应该是tan[(n+1)2+π\4]吧?改一下f(n)=sin[(n+1\2)π+π\4]+cos[(n-1\2)π+π\4]+tan[(n+1)π+π\4],正余弦周期2π正切周期πf(2011)=sin(2011π+1\2π+π\4)+cos(1005π+π\4)+tan(1006π+π\4)=sin(2010π+3\2π+π\4)+cos(1004π+π+π\4)+tanπ\4=sin(3\2π+π\4)+cos(π+π\4)+1=-cosπ\4-cosπ\4+1=1-√2