把函数f(x)=sin(7π/8-x)*cos(x+π/8)的图象向右平移a(a>0)的单位,得函数g(x)图象,且g(X)图象且g(X)图象关于直线x=π/4对称(1)求a的最小值(2)已知当x∈[(2b+1)π/8,(3b+2)π/8](b∈N)时,过函数f(x)图象上任意两点
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把函数f(x)=sin(7π/8-x)*cos(x+π/8)的图象向右平移a(a>0)的单位,得函数g(x)图象,且g(X)图象且g(X)图象关于直线x=π/4对称(1)求a的最小值(2)已知当x∈[(2b+1)π/8,(3b+2)π/8](b∈N)时,过函数f(x)图象上任意两点
把函数f(x)=sin(7π/8-x)*cos(x+π/8)的图象向右平移a(a>0)的单位,得函数g(x)图象,且g(X)图象
且g(X)图象关于直线x=π/4对称
(1)求a的最小值
(2)已知当x∈[(2b+1)π/8,(3b+2)π/8](b∈N)时,过函数f(x)图象上任意两点的直线的斜率恒大于零,求b的值
把函数f(x)=sin(7π/8-x)*cos(x+π/8)的图象向右平移a(a>0)的单位,得函数g(x)图象,且g(X)图象且g(X)图象关于直线x=π/4对称(1)求a的最小值(2)已知当x∈[(2b+1)π/8,(3b+2)π/8](b∈N)时,过函数f(x)图象上任意两点
解 （1）f(x)=sin(7π/8-x)*cos(x+π/8)=sin[π-（π/8+x)]cos(x+π/8)
=sin(x+π/8)cos(x+π/8)=1/2sin(2x+π/4)
则g(x)=f(x-a)=1/2sin(2x-2a+π/4)
由于g(X)图象关于直线x=π/4对称
所以 2*π/4-2a+π/4=kπ+π/2 (k∈Z)
a=-kπ/2+π/8 (k∈Z)
则a的最小值为π/8
（2）f(x)=1/2sin(2x+π/4)
由于x∈[(2b+1)π/8,(3b+2)π/8](b∈N)时,过函数f(x)图象上任意两点的直线的斜率恒大于零
因此f(x)在∈[(2b+1)π/8,(3b+2)π/8](b∈N)为增函数
函数f(x)=1/2sin(2x+π/4)在[kπ-3π/8,kπ+π/8 ] (k∈Z)为增函数
则 (2b+1)π/8≥kπ-3π/8 且(3b+2)π/8≤kπ+π/8 (k∈Z)
又 b∈N 因此b=2
8
f(x)=1/2{Sin[(7π/8-x)+(x+π/8)]+Sin[(7π/8-x)-(x+π/8)]}
=1/2Sin(3π/4-2x)
g(x)=1/2Sin(3π/4-2x-a)
3π/4-2(π/4)-a=2kπ+π/2
(1) a=3π/4
(2)f(x)=1/2[Sin(π)+Sin(3π/4-2x)]
x=(2b+1)π/8时，f(x...
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f(x)=1/2{Sin[(7π/8-x)+(x+π/8)]+Sin[(7π/8-x)-(x+π/8)]}
=1/2Sin(3π/4-2x)
g(x)=1/2Sin(3π/4-2x-a)
3π/4-2(π/4)-a=2kπ+π/2
(1) a=3π/4
(2)f(x)=1/2[Sin(π)+Sin(3π/4-2x)]
x=(2b+1)π/8时，f(x)=-1/2
则1/2Sin(3π/4-2x)=-1/2
3π/4-2(2b+1)π/8=-1/2π
2b+1=5
b=2
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