Помогите с 1 и 4 пожалуйста
1) cos(-π/3)=cos(π/3)=1/2 sin(π/6)=1/2 tg(π/4)=1 sinπ=0 ctg(π/6)=√3 (4·(1/2)+2·(1/2)-5):(0+2√3)=-3/2√3=-√3/2 4) tg(π+β)=tgβ ctg((π/2)-β)=tgβ ctg((3π/2)-β)=tgβ tg(π-β)=-tgβ tg(π+β)·ctg((π/2)-β)-ctg((3π/2)-β)·tg(π-β)=tgβ·tgβ-tgβ·(-tgβ)=2tg²β