作业帮把函数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：2024/11/20 02:34:09
时,过函数f(x)图象上任意两点](/uploads/image/z/12580734-30-4.jpg?t=%E6%8A%8A%E5%87%BD%E6%95%B0f%28x%29%3Dsin%287%CF%80%2F8-x%29%2Acos%28x%2B%CF%80%2F8%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E5%90%91%E5%8F%B3%E5%B9%B3%E7%A7%BBa%28a%3E0%29%E7%9A%84%E5%8D%95%E4%BD%8D%2C%E5%BE%97%E5%87%BD%E6%95%B0g%28x%29%E5%9B%BE%E8%B1%A1%2C%E4%B8%94g%28X%29%E5%9B%BE%E8%B1%A1%E4%B8%94g%28X%29%E5%9B%BE%E8%B1%A1%E5%85%B3%E4%BA%8E%E7%9B%B4%E7%BA%BFx%3D%CF%80%2F4%E5%AF%B9%E7%A7%B0%281%29%E6%B1%82a%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%282%29%E5%B7%B2%E7%9F%A5%E5%BD%93x%E2%88%88%5B%282b%2B1%29%CF%80%2F8%2C%283b%2B2%29%CF%80%2F8%5D%28b%E2%88%88N%29%E6%97%B6%2C%E8%BF%87%E5%87%BD%E6%95%B0f%28x%29%E5%9B%BE%E8%B1%A1%E4%B8%8A%E4%BB%BB%E6%84%8F%E4%B8%A4%E7%82%B9)
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
g(A2)=sinA=sin[(A-π6)+π6]= 12cos(A-π6)+32sin(A-π6)代入可求答案．（1）由题意可知将函数g（x）=sin2x的图象向右平移 π