函数f(x)=sin(wx z)的图像如图所示,如果x1和x2属于(-)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/10 15:55:15
![函数f(x)=sin(wx z)的图像如图所示,如果x1和x2属于(-)](/uploads/image/f/2325020-68-0.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28wx+z%29%E7%9A%84%E5%9B%BE%E5%83%8F%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E5%A6%82%E6%9E%9Cx1%E5%92%8Cx2%E5%B1%9E%E4%BA%8E%28-%29)
f(x)=sin²x=(-1/2)(1-2sin²x)+1/2=-(1/2)cos2x+1/2所以f(x)周期是:π
①原式=f(x)=2cos2x+sinx^2=2cos2x+1-cos2x/2=3/2cos2x+1/2故f(π/3)=3/2*cos2π/3+1/2=-3/4+1/2=-1/4②依f(x)=3/2c
f(x)=sin²x+sinxcosx=[1-cos(2x)]/2+sin(2x)/2=sin(2x)/2-cos(2x)/2+1/2=(√2/2)sin(2x-π/4)+1/2最小正周期T
其图像经过点M(π/3,1/2)代入f(x)=sin(x+φ)1/2=sin(π/3+φ)∵0<φ<π∴π/3<π/3+φ<4π/3∵1/2=sin(π/3+φ)∴π/3+φ=
1.sin(x-π/6)2.可知半周期为2π/3,又在区间(0,π/3)上是增函数,故ω>0,2π/ω=4π/3从而ω=1.5
1、由于函数g(x)=sin(2(x-a)+π/3)为偶函数,所以g(x)的图像关于y轴对称,即函数g(x)当x=0时取得最值,所以g(0)=±1,解得sin(π/3-2a)=±1,sin(2a-π/
∵f(x)=2sin(π-x)cosx=2sinxcosx=sin2x1、最小正周期T=2π/2=π.2、∵-π/6≤x≤π/2∴-π/3≤2x≤π,∴-√3/2≤f(x)≤1,∴最大值1,最小值-√
f(x)=sin2x-2sin^2x=sin2x+cos2x-1=√2sin(2x+π/4)-1.(1)T=2π/2=π.(2).当2x+π/4=2kπ+π/2,k∈Z,即x=kπ+π/8,k∈Z时,
你啊,要好好学习了!还没有悬赏分?把对称轴即x=∏/8代入原式子,即sin(∏/4+φ)=1或者-1,再用(-π
f(x)=cosx+sinxf(x)=√2sin(x+π/4)(1)递增区间:2kπ-π/2≤x+π/4≤2kπ+π/2得:2kπ-3/4π≤x≤2kπ+π/4递增区间是:[2kπ-3π/4,2kπ+
f(x)=sinx-sin(x-π3)=12sinx+32cosx=sin(x+π3)∴函数f(x)=sinx-sin(x-π3)的最大值为1故答案为:1
f(x)=sin2x+cos2x-1=√2sin(2x+π/4)-1.1、最小正周期是π,最大值时2x+π/4=2kπ+π/2,即x=kπ+π/4,k是整数.再问:已知函数f(x)=2sin(∏-X)
因为f(x)=sinx+cosx=√2sin(x+π/4)第一题T=2π/1=2π第二题当sin(x+π/4)=1时,为最大值,即f(x)=√2sin(x+π/4)=-1时,为最小值,即f(x)=-√
1)由三角函数和差化积公式:f(x)=2sin(x+x+π/3)/2cos(x-x-π/3)/2=2sin(x+π/6)cos(π/6)=√3sin(x+π/6)f(x)的最小值为-√3.当x+π/6
函数f(x)=sin(2x+b),(|b|
(1)偶函数,则f(x)=f(-x)即:sin(2x+φ)=sin(-2x+φ),根据积化和差公式sin(2x)*cos(φ)+cos(2x)*sin(φ)=sin(-2x)*cos(φ)+cos(-
∵函数f(x)=sin(ωx+φ)(w>0,0≤φ≤π)是R上的偶函数∴f(-x)=f(x)→sin(-wx+φ)=sin(wx+φ)→-sinωxcosφ=sinωxcosφ∵sinωx不恒等于0,
f(-x)=f(x)所以sin(-2x+a)=sin(2x+a)所以-2x+a=2kπ+2x+a或2x+a=2kπ+π-(2x+a)这是恒等式而-2x+a=2kπ+2x+a,2kπ+4x=0不是恒等式
1:(sinwx)^2+√3sinwxsin(wx+π\2)=(sinwx)^2+√3sinwxcoswx=2[(sinwx)^2+(√3\2)sin2wx]\2=[2(sinwx)^2+√3sin2