【急】已知函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1设函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1(1).求a.b的值【已算得a=1.
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/25 18:11:06
![【急】已知函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1设函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1(1).求a.b的值【已算得a=1.](/uploads/image/z/5223578-50-8.jpg?t=%E3%80%90%E6%80%A5%E3%80%91%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dasinwx%2Bbcoswx%2B1%28ab%E2%89%A00%2Cw%EF%BC%9E0%29%E7%9A%84%E5%91%A8%E6%9C%9F%E4%B8%BA%CF%80%2Cf%28x%29%E5%8F%88%E6%9C%80%E5%A4%A7%E5%80%BC4%2C%E4%B8%94f%28%CF%80%2F6%29%3D%5B%283%E6%A0%B9%E5%8F%B73%29%2F2%5D%2B1%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dasinwx%2Bbcoswx%2B1%28ab%E2%89%A00%2Cw%EF%BC%9E0%29%E7%9A%84%E5%91%A8%E6%9C%9F%E4%B8%BA%CF%80%2Cf%28x%29%E5%8F%88%E6%9C%80%E5%A4%A7%E5%80%BC4%2C%E4%B8%94f%28%CF%80%2F6%29%3D%5B%283%E6%A0%B9%E5%8F%B73%29%2F2%5D%2B1%281%29.%E6%B1%82a.b%E7%9A%84%E5%80%BC%E3%80%90%E5%B7%B2%E7%AE%97%E5%BE%97a%3D1.)
【急】已知函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1设函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1(1).求a.b的值【已算得a=1.
【急】已知函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1
设函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1
(1).求a.b的值【已算得a=1.5,b=(3根号3)/2】
(2).α-β≠kπ(k∈Z),且α,β是方程f(x)=0的两个根,求tan(α+β)的值
第(2)问用其次式做,不要用和差化积做~
【急】已知函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1设函数f(x)=asinwx+bcoswx+1(ab≠0,w>0)的周期为π,f(x)又最大值4,且f(π/6)=[(3根号3)/2]+1(1).求a.b的值【已算得a=1.
(1)由已知易得 w=2,a=3/2,b=(3根号3)/2.(2)所以f(x)=3sin(2x+π/3)
由题意知sin(2a+π/3)=0 sin(2β+π/3) ,两式相减并和差化积得2cos(α+β+π/3)sin(α-β)=0
因为 α-β≠kπ(k∈Z) sin(α-β)不等0 cos(α+β+π/3)=0 即1/2cos(α+β)-(根号3)/2sin(α+β)==0 故tan(α+β)=(根号3)/3
a和b一定是正的吗?
哎,高中学过 大学两年都忘了 一干二净
1)由已知易得 w=2,a=3/2,b=(3根号3)/2. (2)所以f(x)=3sin(2x+π/3
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