高中數學必修四求函數解析式 求解答案 詳細過程

f(x)=cosx*(1/2)+sin2x*(√3/2)+2sin(x-π/4)cos(x-π/4)

=(1/2)cos2x+(√3/2)sin2x+sin(2x-π/2)

=(1/2)cos2x+(√3/2)sin2x-cos2x

=(√3/2)sin2x-(1/2)cos2x

=sin(2x-π/6)

T=2π/2=π

對稱軸方程：

2x-π/6=π/2+2kπ

x=π/3+kπ

(2)

-π/12≤x≤π/2

-π/6≤2x≤π

-π/3≤2x-π/6≤5π/6

-√3/2≤sin(2x-π/6)≤1