求函数y=sin^2(x+π/4)+sin(x+π/4)sin(x-π/4)的周期,单调减区间,值域
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求函数y=sin^2(x+π/4)+sin(x+π/4)sin(x-π/4)的周期,单调减区间,值域
y=sin²(x+π/4)+sin(x+π/4)sin(x-π/4)
=sin(x+π/4)(sin(x+π/4)+sin(x-π/4))
=sin(x+π/4)(2sin(x)cos(π/4))
=(√2)sin(x+π/4)sin(x)
=(-√2/2)[cos(2x+π/4)-cos(π/4)]
=(-√2/2)cos(2x+π/4)+1/2
=(-√2/2)cos[2(x+π/8)]+1/2
频率ω=2,相位φ=π/8,
∴周期(2π)/ω=(2π)/2=π.
一个周期[-π/2-π/8,π/2-π/8)内,
在(-π/2-π/8,-π/8),单调减少;在(-π/8,π/2-π/8),单调增加;
所以定义域内,
在(kπ-π/2-π/8,kπ-π/8),单调减少;在(kπ-π/8,kπ+π/2-π/8),单调增加;
即在(kπ-5π/8,kπ-π/8),单调减少;在(kπ-π/8,kπ+3π/8),单调增加;
-1≤cos(2x+π/4)≤1,所以值域:[(1-√2)/2,(1+√2)/2].
=sin(x+π/4)(sin(x+π/4)+sin(x-π/4))
=sin(x+π/4)(2sin(x)cos(π/4))
=(√2)sin(x+π/4)sin(x)
=(-√2/2)[cos(2x+π/4)-cos(π/4)]
=(-√2/2)cos(2x+π/4)+1/2
=(-√2/2)cos[2(x+π/8)]+1/2
频率ω=2,相位φ=π/8,
∴周期(2π)/ω=(2π)/2=π.
一个周期[-π/2-π/8,π/2-π/8)内,
在(-π/2-π/8,-π/8),单调减少;在(-π/8,π/2-π/8),单调增加;
所以定义域内,
在(kπ-π/2-π/8,kπ-π/8),单调减少;在(kπ-π/8,kπ+π/2-π/8),单调增加;
即在(kπ-5π/8,kπ-π/8),单调减少;在(kπ-π/8,kπ+3π/8),单调增加;
-1≤cos(2x+π/4)≤1,所以值域:[(1-√2)/2,(1+√2)/2].
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