Question

求数学大神17题

Answer 1

f(x)=[cosxcos(π/3)-sinxsin(π/3)]² - [sinxcos(π/6)+cosxsin(π/6)]²
=[(1/2)cosx - (√3/2)sinx]² - [(√3/2)sinx + (1/2)cosx]²
=(1/4)cos²x - (√3/2)cosxsinx + (3/4)sin²x - (3/4)sin²x - (√3/2)sinxcosx - (1/4)cos²x
=(-√3/2)·2sinxcosx
=(－√3/2)·sin2x
(1)∵f(x)的单调递增区间为：
2kπ + π/2≤2x≤2kπ + 3π/2
∴kπ + π/4≤x≤kπ + 3π/4，(k∈Z)
∵已知区间是[0，π/2]
∴当k=0时，f(x)的单调递增区间是[π/4，π/2]
(2)由已知：g(x)=(－√3/2)·sin[2(x - π/6)]
=(-√3/2)·sin(2x - π/3)
∵0≤x≤π/2
∴0≤2x≤π
则－π/3≤2x － π/3≤2π/3
∴sin(2x － π/3)∈[－√3/2，1]
∴g(x)∈[－√3 /2，3/4]