Karta wzorów · dział 9 z 17

Trygonometria

Kąt i bok leżący naprzeciw niego mają tu ten sam kolor — widać parę, o którą chodzi, bez wodzenia palcem po rysunku.

Kąt ostry w trójkącie prostokątnym

$\sin \htmlClass{rZ}{\alpha} = \dfrac{\htmlClass{rZ}{a}}{\htmlClass{rN}{c}}$
$\cos \htmlClass{rZ}{\alpha} = \dfrac{\htmlClass{rC}{b}}{\htmlClass{rN}{c}}$
$\operatorname{tg} \htmlClass{rZ}{\alpha} = \dfrac{\htmlClass{rZ}{a}}{\htmlClass{rC}{b}}$

a — przyprostokątna leżąca naprzeciw kąta α, b — przyprostokątna przy kącie α, c — przeciwprostokątna.

$\sin \htmlClass{rZ}{\alpha} = \dfrac{\htmlClass{rF}{y}}{\htmlClass{rN}{r}}$
$\cos \htmlClass{rZ}{\alpha} = \dfrac{\htmlClass{rY}{x}}{\htmlClass{rN}{r}}$
$\operatorname{tg} \htmlClass{rZ}{\alpha} = \dfrac{\htmlClass{rF}{y}}{\htmlClass{rY}{x}}$
$\htmlClass{rN}{r} = |OM| = \sqrt{\htmlClass{rY}{x}^2 + \htmlClass{rF}{y}^2} > 0$

M = (x, y) jest punktem końca ramienia kąta α, a r — jego odległością od początku układu. Tangens wymaga x ≠ 0.

$\sin^2 \htmlClass{rZ}{\alpha} + \cos^2 \htmlClass{rZ}{\alpha} = 1$
$\operatorname{tg} \htmlClass{rZ}{\alpha} = \dfrac{\sin \htmlClass{rZ}{\alpha}}{\cos \htmlClass{rZ}{\alpha}}$

Druga równość dla α ≠ ½π + kπ, gdzie k jest liczbą całkowitą.

	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
	$0$	$\tfrac{1}{6}\pi$	$\tfrac{1}{4}\pi$	$\tfrac{1}{3}\pi$	$\tfrac{1}{2}\pi$
$\sin \htmlClass{rZ}{\alpha}$	$0$	$\tfrac{1}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{\sqrt{3}}{2}$	$1$
$\cos \htmlClass{rZ}{\alpha}$	$1$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{1}{2}$	$0$
$\operatorname{tg} \htmlClass{rZ}{\alpha}$	$0$	$\tfrac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$	nie istnieje

$\sin(\htmlClass{rZ}{\alpha} + \htmlClass{rC}{\beta}) = \sin \htmlClass{rZ}{\alpha} \cos \htmlClass{rC}{\beta} + \cos \htmlClass{rZ}{\alpha} \sin \htmlClass{rC}{\beta}$
$\sin(\htmlClass{rZ}{\alpha} - \htmlClass{rC}{\beta}) = \sin \htmlClass{rZ}{\alpha} \cos \htmlClass{rC}{\beta} - \cos \htmlClass{rZ}{\alpha} \sin \htmlClass{rC}{\beta}$
$\cos(\htmlClass{rZ}{\alpha} + \htmlClass{rC}{\beta}) = \cos \htmlClass{rZ}{\alpha} \cos \htmlClass{rC}{\beta} - \sin \htmlClass{rZ}{\alpha} \sin \htmlClass{rC}{\beta}$
$\cos(\htmlClass{rZ}{\alpha} - \htmlClass{rC}{\beta}) = \cos \htmlClass{rZ}{\alpha} \cos \htmlClass{rC}{\beta} + \sin \htmlClass{rZ}{\alpha} \sin \htmlClass{rC}{\beta}$
$\operatorname{tg}(\htmlClass{rZ}{\alpha} + \htmlClass{rC}{\beta}) = \dfrac{\operatorname{tg} \htmlClass{rZ}{\alpha} + \operatorname{tg} \htmlClass{rC}{\beta}}{1 - \operatorname{tg} \htmlClass{rZ}{\alpha} \cdot \operatorname{tg} \htmlClass{rC}{\beta}}$
$\operatorname{tg}(\htmlClass{rZ}{\alpha} - \htmlClass{rC}{\beta}) = \dfrac{\operatorname{tg} \htmlClass{rZ}{\alpha} - \operatorname{tg} \htmlClass{rC}{\beta}}{1 + \operatorname{tg} \htmlClass{rZ}{\alpha} \cdot \operatorname{tg} \htmlClass{rC}{\beta}}$

Wzory na tangens wymagają, by α, β oraz α ± β były różne od ½π + kπ dla całkowitego k.

