已知f(n)=sin[(n+1\2)π+π\(zhòng)4]+cos[(n-1\2)+π\(zhòng)4]+tan[(n+1)π+π\(zhòng)4],求f(2011)的值

數(shù)學(xué)人氣：242 ℃時(shí)間：2020-05-22 12:45:00

優(yōu)質(zhì)解答

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