Question

高一数学 19题

Answer 1

1)
tan(2α-2β)=2tan(α-β)/(1-(tan(α-β))^2)
=4/3
tan(2α-β)=(tan(2α-2β)+tanβ)/(1-(tan(2α-2β)tanβ)
=1
α,β∈(0,π)
α-β>0
2α-β>0
π/2<β<π
π/2<α<π
0<2α-β<(3π/2)
2α-β=π/4或2α-β=5π/4
2)
cos(α-(β/2))=-1/9
sin((α/2)-β)=2/3
π/2<α<π
0<β<π/2
cos((α/2)-β)=(√5)/3
sin(α-(β/2))=(4√5)/9
cos((α/2)+(β/2))=cos[(α-(β/2))-((α/2)-β)]
cos(α+β)=[2os((α/2)+(β/2))]^2-1