$\sin 2\htmlClass{rZ}{\alpha} = 2 \sin \htmlClass{rZ}{\alpha} \cos \htmlClass{rZ}{\alpha}$
$\cos 2\htmlClass{rZ}{\alpha} = \cos^2 \htmlClass{rZ}{\alpha} - \sin^2 \htmlClass{rZ}{\alpha}$
$\cos 2\htmlClass{rZ}{\alpha} = 2\cos^2 \htmlClass{rZ}{\alpha} - 1$
$\cos 2\htmlClass{rZ}{\alpha} = 1 - 2\sin^2 \htmlClass{rZ}{\alpha}$
$\operatorname{tg} 2\htmlClass{rZ}{\alpha} = \dfrac{2 \operatorname{tg} \htmlClass{rZ}{\alpha}}{1 - \operatorname{tg}^2 \htmlClass{rZ}{\alpha}}$

Ostatni wzór o ile tangens α istnieje i tg²α ≠ 1.

$\sin(90^\circ - \htmlClass{rZ}{\alpha}) = \cos \htmlClass{rZ}{\alpha}$
$\cos(90^\circ - \htmlClass{rZ}{\alpha}) = \sin \htmlClass{rZ}{\alpha}$
$\sin(90^\circ + \htmlClass{rZ}{\alpha}) = \cos \htmlClass{rZ}{\alpha}$
$\cos(90^\circ + \htmlClass{rZ}{\alpha}) = -\sin \htmlClass{rZ}{\alpha}$
$\sin(180^\circ - \htmlClass{rZ}{\alpha}) = \sin \htmlClass{rZ}{\alpha}$
$\cos(180^\circ - \htmlClass{rZ}{\alpha}) = -\cos \htmlClass{rZ}{\alpha}$
$\operatorname{tg}(180^\circ - \htmlClass{rZ}{\alpha}) = -\operatorname{tg} \htmlClass{rZ}{\alpha}$
$\sin(180^\circ + \htmlClass{rZ}{\alpha}) = -\sin \htmlClass{rZ}{\alpha}$
$\cos(180^\circ + \htmlClass{rZ}{\alpha}) = -\cos \htmlClass{rZ}{\alpha}$
$\operatorname{tg}(180^\circ + \htmlClass{rZ}{\alpha}) = \operatorname{tg} \htmlClass{rZ}{\alpha}$

$\sin \htmlClass{rZ}{\alpha} + \sin \htmlClass{rC}{\beta} = 2 \sin \dfrac{\htmlClass{rZ}{\alpha}+\htmlClass{rC}{\beta}}{2} \cos \dfrac{\htmlClass{rZ}{\alpha}-\htmlClass{rC}{\beta}}{2}$
$\sin \htmlClass{rZ}{\alpha} - \sin \htmlClass{rC}{\beta} = 2 \cos \dfrac{\htmlClass{rZ}{\alpha}+\htmlClass{rC}{\beta}}{2} \sin \dfrac{\htmlClass{rZ}{\alpha}-\htmlClass{rC}{\beta}}{2}$
$\cos \htmlClass{rZ}{\alpha} + \cos \htmlClass{rC}{\beta} = 2 \cos \dfrac{\htmlClass{rZ}{\alpha}+\htmlClass{rC}{\beta}}{2} \cos \dfrac{\htmlClass{rZ}{\alpha}-\htmlClass{rC}{\beta}}{2}$
$\cos \htmlClass{rZ}{\alpha} - \cos \htmlClass{rC}{\beta} = -2 \sin \dfrac{\htmlClass{rZ}{\alpha}+\htmlClass{rC}{\beta}}{2} \sin \dfrac{\htmlClass{rZ}{\alpha}-\htmlClass{rC}{\beta}}{2}$

$\sin \htmlClass{rZ}{\alpha} \cdot \sin \htmlClass{rC}{\beta} = -\tfrac{1}{2}\left[\cos(\htmlClass{rZ}{\alpha}+\htmlClass{rC}{\beta}) - \cos(\htmlClass{rZ}{\alpha}-\htmlClass{rC}{\beta})\right]$
$\cos \htmlClass{rZ}{\alpha} \cdot \cos \htmlClass{rC}{\beta} = \tfrac{1}{2}\left[\cos(\htmlClass{rZ}{\alpha}+\htmlClass{rC}{\beta}) + \cos(\htmlClass{rZ}{\alpha}-\htmlClass{rC}{\beta})\right]$
$\sin \htmlClass{rZ}{\alpha} \cdot \cos \htmlClass{rC}{\beta} = \tfrac{1}{2}\left[\sin(\htmlClass{rZ}{\alpha}+\htmlClass{rC}{\beta}) + \sin(\htmlClass{rZ}{\alpha}-\htmlClass{rC}{\beta})\right]$

$\sin(\htmlClass{rZ}{\alpha} + \htmlClass{rC}{k} \cdot 360^\circ) = \sin \htmlClass{rZ}{\alpha}$
$\cos(\htmlClass{rZ}{\alpha} + \htmlClass{rC}{k} \cdot 360^\circ) = \cos \htmlClass{rZ}{\alpha}$
$\operatorname{tg}(\htmlClass{rZ}{\alpha} + \htmlClass{rC}{k} \cdot 180^\circ) = \operatorname{tg} \htmlClass{rZ}{\alpha}$

Dla całkowitego k. Wzór na tangens wymaga α ≠ 90° + m · 180° dla całkowitego m.