Trigonometrinės funkcijos

${\displaystyle \alpha \,}$	${\displaystyle {\frac {\pi }{12}}=15^{\circ }}$	${\displaystyle {\frac {\pi }{10}}=18^{\circ }}$	${\displaystyle {\frac {\pi }{8}}=22,5^{\circ }}$	${\displaystyle {\frac {\pi }{5}}=36^{\circ }}$	${\displaystyle {\frac {3\,\pi }{10}}=54^{\circ }}$	${\displaystyle {\frac {3\,\pi }{8}}=67,5^{\circ }}$	${\displaystyle {\frac {2\,\pi }{5}}=72^{\circ }}$
${\displaystyle \sin \alpha \,}$	${\displaystyle {\frac {{\sqrt {3}}-1}{2\,{\sqrt {2}}}}}$	${\displaystyle {\frac {{\sqrt {5}}-1}{4}}}$	${\displaystyle {\frac {\sqrt {2-{\sqrt {2}}}}{2}}}$	${\displaystyle {\frac {\sqrt {5-{\sqrt {5}}}}{2\,{\sqrt {2}}}}}$	${\displaystyle {\frac {{\sqrt {5}}+1}{4}}}$	${\displaystyle {\frac {\sqrt {2+{\sqrt {2}}}}{2}}}$	${\displaystyle {\frac {\sqrt {5+{\sqrt {5}}}}{2\,{\sqrt {2}}}}}$
${\displaystyle \cos \alpha \,}$	${\displaystyle {\frac {{\sqrt {3}}+1}{2\,{\sqrt {2}}}}}$	${\displaystyle {\frac {\sqrt {5+{\sqrt {5}}}}{2\,{\sqrt {2}}}}}$	${\displaystyle {\frac {\sqrt {2+{\sqrt {2}}}}{2}}}$	${\displaystyle {\frac {{\sqrt {5}}+1}{4}}}$	${\displaystyle {\frac {\sqrt {5-{\sqrt {5}}}}{2\,{\sqrt {2}}}}}$	${\displaystyle {\frac {\sqrt {2-{\sqrt {2}}}}{2}}}$	${\displaystyle {\frac {{\sqrt {5}}-1}{4}}}$
${\displaystyle \operatorname {tg} \,\alpha }$	${\displaystyle 2-{\sqrt {3}}}$	${\displaystyle {\sqrt {1-{\frac {2}{\sqrt {5}}}}}}$	${\displaystyle {\sqrt {\frac {{\sqrt {2}}-1}{{\sqrt {2}}+1}}}}$	${\displaystyle {\sqrt {5-2\,{\sqrt {5}}}}}$	${\displaystyle {\sqrt {1+{\frac {2}{\sqrt {5}}}}}}$	${\displaystyle {\sqrt {\frac {{\sqrt {2}}+1}{{\sqrt {2}}-1}}}}$	${\displaystyle {\sqrt {5+2\,{\sqrt {5}}}}}$
${\displaystyle \operatorname {ctg} \,\alpha }$	${\displaystyle 2+{\sqrt {3}}}$	${\displaystyle {\sqrt {5+2\,{\sqrt {5}}}}}$	${\displaystyle {\sqrt {\frac {{\sqrt {2}}+1}{{\sqrt {2}}-1}}}}$	${\displaystyle {\sqrt {1+{\frac {2}{\sqrt {5}}}}}}$	${\displaystyle {\sqrt {5-2\,{\sqrt {5}}}}}$	${\displaystyle {\sqrt {\frac {{\sqrt {2}}-1}{{\sqrt {2}}+1}}}}$	${\displaystyle {\sqrt {1-{\frac {2}{\sqrt {5}}}}}}$

${\displaystyle \operatorname {tg} {\frac {\pi }{120}}=\operatorname {tg} 1,5^{\circ }={\sqrt {\frac {8-{\sqrt {2(2-{\sqrt {3}})(3-{\sqrt {5}})}}-{\sqrt {2(2+{\sqrt {3}})(5+{\sqrt {5}})}}}{8+{\sqrt {2(2-{\sqrt {3}})(3-{\sqrt {5}})}}+{\sqrt {2(2+{\sqrt {3}})(5+{\sqrt {5}})}}}}}}$

${\displaystyle \cos {\frac {\pi }{240}}={\frac {1}{16}}\left({\sqrt {2-k}}\left({\sqrt {2(5+{\sqrt {5}})}}+{\sqrt {3}}-{\sqrt {15}}\right)+{\sqrt {2+k}}\left({\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1\right)\right)}$, kur ${\displaystyle k={\sqrt {2+{\sqrt {2}}}}}$ .

