求F(X)=SIN(2X-π/4)-2√2SIN^2A的最小正周期及值域

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求F(X)=SIN(2X-π/4)-2√2SIN^2A的最小正周期及值域

求F(X)=SIN(2X-π/4)-2√2SIN^2A的最小正周期及值域
求F(X)=SIN(2X-π/4)-2√2SIN^2A的最小正周期及值域

求F(X)=SIN(2X-π/4)-2√2SIN^2A的最小正周期及值域
函数是下面的吧?
F(x)=sin(2x-π/4)-2√2(sinx)^2
=sin2xcos(π/4)-cos2xsin(π/4)-√2(1-cos2x)
=(√2/2)sin2x-(√2/2)cos2x+√2cos2x-√2
=(√2/2)sin2x+(√2/2)cos2x-√2
=sin(2x+π/4)-√2
所以最小正周期是T=2π/2=π
因为-1≤sin(2x+π/4)≤1
所以-1-√2≤sin(2x+π/4)-√2≤1-√2
故值域是[-1-√2,1-√2]

周期为π,值域为-1-2√2