Sin(-930)+√3·cos(-210)+tg315=

Question

Sin(-930)+√3·cos(-210)+tg315=

Answer 1

-sin(6p-5p/6+√3*-cos(p+p/6)+tg(3p/2+45)=
-sin(5p/6)+√3*-cos(p/6)-ctg(45)=
-1/2+√3*-√3 /2 -1=
-1/2+(-√9/2)-1=
-0.5+-3/2)-1=-1/2-1.5-1=-0.5-1.5-1=-2-1=-3