求值若αβ∈（0，π/2），cos[α
若α β ∈（0，π/2），c[α -(β/2)]=√3/2,sin[α/2-β)=-1/2,则cos(α +β)的值
cos[α -(β/2)]=√3/2 ===>α -(β/2) =π/6 或 -π/6 2α -β =π/3 或 -π/3 sin[α/2-β)=-1/2 =====>α/2-β = - π/6 α -2β = - π/3 如果,2α -β =π/3 ,α -2β = - π/3 ==>α =β =π/3 如故2α -β = -π/3 ,α -2β = - π/3 ==>α=- π/9 β =π/9与已知α β ∈（0，π/2）矛盾 所以,α =β =π/3 ==>cos(α +β) =-1/2