已知函数fx=2cos^2x+2sinxcosx 记函数gx=f(x-π/8)f(x+π/8) 求值域

解: f(x)=2cos^2x+2sinxcosx.
=1+cos2x+sin2x.
f(x)=√2sin(2x+π/4)+1..
f(x-π/8)=√2sin[2(x-π/8)+π/4}+1.
=√2sin(2x-π/4+π/4)+1.
∴f(x-π/8)=√2sin2x+1.
f(x+π/8)=√2sin[2(x+π/8)+π/4]+1.
=√2sin(2x+π/4+π/4)+1.
=√2sin(2x+π/2)+1.
∴f(x+π/8)=√2cos2x+1.
g(x)=f(x-π/8)*f(x+π/8).
=(√2sin2x+1)*(√2cos2x+1).
=2sin2xcos2x+√2(sin2x+cos2x)+1.
=sin4x+√2*√2sin(2x+π/4)+1.
∴g(x)=sin4x+2sin(2x+π/4)+1. (*)
由(*)可见,g(x)的值域取决于2sin(2x+π/4)的大小和符号.∵-1≤sin(2x+π/4)≤1.
当2x+π/4=π/2, x=π/8时, sin(2x+π/4)=1, sin4x=sinπ/2=1,
此时,.函数g(x)取得最大值,g(x)max=1+2+1=4;
当2x+π/4=3π/2,x=5π/8, sin(2x+π/4)=-1, sin4x=sin5π/2=sin(2π+π/2)=sinπ/2=1.
此时.g(x)取得最小值,g(x)min=1+2*(-1)+1=0.
∴g(x)的值域为:[0.4].
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