已知f(x)=2cos²x-(sinx+cosx)²,若f(α/2)=√2；α∈[-π/2,π/2]；求tan(α+π/3)的值
f(x)=2cos²x-(sin²x+2sinxcosx+cos²x)=cos²x-sin²x-2sinxcosx=cos2x-sin2x=(√2)cos(2x+π/4)；
f(α/2)=(√2)cos(α+π/4)=√2；故cos(α+π/4)=1；于是得α+π/4=0,α=-π/4.
故tan(α+π/3)=tan(-π/4+π/3)=tan(π/3-π/4)=(tan(π/3)-tan(π/4)/[1+tan(π/3)tan(π/4)]=(√3-1)/(1+√3)
=(√3-1)²/2=(4-2√3)/2=2-√3.