函数f(X)=sin(2x-π/4)-2根号2sin^2x的最小值

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函数f(X)=sin(2x-π/4)-2根号2sin^2x的最小值
函数f(X)=sin(2x-π/4)-2根号2sin^2x的最小值

函数f(X)=sin(2x-π/4)-2根号2sin^2x的最小值
f(X)=sin(2x-π/4)-2根号2sin^2x
=√2/2sin2x-√2/2cos2x-√2(1-cos2x)
=√2/2sin2x+√2/2cos2x-√2
=sin(2x+π/4)-√2
当sin(2x+π/4)=-1时,函数有最小值 ymin=-1-√2

T=2π/2=π sin最大是1 所以最大值=1+√2 递增则sin递增所以2kπ-f(x)=1+√2sin(2x-π/4) 由于y=sinxr的周期为2π，故原函数的最小

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