${\displaystyle \cos {\frac {\pi }{17}}={\frac {1}{8}}{\sqrt {2\left(2{\sqrt {{\sqrt {\frac {17k}{2}}}-{\sqrt {\frac {k}{2}}}-4{\sqrt {2(17+{\sqrt {17}})}}+3{\sqrt {17}}+17}}+{\sqrt {2k}}+{\sqrt {17}}+15\right)}}}$, kur ${\displaystyle k=17-{\sqrt {17}}}$ .

Kadangi sinusas ir kosinusas yra atitinkamai taško, atitinkančio kampo α apskritimą, ordinatė ir abscisė, tai pagal Pitagoro teoremą:

${\displaystyle \sin ^{2}\alpha +\cos ^{2}\alpha =1.\qquad \qquad \,}$

Abi šios lygties puses padalijus iš sinuso kvadrato arba kosinuso kvadrato, gaunama:

${\displaystyle 1+\mathop {\mathrm {tg} } \,^{2}\alpha ={\frac {1}{\cos ^{2}\alpha }},\qquad \qquad \,}$

${\displaystyle 1+\mathop {\mathrm {ctg} } \,^{2}\alpha ={\frac {1}{\sin ^{2}\alpha }}.\qquad \qquad \,}$

Funkcijos ${\displaystyle y=\sin \alpha }$, ${\displaystyle y=\cos \alpha }$, ${\displaystyle y=\sec \alpha }$ ir ${\displaystyle y=\csc \alpha }$ yra periodinės funkcijos su periodu ${\displaystyle 2\pi }$ . O funkcijos ${\displaystyle y=\operatorname {tg} \alpha }$ ir ${\displaystyle y=\operatorname {ctg} \alpha }$ yra periodinės su periodu ${\displaystyle \pi }$

Kosinusas yra lyginė funkcija, nes

${\displaystyle \cos(-\alpha )=\cos \alpha .}$

Sinusas yra nelyginė funkcija, nes

${\displaystyle \sin(-\alpha )=-\sin \alpha .}$

Tangentas ir kotangentas yra nelyginės funkcijos, t. y.

${\displaystyle {\text{tg}}(-\alpha )=-{\text{tg}}\;\alpha$ ;}

${\displaystyle {\text{ctg}}(-\alpha )=-{\text{ctg}}\;\alpha .}$

${\displaystyle \cos x=\sin {\Big (}x+{\frac {\pi }{2}}{\Big )}.}$

Į formulę

${\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta \quad (1)}$

įstačius ${\displaystyle {\frac {\pi }{2}}}$ vietoje ${\displaystyle \alpha }$ ir įstačius ${\displaystyle \alpha }$ vietoje ${\displaystyle \beta }$ gausime

${\displaystyle \cos({\frac {\pi }{2}}-\alpha )=\cos {\frac {\pi }{2}}\cos \alpha +\sin {\frac {\pi }{2}}\sin \alpha =\sin \alpha .}$

Gautoje formulėje

${\displaystyle \sin \alpha =\cos({\frac {\pi }{2}}-\alpha )\quad (2)}$

įstačius ${\displaystyle \alpha +\beta }$ vietoje ${\displaystyle \alpha ,}$ gausime

${\displaystyle \sin(\alpha +\beta )=\cos({\frac {\pi }{2}}-\alpha -\beta )=\cos(({\frac {\pi }{2}}-\alpha )-\beta ).}$

Toliau į (1) formulę įstačius ${\displaystyle {\frac {\pi }{2}}-\alpha }$ vietoje ${\displaystyle \alpha ,}$ gausime

${\displaystyle \sin(\alpha +\beta )=\cos(({\frac {\pi }{2}}-\alpha )-\beta )=}$

${\displaystyle =\cos({\frac {\pi }{2}}-\alpha )\cos \beta +\sin({\frac {\pi }{2}}-\alpha )\sin \beta =}$

${\displaystyle =\sin \alpha \cos \beta +\cos \alpha \sin \beta .}$

Pasinaudojome formule

${\displaystyle \sin({\frac {\pi }{2}}-\alpha )=\cos \alpha ,\quad (3)}$

kuri išplaukia iš formulės (2) įstačius į ją ${\displaystyle {\frac {\pi }{2}}-\alpha }$ vietoje ${\displaystyle \alpha$ ;} tada

[${\displaystyle \sin \alpha =\cos({\frac {\pi }{2}}-\alpha )\quad (2)}$]

${\displaystyle \sin({\frac {\pi }{2}}-\alpha )=\cos({\frac {\pi }{2}}-({\frac {\pi }{2}}-\alpha ))=\cos \alpha .}$

Taigi, gavome formulę (3).