这道高中数学题，怎么化简的

Question

如题所述

Answer 1

你参考看看～

追问

谢谢你的答案

Answer 2

公式：2sinAsinB=cos(A-B)-cos(A+B)
(1). f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cos2xcos(π/3)+sin2xsin(π/3)+cos(π/2)-cos2x
=(1/2)cos2x+(√3/2)sin2x+0-cos2x
=-(1/2)cos2x+(√3/2)sin2x
=-[cos2xcos(π/3)-sin2xsin(π/3)]
=-cos(2x+π/3)=-sin[π/2-(2x+π/3)]
=-sin(π/6-2x)=sin(2x-π/6)
故最小正周期T=2π/2=π;
对称轴：由2x-π/6=π/2+2kπ；得对称轴x=π/3+kπ(k∈Z);
(2). f(x)在[-π/12，π/2]上的值域。
【用五点作图法，能很快确定其值域】
x=-π/12时f(-π/12)=sin(-π/6-π/6)=sin(-π/3)=-√3/2;
x=π/3时f(π/3)=sin(2π/3-π/6)=sin(π/2)=1;
∴f(x)在[-π/12，π/2]上的值域为[-√3/2，1].

Answer 3

利用二倍角公式化简，消去