有两个函数f(x)=asin(kx+π
有两个函数f(x)=asin(kx+π/3),g(x)=bcos(2kx-π/3)（k＞0），它们的有两个函数f(x)=asin(kx+π/3),g(x)=bcos(2kx-π/3)（k＞0），它们的周期之和为3π/2，且f（π/2)=g（π/2),f（π/4)=（-根号3）×g（π/4）+1.求k，a，b。
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)
T1=2π/k，T2=2π/(2k) T1+T2=3π/2，即：2π/k+π/k=3π/2, k=2 f(π/2)=g(π/2),f(π/4)=-sqrt(3)*g(π/4)+1，则： asin(π+π/3)=bcos(2π-π/3)，asin(π/2+π/3)=-sqrt(3)*bcos(π-π/3)+1 -asin(π/3)=bcos(π/3)，-acos(π/3)=-sqrt(3)*bcos(π/3)+1 -sqrt(3)*a=b，-a=-sqrt(3)*b+2 故： a=-1/2 b=sqrt(3)/2
k=2,a=-1,b=根号3
答：设有函数f(x)=asin(kx-π/3)和函数g(x）=bcos(2kx-π/6),(a>0,b>0,k>o),若它们的最小正周期之和 为（3π）/2，且f(...详情>>
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