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

来源：学生作业帮助网编辑：作业帮时间：2024/05/07 04:21:18

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

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...

全部展开

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

收起

判断下列函数的奇偶性 f(x)=[sin(π/2+x)cos(π/2-x)tan(-x+3π)]/[sin(7π-x)tan(8π-x)] 已知函数f(x)=sin(x+π/6)cos(x+π/6)如何把这种函数化简设函数 f(x)=sin(2x+y),(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f x=SIN(2X+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 已知函数F(X)=SIN(2X+φ)(-π 已知函数f(x)=cos2x/[sin(π/4-x)] 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+ φ)(-π 如果函数f(x)=sin(πx+a)(0 函数f(x)=sin(πx²)(-1 设函数f(x)=sin(2x+φ)(-π 函数f(x)=sin(2x+a) -π 如果函数f(x)=sin(π x+A)(0