已知函数f(x)=asin(kx+π/3)和φ(x)=btan(kx-π/3),k>0若它们的最小正周期之和是3π/2,且f(π/2)=φ(π/2),f(π/4)=-√3φ(π/4)+1,求f(x)和φ(x)的解析式
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![已知函数f(x)=asin(kx+π/3)和φ(x)=btan(kx-π/3),k>0若它们的最小正周期之和是3π/2,且f(π/2)=φ(π/2),f(π/4)=-√3φ(π/4)+1,求f(x)和φ(x)的解析式](/uploads/image/z/5243831-71-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dasin%28kx%2B%CF%80%2F3%29%E5%92%8C%CF%86%28x%29%3Dbtan%28kx-%CF%80%2F3%29%2Ck%3E0%E8%8B%A5%E5%AE%83%E4%BB%AC%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B9%8B%E5%92%8C%E6%98%AF3%CF%80%2F2%2C%E4%B8%94f%28%CF%80%2F2%29%3D%CF%86%28%CF%80%2F2%29%2Cf%28%CF%80%2F4%29%3D-%E2%88%9A3%CF%86%28%CF%80%2F4%29%2B1%2C%E6%B1%82f%28x%29%E5%92%8C%CF%86%28x%29%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F)
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iU}ۖ@(wntGZ1 已知函数f(x)=asin(kx+π/3)和φ(x)=btan(kx-π/3),k>0若它们的最小正周期之和是3π/2,且f(π/2)=φ(π/2),f(π/4)=-√3φ(π/4)+1,求f(x)和φ(x)的解析式 已知函数f(x)=asin(kx+π/3)和φ(x)=btan(kx-π/3),k>0若它们的最小正周期之和是3π/2,且f(π/2)=φ(π/2),f(π/4)=-√3φ(π/4)+1,求f(x)和φ(x)的解析式 正在百度搜同一题······
已知函数f(x)=asin(kx+π/3)和φ(x)=btan(kx-π/3),k>0
若它们的最小正周期之和是3π/2,且f(π/2)=φ(π/2),f(π/4)=-√3φ(π/4)+1,求f(x)和φ(x)的解析式
【解】函数f(x) 和φ(x) 的最小正周期之和是3π/2,
则2π/k+π/k=3π/2,k=2.
由f(π/2)=φ(π/2)可得,asin(π+π/3) =btan(π-π/3),
-√3a/2=-√3b,a/2= b.
由f(π/4)=-√3φ(π/4)+1可得,asin(π/2+π/3) =-√3btan(π/2-π/3)+1,
即a/2=-b+1.
由此解得:b=1/2,a=1.
∴f(x)=sin(2x+π/3),φ(x)= 1/2tan(2x-π/3).
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