已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/10 14:03:16
已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)
(1)求最小正周期
(2)求函数单调增区间
(3)若x属于【0,pai/2】时,F(x)的最小值是-2,求a
(1)求最小正周期
(2)求函数单调增区间
(3)若x属于【0,pai/2】时,F(x)的最小值是-2,求a
![已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)](/uploads/image/z/8224845-69-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x%2B%CF%80%2F6%29%2Bsin%282x%E2%80%94%CF%80%2F6%29%E2%80%94cos2x%2Ba%28a%E4%B8%BA%E5%AE%9E%E6%95%B0%2C%E5%B1%9E%E4%BA%8ER%29)
第一个问题:
f(x)=sin2xcos(π/6)+cos2xsin(π/6)+sin2xcos(π/6)-cos2xsin(π/6)-cos2x+a
=2sin2xcos(π/6)-cos2x+a=2[sin2xcos(π/6)-cos2xsin(π/6)]+a
=2sin(2x-π/6)+a.
∴函数f(x)的最小正周期为2π/2=π.
第二个问题:
∵f(x)=2sin(2x-π/6)+a.∴当 2kπ-π/2≦2π-π/6≦2kπ+π/2 时,f(x)单调递增.
由2kπ-π/2≦2x-π/6≦2kπ+π/2,得:2kπ-3π/6+π/6≦2x≦2kπ+3π/6+π/6,
∴2kπ-2π/6≦2x≦2kπ+4π/6,∴kπ-π/6≦x≦kπ+π/3.
即函数f(x)的单调增区间是[kπ-π/6,kπ+π/3],其中k为整数.
第三个问题:
∵0≦x≦π/2 ,∴0≦2x≦π,∴-π/6≦2x-π/6≦π-π/6,
∴f(x)的最小值为2sin(-π/6)+a=-1+a=-2,∴a=-1.
f(x)=sin2xcos(π/6)+cos2xsin(π/6)+sin2xcos(π/6)-cos2xsin(π/6)-cos2x+a
=2sin2xcos(π/6)-cos2x+a=2[sin2xcos(π/6)-cos2xsin(π/6)]+a
=2sin(2x-π/6)+a.
∴函数f(x)的最小正周期为2π/2=π.
第二个问题:
∵f(x)=2sin(2x-π/6)+a.∴当 2kπ-π/2≦2π-π/6≦2kπ+π/2 时,f(x)单调递增.
由2kπ-π/2≦2x-π/6≦2kπ+π/2,得:2kπ-3π/6+π/6≦2x≦2kπ+3π/6+π/6,
∴2kπ-2π/6≦2x≦2kπ+4π/6,∴kπ-π/6≦x≦kπ+π/3.
即函数f(x)的单调增区间是[kπ-π/6,kπ+π/3],其中k为整数.
第三个问题:
∵0≦x≦π/2 ,∴0≦2x≦π,∴-π/6≦2x-π/6≦π-π/6,
∴f(x)的最小值为2sin(-π/6)+a=-1+a=-2,∴a=-1.
已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—cos2x+a(a为实数,属于R)
已知函数f(x)=sin(2x+pai/6)+sin(2x-pai/6)+cos2x+a(a属于R,a为常数)
已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a(a∈R,a为常数)
已知函数f(x)=sin(x+π/6)+sin(x- π /6)+cosx+a (x属于R,a为常数) ①求函数f(x)
已知函数f(x)=sin(2x+π6)+2sin2(x+π6)−2cos2x+a−1(a∈R,a为常数)
已知函数f(χ)=sin(2x+π/6 )+sin(2x- π/6)+cos2x+1(x∈R),
已知函数f(x)=sin(x+6分之π)+sin(x-6分之π)+cosx+a(a属于R,a是常数).(1)求函数f(x
已知函数f(x)=2SIN(2X+π/6)+a a属于常数
已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+1(X属于R)求函数f(x)的最大值及此是
已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a,(1)求函数的最小正周期及单调
已知函数f(x)=sin(2x+π6)+sin(2x-π6)+2cos2x(x∈R).
已知函数f(x)=2sin^2(π/4+x)-(根号3乘以cos2x),x属于[π/4,π/2].1:将f(x)化简成A