有两个函数f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),已知它们的周期和为3π/2,且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)+1,求a、b、k的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 03:54:39
![有两个函数f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),已知它们的周期和为3π/2,且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)+1,求a、b、k的值](/uploads/image/z/5483980-28-0.jpg?t=%E6%9C%89%E4%B8%A4%E4%B8%AA%E5%87%BD%E6%95%B0f%28x%29%3Dasin%28kx%2B%CF%80%EF%BC%8F3%29%2Cg%28x%29%3Dbtan%28kx-%CF%80%EF%BC%8F3%29%28k%3E0%29%2C%E5%B7%B2%E7%9F%A5%E5%AE%83%E4%BB%AC%E7%9A%84%E5%91%A8%E6%9C%9F%E5%92%8C%E4%B8%BA3%CF%80%EF%BC%8F2%2C%E4%B8%94f%EF%BC%88%CF%80%EF%BC%8F2%EF%BC%89%EF%BC%9Dg%EF%BC%88%CF%80%EF%BC%8F2%EF%BC%89%2Cf%EF%BC%88%CF%80%EF%BC%8F4%EF%BC%89%EF%BC%9D%EF%BC%8D%E2%88%9A3g%EF%BC%88%CF%80%EF%BC%8F4%EF%BC%89%EF%BC%8B1%2C%E6%B1%82a%E3%80%81b%E3%80%81k%E7%9A%84%E5%80%BC)
有两个函数f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),已知它们的周期和为3π/2,且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)+1,求a、b、k的值
有两个函数f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),已知它们的周期和为3π/2,
且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)+1,求a、b、k的值
有两个函数f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),已知它们的周期和为3π/2,且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)+1,求a、b、k的值
f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),
已知它们的周期和为2π/k+π/k=3π/2,
∴k=2.
又f(π/2)=g(π/2),f(π/4)=-√3g(π/4)+1,
∴-a(√3)/2=-b√3,a/2=-√3*b(1-√3)/(1+√3)+1,
化简得a=2b,b=b(2√3-3)+1,
解得b=(2+√3)/2,a=2+√3.
F(x)周期为2π/k,G(x)周期为π/k,
3π/k=3π/2,k=2
F(x)=asin(2x+π/3),G(x)=btan(2x-π/3)
代入已知条件F(π/2)=G(π/2),F(π/4)=-√3G(π/4)+1,
再由诱导公式化简
得a=2b,a=-2b+2
a=1,b=1/2 ,k=2
T1+T2=2π/|k|+2π/|2k|=3π/k=3π/2
k=2
f(x)=asin(2x-π/3)
g(x)=bcos(4x-π/6)
f(π/2)=asin(2π/3)=asin(π-π/3)=asin(π/3)=acos(π/2-π/6)=acos(π/6)=g(π/2)=bcos(-π/6)=bcos(π/6)
所以a=b
f(π/4)=...
全部展开
T1+T2=2π/|k|+2π/|2k|=3π/k=3π/2
k=2
f(x)=asin(2x-π/3)
g(x)=bcos(4x-π/6)
f(π/2)=asin(2π/3)=asin(π-π/3)=asin(π/3)=acos(π/2-π/6)=acos(π/6)=g(π/2)=bcos(-π/6)=bcos(π/6)
所以a=b
f(π/4)=asin(π/6)=-√3g(π/4)-1=-√3bcos(5π/6)-1
所以a/2=-√3*b(-√3/2)-1=3b/2-1
a=3b-2
a=b
所以a=b=1
f(x)=sin(2x-π/3)
g(x)=cos(4x-π/6)
收起