Question

Используя формулы приведения, вычислите значения тригонометрических функции: 1) sin 150°, cos 120°, tg135°, ctg150°.2) sin 225°, cos 210°, tg 210°, ctg 240°.3) sin 315°, cos 300°, tg 315°, ctg 300°.4) sin(-135°), cos(-240°), tg(-300°), ctg(-225)​

Answer 1

$1)\\\sin 150^\circ = \sin (180^\circ - 30^\circ) = \sin 30^\circ = \dfrac 12\\\\\cos 120^\circ = \cos (180^\circ - 60^\circ) = - \cos 60^\circ = -\dfrac 12\\\\tg\ 135^\circ = tg (180^\circ - 45^\circ) = - tg\ 45^\circ = -1\\\\ctg\ 150^\circ = ctg (180^\circ - 30^\circ) = - ctg\ 30^\circ = -\sqrt3$

$2)\\\sin 225^\circ = \sin (180^\circ + 45^\circ) = - \sin 45^\circ = -\dfrac{\sqrt2}2\\\\\cos 210^\circ = \cos(180^\circ+30^\circ)=-\cos 30^\circ=-\dfrac{\sqrt3}2\\\\tg\ 210^\circ=tg(180^\circ+30^\circ)=tg\ 30^\circ=\dfrac1{\sqrt3}\\\\ctg\ 240^\circ=ctg(180^\circ+60^\circ)=ctg\ 60^\circ=\dfrac 1{\sqrt3}$

$3)\\ \sin 315^\circ=\sin(270^\circ+45^\circ) =-\cos45^\circ=-\dfrac{\sqrt2}2\\\\\cos 300^\circ=\cos(270^\circ+30^\circ)=\sin 30^\circ=\dfrac12\\\\tg\ 315^\circ=tg(270^\circ+45^\circ)=-ctg\ 45^\circ=-1\\\\ctg\ 300^\circ=ctg(270^\circ+30^\circ)=-tg\ 30^\circ=-\dfrac 1{\sqrt3}$

$4)\\ \sin(-135^\circ)=-\sin 135^\circ=-\sin(180^\circ-45^\circ)=-\sin 45^\circ=-\dfrac{\sqrt2}2\\\\\cos(-240^\circ)=\cos 240^\circ=\cos(180^\circ+60^\circ)=-\cos60^\circ=-\dfrac12\\\\ tg(-300^\circ) =tg(-360^\circ+60^\circ)=tg\ 60^\circ=\sqrt3\\\\ctg(-225^\circ)=-ctg\ 225^\circ=-ctg(180^\circ+45^\circ)=-ctg\ 45^\circ=-1$