Question

Даю 40 баллов помогите пожалуйста

Answer 1

Ответ:

Применяем формулы приведения и периодичность

тригонометрических функций .

$1)\ \ sin240^\circ =sin(180^\circ +60^\circ )=-sin60^\circ =-\dfrac{\sqrt3}{2}\\\\2)\ \ tg300^\circ =tg(360^\circ -60^\circ )=tg(-60^\circ )=-tg60^\circ =-\sqrt3\\\\3)\ \ cos330^\circ =cos(360^\circ -30^\circ )=cos(-30^\circ )=cos30^\circ =\dfrac{\sqrt3}{2}\\\\4)\ \ ctg315^\circ =ctg(360^\circ -45^\circ )=ctg(-45^\circ )=-ctg45^\circ =-1$

$5)\ \ \displaystyle cos\frac{4\pi }{3}=cos\Big(\pi +\frac{\pi}{3}\Big)=-cos\frac{\pi}{3}=-\frac{1}{2}\\\\\\6)\ \ sin\Big(-\frac{11\pi }{6}\Big)=-sin\Big(2\pi -\frac{\pi}{6}\Big)=sin\frac{\pi }{6}=\frac{1}{2}\\\\\\7)\ \ tg\frac{7\pi }{6}=tg\Big(\pi +\frac{\pi}{6}\Big)=tg\frac{\pi}{6}=\frac{\sqrt3}{3}\\\\\\8)\ \ ctg\Big(-\frac{3\pi }{4}\Big)=-ctg\frac{3\pi }{4}=-ctg\Big(\pi -\frac{\pi}{4}\Big)=ctg\frac{\pi }{4}=1$