Question

Cos(-570)+√3sin(-300)+2tg(-135)

Answer 1

begin{lgathered}cos(-570)+ sqrt{3}sin(-300)+2tg(-135)= \ =cos(2cdot360-570)+ sqrt{3}sin(360-300)+2tg(180-135)= \ =cos150+ sqrt{3}sin60+2tg45=- frac{sqrt{3}}{2} +sqrt{3}cdotfrac{sqrt{3}}{2}+2cdot1= \ =- 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​​​​​​​

Answer 2

Вот, держи! Ответик :3