Question

高一数学题，要详细过程

详见图

Answer 1

f(x) = 2[cos(x/2)]^2 = cosx + 1

g(x) = [sin(x/2) + cos(x/2)]^2 = [sin(x/2)]^2 + [cos(x/2)]^2 + 2sin(x/2)*cos(x/2) = 1 + sinx

所以：

1. f(π/2 - x) = cos(π/2 - x) + 1 = sinx + 1 = g(x)

2. h(x) = f(x) - g(x)

          = (cosx + 1) - (1 + sinx)

          = cosx - sinx

          =√2 * [(√2 /2)*cosx - (√2 /2)*sinx]

          =√2 * [cos(π/4)*cosx - sin(π/4)*sinx]

          =√2 * cos(π/4 + x)

          因为 π/4 + x ∈[π/4, 5π/4]，所以，h(x) ∈[-1, √2 /2]

   当 x = 3π/4 时，h(x) = cosπ = -1