f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)cos(x+π/8)
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/25 08:23:32
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)cos(x+π/8)
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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