f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(
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f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(2x+π/4)=√2sin(2x+π/2)=√2cos2xf(x)=√2cos2x的最小值即cos2x=-1时f(x) 的最小值为 -√2递增区间为[-π/2+kπ,kπ] (k∈Z),不是tan4/π=b/a=1吗,为啥是√2sin(2x+2/π)呢?
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)cos(x+π/8)
=cos(2x+π/4)+sin(2x+π/4)
=√2*[√2/2*cos(2x+π/4)+√2/2*sin(2x+π/4)]
=√2*[sinπ/4*cos(2x+π/4)+cosπ/4*sin(2x+π/4)]
=√2*sin(π/4+2x+π/4)
=√2*sin(π/2+2x)
=√2*cos2x
所以f(x)最大值为:√2,
单调增区间:2kπ-π
=cos(2x+π/4)+sin(2x+π/4)
=√2*[√2/2*cos(2x+π/4)+√2/2*sin(2x+π/4)]
=√2*[sinπ/4*cos(2x+π/4)+cosπ/4*sin(2x+π/4)]
=√2*sin(π/4+2x+π/4)
=√2*sin(π/2+2x)
=√2*cos2x
所以f(x)最大值为:√2,
单调增区间:2kπ-π
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