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(π/2)=g(π/2),f(π/4)=-√3g(π/4)-1,求这两个函数的解析式.
人气:218 ℃ 时间:2019-12-20 23:01:38
解答
T1+T2=2π/|k|+2π/|2k|=3π/k=3π/2k=2f(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=bf(π/4)=asin(π/6)...
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