已知向量a={cos(x+π/8),sin方(x+π/8)},b向量={sin(x+π/8),1}函数f(x)=2ab向
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已知向量a={cos(x+π/8),sin方(x+π/8)},b向量={sin(x+π/8),1}函数f(x)=2ab向量-1
1 求函数f(x)解析式,并写出函数f(x)周期
2 求函数y=f(-1/2x)的单调递增区间
1 求函数f(x)解析式,并写出函数f(x)周期
2 求函数y=f(-1/2x)的单调递增区间
![已知向量a={cos(x+π/8),sin方(x+π/8)},b向量={sin(x+π/8),1}函数f(x)=2ab向](/uploads/image/z/4749444-36-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%7Bcos%28x%2B%CF%80%2F8%EF%BC%89%2Csin%E6%96%B9%EF%BC%88x%2B%CF%80%2F8%EF%BC%89%EF%BD%9D%2Cb%E5%90%91%E9%87%8F%3D%EF%BD%9Bsin%28x%2B%CF%80%2F8%EF%BC%89%2C1%EF%BD%9D%E5%87%BD%E6%95%B0f%28x%29%3D2ab%E5%90%91)
(1)f(x)=2[cos(x+π/8)sin(x+π/8)+sin^2(x+π/8)]-1
=2sin(x+π/8)cos(x+π/8)+2sin^2(x+π/8)-1
=sin(2x+π/4)-cos(2x+π/4)
=√2sin[(2x+π/4)-π/4]
=√2sin2x,
T=2π/2=π
(2)因为f(x)=√2sin2x,所以y=f(-1/2x)=√2sin(-x)=-√2sinx,当x∈(π/2+2kπ,3π/2+2kπ),k∈Z时,y=f(-1/2x))=-√2sinx单调递增,故函数y=f(-1/2x)的单调递增区间是(π/2+2kπ,3π/2+2kπ),k∈Z.
=2sin(x+π/8)cos(x+π/8)+2sin^2(x+π/8)-1
=sin(2x+π/4)-cos(2x+π/4)
=√2sin[(2x+π/4)-π/4]
=√2sin2x,
T=2π/2=π
(2)因为f(x)=√2sin2x,所以y=f(-1/2x)=√2sin(-x)=-√2sinx,当x∈(π/2+2kπ,3π/2+2kπ),k∈Z时,y=f(-1/2x))=-√2sinx单调递增,故函数y=f(-1/2x)的单调递增区间是(π/2+2kπ,3π/2+2kπ),k∈Z.
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