f(x)=sin(x-π/4)+cos(x+π/4)的单调递增区间

f(x)=sin(x-π/4)+cos(x+π/4)的单调递增区间
f(x)=sin(x-π/4)+cos(x+π/4)的单调递增区间

f(x)=sin(x-π/4)+cos(x+π/4)的单调递增区间
f(x)=sin(x-π/4)+sin[π/2-(x+π/4)]
=sin(x-π/4)+sin(π/4-x)
=sin(x-π/4)-sin(x-π/4)
=0
所以是常函数
所以没有增区间

f(x)=sin(x-π/4)+cos(x+π/4)=sinx*根2/2-根2/2*cosx+根2/2*cosx-根2/2*sinx=0
整个实数区间不增不减

f(x)=sin(x-π/4)+cos(x+π/4)
=sin(x-π/4)+sin(π/2-x-π/4)
=sin(x-π/4)+sin(π/4-x)
=sin(x-π/4)-sin(x-π/4)
=0
题目有问题呢？
若为f(x)=sin(x-π/4)+cos(x-π/4)
=√2*[(√2/2)sin(x-π/4)+(√2/2)cos(...

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f(x)=sin(x-π/4)+cos(x+π/4)
=sin(x-π/4)+sin(π/2-x-π/4)
=sin(x-π/4)+sin(π/4-x)
=sin(x-π/4)-sin(x-π/4)
=0
题目有问题呢？
若为f(x)=sin(x-π/4)+cos(x-π/4)
=√2*[(√2/2)sin(x-π/4)+(√2/2)cos(x-π/4)]
=√2sin(x-π/4+π/4)
=√2sinx
所以单增区间x∈[2kπ-π/2, 2kπ+π/2] k∈Z

收起

f(X)=0
????