Question

已知0≤α≤π，0≤β≤π/4，且α+β=2π/3 求y=(1+cos2α)/(cotα/2-tanα/2)-cos^2(π/4-β)的最大

已知0≤α≤π，0≤β≤π/4，且α+β=2π/3 求y=(1+cos2α)/(cotα/2-tanα/2)-cos^2(π/4-β)的最大值 并求出相应的α,β的值

Answer 1

0≤α≤π，0≤β≤π/4，-π/4≤-β≤0，-π/4≤α-β≤π
y=(1+cos2α)/(cotα/2-tanα/2)-cos^2(π/4-β)
=2(cosα)^2/[cosα/(1/2*sinα)]-1/2*[1+cos(π/2-2β)]
=sinαcosα-1/2*sin2β-1/2
=1/2*(sin2α-sin2β)-1/2
=cos(α+β)sin(α-β)=-1/2*sin(α-β)
ymax= √2/4(此时α-β=-π/4)