有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/2π,且f1(π/2)=f2(π/2),f1(π/4)=-根
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/15 05:46:22
![有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/2π,且f1(π/2)=f2(π/2),f1(π/4)=-根](/uploads/image/z/4337469-45-9.jpg?t=%E6%9C%89%E4%B8%A4%E4%B8%AA%E5%87%BD%E6%95%B0f1%28x%29%3Dasin%28wx%2B%CF%80%2F3%29%2Cf2%28x%29%3Dbtan%28wx-%CF%80%2F3%29%2Cw%3E0%2C%E5%B7%B2%E7%9F%A5%E4%BB%96%E4%BB%AC%E7%9A%84%E5%91%A8%E6%9C%9F%E4%B9%8B%E5%92%8C%E4%B8%BA3%2F%E6%9C%89%E4%B8%A4%E4%B8%AA%E5%87%BD%E6%95%B0f1%EF%BC%88x%EF%BC%89%3Dasin%EF%BC%88wx%2B%CF%80%2F3%EF%BC%89%2Cf2%EF%BC%88x%EF%BC%89%3Dbtan%EF%BC%88wx-%CF%80%2F3%EF%BC%89%2Cw%EF%BC%9E0%2C%E5%B7%B2%E7%9F%A5%E4%BB%96%E4%BB%AC%E7%9A%84%E5%91%A8%E6%9C%9F%E4%B9%8B%E5%92%8C%E4%B8%BA3%2F2%CF%80%2C%E4%B8%94f1%28%CF%80%2F2%EF%BC%89%3Df2%EF%BC%88%CF%80%2F2%EF%BC%89%2Cf1%EF%BC%88%CF%80%2F4%EF%BC%89%3D-%E6%A0%B9)
有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/2π,且f1(π/2)=f2(π/2),f1(π/4)=-根
有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/
有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/2π,且f1(π/2)=f2(π/2),f1(π/4)=-根号3*f2(π/4)+1.1求a,b,w的值
有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),w>0,已知他们的周期之和为3/2π,且f1(π/2)=f2(π/2),f1(π/4)=-根
有两个函数f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),
w>0,已知他们的周期之和为3/2π,w=2
且f1(π/2)=f2(π/2),a=-2b
f1(π/4)=-根号3*f2(π/4)+1
a = 根号3+1 b=-(根号3+1)/2 w=2
f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),
w>0,周期之和为3/2π,
根据周期公式,正弦函数周期T1=2π/w,正切T2=π/w,两者和已知,很容易求出w=2
f1(π/2)=f2(π/2),即asin(4π/3)=btan(2π/3),即a根号3/2=-b根号3,即a=-2b
f1(π/4)=-根号3*f2(π/4)+1,即...
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f1(x)=asin(wx+π/3),f2(x)=btan(wx-π/3),
w>0,周期之和为3/2π,
根据周期公式,正弦函数周期T1=2π/w,正切T2=π/w,两者和已知,很容易求出w=2
f1(π/2)=f2(π/2),即asin(4π/3)=btan(2π/3),即a根号3/2=-b根号3,即a=-2b
f1(π/4)=-根号3*f2(π/4)+1,即asin(5π/6)=-根号3*btan(π/6)+1,即-a/2=-b+1
两式联立,a=-1,b=1/2
看看和答案一样不?有正确答案麻烦告知。
f1(π/4)=-根号3*f2(π/4)+1
a = 根号3+1 b=-(根号3+1)/2 w=2
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