Question

cos315°+ctg225°-6sin300°

Answer 1

Cos315°+ctg225°-6sin300°=Cos(360°-45°)+ctg(180°+45°)-6sin(360°-60°)=

=Cos45°+ctg45°-6sin(-60°)=1/√2+1+6√3/2=(√2+2+6√3)/2