已知x+y=60°,其中x≥0°及y≥0°.求tanx+tany的最大值和最小值

Question

已知x+y=60°,其中x≥0°及y≥0°.求tanx+tany的最大值和最小值

Answer 1

已知x+y=60°,其中x≥0°及y≥0°.求tanx+tany的最大值和最小值
z=tanx+tany=tanx+tan(π/3-x)=sinx/cosx+sin(π/3-x)/cos(π/3-x)
=[sinxcos(π/3-x)+cosxsin(π/3-x)]/cosxcos(π/3-x)
=sin(π/3)/cosxcos(π/3-x)=(√3)/[2cosxcos(π/3-x)]
=(√3)/[cos(π/3)+cos(2x-π/3)]=(2√3)/[1+2cos(2x-π/3)]
0≤x≤π/3,0≤2x≤2π/3,-π/3≤2x-π/3≤π/3,故1/2≤cos(2x-π/3)≤1,1≤2cos(2x-π/3)≤2
∴2(√3)/3≤tanx+tany≤√3
当x=y=30º时获得最小值2(√3)/3；当x=60º,y=0º,或x=0º,y=60º时获得最大值√3.