Question

tg(-225°)×cos120°=
2/3sin300°-tg330°=
cos^2 7/4п×ctg^2 2/3п=​

Answer 1

Ответ:

$1)\ \ tg(-225^\circ )\cdot cos120^\circ =-tg(270^\circ -45^\circ )\cdot cos(180^\circ -60^\circ )=\\\\=-tg45^\circ \cdot (-cos60^\circ )=\dfrac{1}{2}\\\\\\2)\ \dfrac{2}{3}\cdot sin300^\circ -tg330^\circ =\dfrac{2}{3}\cdot sin(360^\circ -60^\circ )-tg(360^\circ -30^\circ )=\\\\=\dfrac{2}{3}\cdot (-sin60^\circ )+tg30^\circ =-\dfrac{2}{3}\cdot \dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt3}=-\dfrac{1}{\sqrt3}-\dfrac{1}{\sqrt3}=-\dfrac{2}{\sqrt3}=-\dfrac{2\sqrt3}{3}$

$3)\ \ cos^2\dfrac{7\pi }{4}\cdot ctg^2\dfrac{2\pi}{3}=cos^2\Big(2\pi -\dfrac{\pi}{4}\Big)\cdot ctg^2\Big(\pi -\dfrac{\pi}{3}\Big)=cos^2\dfrac{\pi}{4}\cdot ctg^2\dfrac{\pi}{3}=\\\\=\Big(\dfrac{\sqrt2}{2}\Big)^2\cdot \Big(\dfrac{1}{\sqrt3}\Big)^2=\dfrac{1}{2}\cdot \dfrac{1}{3}=\dfrac{1}{6}$