1.已知f(cosx)=cos17x,求证f(sinx)=sin17x2.已知x属于R,n属于Z,且f(sinx)=sin(4n+1)x,求f(cosx)小妹定当感恩不尽`.
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![1.已知f(cosx)=cos17x,求证f(sinx)=sin17x2.已知x属于R,n属于Z,且f(sinx)=sin(4n+1)x,求f(cosx)小妹定当感恩不尽`.](/uploads/image/z/3903067-19-7.jpg?t=1.%E5%B7%B2%E7%9F%A5f%28cosx%29%3Dcos17x%2C%E6%B1%82%E8%AF%81f%EF%BC%88sinx%EF%BC%89%3Dsin17x2.%E5%B7%B2%E7%9F%A5x%E5%B1%9E%E4%BA%8ER%2Cn%E5%B1%9E%E4%BA%8EZ%2C%E4%B8%94f%28sinx%29%3Dsin%284n%2B1%29x%2C%E6%B1%82f%28cosx%29%E5%B0%8F%E5%A6%B9%E5%AE%9A%E5%BD%93%E6%84%9F%E6%81%A9%E4%B8%8D%E5%B0%BD%60.)
1.已知f(cosx)=cos17x,求证f(sinx)=sin17x2.已知x属于R,n属于Z,且f(sinx)=sin(4n+1)x,求f(cosx)小妹定当感恩不尽`.
1.已知f(cosx)=cos17x,求证f(sinx)=sin17x
2.已知x属于R,n属于Z,且f(sinx)=sin(4n+1)x,求f(cosx)
小妹定当感恩不尽`.
1.已知f(cosx)=cos17x,求证f(sinx)=sin17x2.已知x属于R,n属于Z,且f(sinx)=sin(4n+1)x,求f(cosx)小妹定当感恩不尽`.
第一题目
f(cosx)=cos17x 可以推出->
f[cos(pi/2-x)]=cos(17*(pi/2-x))=cos(pi/2-17*x)=sin17x ;
f[cos(pi/2-x)]=f(sinx);
即f(sinx)=sin17x
第二题
由f(sinx)=sin(4n+1)x对于任何实数x均成立,且cosx=sin( -x),得
f(cosx)=f〔sin( -x)〕=sin〔(4n+1)·( -x)〕
=sin〔2nπ+ -(4n+1)x〕=sin〔 -(4n+1)x〕
=cos(4n+1)x.