(1)函数y=sin(x+π/4),x∈(-π/2,π/2)的值域是 (2)函数y=1/2sin(π/4-2π/3)的单调区间是(3)已知f(x)=sin(x+θ)+根号3cos(x-θ)为偶函数,则θ=
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 01:38:44
![(1)函数y=sin(x+π/4),x∈(-π/2,π/2)的值域是 (2)函数y=1/2sin(π/4-2π/3)的单调区间是(3)已知f(x)=sin(x+θ)+根号3cos(x-θ)为偶函数,则θ=](/uploads/image/z/3647657-65-7.jpg?t=%EF%BC%881%EF%BC%89%E5%87%BD%E6%95%B0y%3Dsin%28x%2B%CF%80%2F4%29%2Cx%E2%88%88%28-%CF%80%2F2%2C%CF%80%2F2%29%E7%9A%84%E5%80%BC%E5%9F%9F%E6%98%AF+%EF%BC%882%EF%BC%89%E5%87%BD%E6%95%B0y%3D1%2F2sin%28%CF%80%2F4-2%CF%80%2F3%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%E6%98%AF%EF%BC%883%EF%BC%89%E5%B7%B2%E7%9F%A5f%28x%29%3Dsin%28x%2B%CE%B8%29%2B%E6%A0%B9%E5%8F%B73cos%28x-%CE%B8%29%E4%B8%BA%E5%81%B6%E5%87%BD%E6%95%B0%2C%E5%88%99%CE%B8%3D)
(1)函数y=sin(x+π/4),x∈(-π/2,π/2)的值域是 (2)函数y=1/2sin(π/4-2π/3)的单调区间是(3)已知f(x)=sin(x+θ)+根号3cos(x-θ)为偶函数,则θ=
(1)函数y=sin(x+π/4),x∈(-π/2,π/2)的值域是 (2)函数y=1/2sin(π/4-2π/3)的单调区间是
(3)已知f(x)=sin(x+θ)+根号3cos(x-θ)为偶函数,则θ=
(1)函数y=sin(x+π/4),x∈(-π/2,π/2)的值域是 (2)函数y=1/2sin(π/4-2π/3)的单调区间是(3)已知f(x)=sin(x+θ)+根号3cos(x-θ)为偶函数,则θ=
(1)因为x∈(-π/2,π/2),则x+π/4∈(-π/4,3π/4)
所以由正弦函数的单调性可知:
函数y=sin(x+π/4),x∈(-π/2,π/2)的值域是 ( -√2/2,1 ]
(2)函数y=1/2sin(π/4-2π/3)应该是 函数y=1/2sin(x/4-2π/3) 吧!
当x/4-2π/3∈[2kπ-π/2,2kπ+π/2]即x∈[8kπ+2π/3,8kπ+14π/3]时,该函数是增函数;
当x/4-2π/3∈[2kπ+π/2,2kπ+3π/2]即x∈[8kπ+14π/3,8kπ+26π/3]时,该函数是减函数;
所以函数y=1/2sin(x/4-2π/3) 的单调增区间为[8kπ+2π/3,8kπ+14π/3];
单调减区间为[8kπ+14π/3,8kπ+26π/3],其中k∈Z
(3)已知f(x)=sin(x+θ)+根号3cos(x-θ)为偶函数,
则f(x)=2[1/2 *sin(x+θ)+√3/2 *cos(x-θ)]
=2sin(x+θ+π/3)
且对于任意实数x,都有f(-x)=f(x)
则2sin(-x+θ+π/3)=2sin(x+θ+π/3)
即sin(-x+θ+π/3)=sin(x+θ+π/3)
令x=π/3,代入上式,得:sinθ=sin(θ+2π/3)
即sinθ=sinθ*cos(2π/3)+cosθ*sin(2π/3)=-1/2 *sinθ+√3/2 *cosθ
则3/2 *sinθ-√3/2 *cosθ=0
即√3/2 *sinθ-1/2 *cosθ=0
sin(θ-π/6)=0
则θ-π/6=kπ
解得θ=kπ+ π/6,k∈Z
(注:当然,题目若θ有限制条件,那么θ取具体的一个值,比如θ是锐角,那么θ=π/6;
否则有无限多个解)