaystyle \operatorname {tg} \theta ={\frac {e^{i\theta }-e^{-i\theta }}{i(e^{i\theta }+e^{-i\theta })}}\,}
arctg
x
=
i
2
ln
(
i
+
x
i
−
x
)
{\displaystyle \operatorname {arctg} x={\frac {i}{2}}\ln \left({\frac {i+x}{i-x}}\right)\,}
csc
θ
=
2
i
e
i
θ
−
e
−
i
θ
{\displaystyle \csc \theta ={\frac {2i}{e^{i\theta }-e^{-i\theta }}}\,}
arccsc
x
=
−
i
ln
(
i
x
+
1
−
1
x
2
)
{\displaystyle \operatorname {arccsc} x=-i\ln \left({\tfrac {i}{x}}+{\sqrt {1-{\tfrac {1}{x^{2}}}}}\right)\,}
sec
θ
=
2
e
i
θ
+
e
−
i
θ
{\displaystyle \sec \theta ={\frac {2}{e^{i\theta }+e^{-i\theta }}}\,}
arcsec
x
=
−
i
ln
(
1
x
+
1
−
i
x
2
)
{\displaystyle \operatorname {arcsec} x=-i\ln \left({\tfrac {1}{x}}+{\sqrt {1-{\tfrac {i}{x^{2}}}}}\right)\,}
ctg
θ
=
i
(
e
i
θ
+
e
−
i
θ
)
e
i
θ
−
e
−
i
θ
{\displaystyle \operatorname {ctg} \theta ={\frac {i(e^{i\theta }+e^{-i\theta })}{e^{i\theta }-e^{-i\theta }}}\,}
arcctg
x
=
i
2
ln
(
x
−
i
x
+
i
)
{\displaystyle \operatorname {arcctg} x={\frac {i}{2}}\ln \left({\frac {x-i}{x+i}}\right)\,}
cis
θ
=
e
i
θ
{\displaystyle \operatorname {cis} \,\theta =e^{i\theta }\,}
arccis
x
=
ln
x
i
=
−
i
ln
x
=
arg
x
{\displaystyle \operatorname {arccis} \,x={\frac {\ln x}{i}}=-i\ln x=\operatorname {arg} \,x\,}
Ako je
t
=
tg
(
x
2
)
,
{\displaystyle t=\operatorname {tg} \left({\frac {x}{2}}\right),}
tada vrijedi[ 35]
sin
(
x
)
=
2
t
1
+
t
2
i
cos
(
x
)
=
1
−
t
2
1
+
t
2
i
e
i
x
=
1
+
i
t
1
−
i
t
{\displaystyle \sin(x)={\frac {2t}{1+t^{2}}}{\text{ i }}\cos(x)={\frac {1-t^{2}}{1+t^{2}}}{\text{ i }}e^{ix}={\frac {1+it}{1-it}}}
gdje je eix x ) + i sin(x ), što ponekad skraćeno pišemo kao cis(x ).
↑ Abramowitz and Stegun, p. 73, 4.3.45
↑ Abramowitz and Stegun, p. 78, 4.3.147
↑ Abramowitz and Stegun, p. 72, 4.3.13–15
↑ The Elementary Identities . Inačica izvorne stranice arhivirana 30. srpnja 2017. Pristupljeno 20. prosinca 2011. ↑ Abramowitz and Stegun, p. 72, 4.3.9
↑ Abramowitz and Stegun, p. 72, 4.3.7–8
↑ Abramowitz and Stegun, p. 72, 4.3.16
↑ Abramowitz and Stegun, p. 72, 4.3.17
↑ Abramowitz and Stegun, p. 72, 4.3.18
↑ Abramowitz and Stegun, p. 80, 4.4.42
↑ Abramowitz and Stegun, p. 80, 4.4.43
↑ Abramowitz and Stegun, p. 80, 4.4.36
↑ Bronstein, Manual. 1989. Simplification of Real Elementary Functions. Proceedings of the ACM-SIGSAM 1989 international symposium on Symbolic and algebraic computation : 211 ↑ Abramowitz and Stegun, p. 74, 4.3.48
↑ Abramowitz and Stegun, p. 72, 4.3.24–26
↑ Abramowitz and Stegun, p. 72, 4.3.20–22
↑ Ken Ward's Mathematics Pages, http://www.trans4mind.com/personal_development/mathematics/trigonometry/multipleAnglesRecursiveFormula.htm Arhivirana inačica izvorne stranice od 27. srpnja 2011. (Wayback Machine )
↑ Abramowitz and Stegun, p. 72, 4.3.31–33
↑ Abramowitz and Stegun, p. 72, 4.3.34–39
↑ Warren P. Johnson, "Trigonometric Identities à la Hermite", American Mathematical Monthly
↑ Proof at http://pages.pacificcoast.net/~cazelais/252/lc-trig.pdf Arhivirana inačica izvorne stranice od 3. prosinca 2011. (Wayback Machine )
↑
Eddie Ortiz Muñiz. Veljača 1953. A Method for Deriving Various Formulas in Electrostatics and Electromagnetism Using Lagrange's Trigonometric Identities. American Journal of Physics . 21 (2): 140
↑
Alan Jeffrey and Hui-hui Dai. 2008. Section 2.4.1.6. Handbook of Mathematical Formulas and Integrals . 4th izdanje. Academic Press. ISBN 9780123742889
