满足拉格朗日中值定理，并求满足定理的ξ的值．

首先，函数f(x) = sin(x) + cos(x)在区间[0, π/2]上是连续的。其次，求导数f’(x) = d/dx (sin(x) + cos(x)) = cos(x) - sin(x)。根据拉格朗日中值定理，存在ξ ∈ (0, π/2)，使得f’(ξ) = (f(π/2) - f(0))/(π/2 - 0) = (sin(π/2) + cos(π/2) - (sin(0) + cos(0)))/π = 1/π。解方程cos(ξ) - sin(ξ) = 1/π，可求得ξ的值。