Cos(-570)+√3sin(-300)+2tg(-135)

\begin{lgathered}cos(-570)+ \sqrt{3}sin(-300)+2tg(-135)= \\\ =cos(2\cdot360-570)+ \sqrt{3}sin(360-300)+2tg(180-135)= \\\ =cos150+ \sqrt{3}sin60+2tg45=- \frac{\sqrt{3}}{2} +\sqrt{3}\cdot\frac{\sqrt{3}}{2}+2\cdot1= \\\ =- \frac{\sqrt{3}}{2} +\frac{3}{2}+2=3.5- \frac{\sqrt{3}}{2}\end{lgathered}​cos(−570)+√​3​​​sin(−300)+2tg(−135)=​ =cos(2⋅360−570)+√​3​​​sin(360−300)+2tg(180−135)=​ =cos150+√​3​​​sin60+2tg45=−​2​​√​3​​​​​+√​3​​​⋅​2​​√​3​​​​​+2⋅1=​ =−​2​​√​3​​​​​+​2​​3​​+2=3.5−​2​​√​3​​​​​​​

Вот, держи! Ответик :3