↑ Michael P. Knapp, Sines and Cosines of Angles in Arithmetic Progression Arhivirana inačica izvorne stranice od 19. srpnja 2011. (Wayback Machine )
↑ Abramowitz and Stegun, p. 74, 4.3.47
↑ Abramowitz and Stegun, p. 71, 4.3.2
↑ Abramowitz and Stegun, p. 71, 4.3.1
↑ Abramowitz and Stegun, p. 75, 4.3.89–90
↑ Abramowitz and Stegun, p. 85, 4.5.68–69
↑ Weisstein, Eric W. , "Sine " from MathWorld ↑ Abramowitz and Stegun, p. 77, 4.3.105–110
↑ Abramowitz and Stegun, p. 82, 4.4.52–57
↑ Finney, Ross. 2003. Calculus : Graphical, Numerical, Algebraic . Prentice Hall. Glenview, Illinois. str. 159–161. ISBN 0-13-063131-0 ↑ Abramowitz and Stegun, p. 80, 4.4.26–31
↑ Abramowitz and Stegun, p. 72, 4.3.23
Podrobniji članak o temi: Kut
![{\displaystyle {\begin{aligned}&{}\quad \left({\begin{array}{rr}\cos \phi &-\sin \phi \\\sin \phi &\cos \phi \end{array}}\right)\left({\begin{array}{rr}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{array}}\right)\\[12pt]&=\left({\begin{array}{rr}\cos \phi \cos \theta -\sin \phi \sin \theta &-\cos \phi \sin \theta -\sin \phi \cos \theta \\\sin \phi \cos \theta +\cos \phi \sin \theta &-\sin \phi \sin \theta +\cos \phi \cos \theta \end{array}}\right)\\[12pt]&=\left({\begin{array}{rr}\cos(\theta +\phi )&-\sin(\theta +\phi )\\\sin(\theta +\phi )&\cos(\theta +\phi )\end{array}}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/852d1e76b4190362dde8aefd41ddca69347d1961)
![{\displaystyle {\begin{aligned}e_{0}&=1\\[6pt]e_{1}&=\sum _{1\leq i\leq n}x_{i}&&=\sum _{1\leq i\leq n}\operatorname {tg} \theta _{i}\\[6pt]e_{2}&=\sum _{1\leq i<j\leq n}x_{i}x_{j}&&=\sum _{1\leq i<j\leq n}\operatorname {tg} \theta _{i}\operatorname {tg} \theta _{j}\\[6pt]e_{3}&=\sum _{1\leq i<j<k\leq n}x_{i}x_{j}x_{k}&&=\sum _{1\leq i<j<k\leq n}\operatorname {tg} \theta _{i}\operatorname {tg} \theta _{j}\operatorname {tg} \theta _{k}\\&{}\ \ \vdots &&{}\ \ \vdots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f2c5db750b9b1fba93aa9fefa0a792dd377829e)
![{\displaystyle {\begin{aligned}\sec(\theta _{1}+\cdots +\theta _{n})&={\frac {\sec \theta _{1}\cdots \sec \theta _{n}}{e_{0}-e_{2}+e_{4}-\cdots }}\\[8pt]\csc(\theta _{1}+\cdots +\theta _{n})&={\frac {\sec \theta _{1}\cdots \sec \theta _{n}}{e_{1}-e_{3}+e_{5}-\cdots }}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c82ca2d8da55d938d78aad99295a11f79665fde)
![{\displaystyle {\begin{aligned}\sec(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{1-\operatorname {tg} \alpha \operatorname {tg} \beta -\operatorname {tg} \alpha \operatorname {tg} \gamma -\operatorname {tg} \beta \operatorname {tg} \gamma }}\\[8pt]\csc(\alpha +\beta +\gamma )&={\frac {\sec \alpha \sec \beta \sec \gamma }{\operatorname {tg} \alpha +\operatorname {tg} \beta +\operatorname {tg} \gamma -\operatorname {tg} \alpha \operatorname {tg} \beta \operatorname {tg} \gamma }}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1feb844ca95b6583a9ded365961e15f0a0b41b8d)
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![{\displaystyle |\sin x|={\frac {1}{2}}\prod _{n=0}^{\infty }{\sqrt[{2^{n+1}}]{\left|\operatorname {tg} \left(2^{n}x\right)\right|}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/897fe4518be9ba1bb0234ee4cf9f91ea14bbb1ed)
![{\displaystyle {\begin{aligned}&\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)\\[10pt]&{}\qquad {}+\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}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15e07fe52f1162c9c933e28a6bec4fb9f59528e8)
