(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=

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(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=
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(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)