Помогите решите геометрическое выражения

1)cos(180+60)=-cos60=-1/2
sin240=sin(270-30)=-cos30=-√3/2
cos(-2π/3)=cos2π/3=cos(π-π/3)=-cosπ/3=-1/2
2)1-tg²x/2=(cos²x/2-sin²x/2)/cos²x/2=cosx/cos²x/2
1+tg²x/2=(cos²x/2+sin²x/2)/cos²x/2=1/cos²x/2
(1-tg²x/2)/(1+tg²x/2)=cosx/cos²x/2:1/cos²x/2=cosx
cosx=cosx
3)sin120+cosπ/8-tg480=sin60+cosπ/8+tg60=√3/2+√3+ $\sqrt{(1+cos \pi /4)} 2$ =
=3√3/2+ $\sqrt{2+ \sqrt{2} } /2=1/2*(3 \sqrt{3} + \sqrt{2+ \sqrt{2} })$