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若函数f（x）=sin[(π/4)x] 则f(x+2)-f(x)=A. √2f(x+1) B.√2f(x-1)C.√2f(x+3) D.√2f(x-3) 急求解答过程不会啊，
谢谢，高手们
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huluyu：明白了，谢谢二位兄台！！ 呵呵(2009-02-05 00:13)
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teacher2：解答：f(x)=sin(πx/4) f(x+2)=sin(πx/4+π/2)=cos(πx/4) f(x+2)-f(x) =cos(πx/4)-sin(πx/4) =-√2sin(πx/4-π/4) =-√2f(x-1) 因为函数的周期是8， 所以f(x-1)=-f...(2009-02-04 21:34)
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huanyinghj：解： f(x+2)=sin[(π/4)(x+2)]=sin[(π/4)x+π/2]=cos(π/4)x f(x+2)-f(x)=cos(π/4)x - sin(π/4)x =-√2sin(π/4)(x-1)=√2sin[(π/4)(x-1)+π]=√2sin[(π/4)(x...(2009-02-04 21:34)
