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已知f(n)=sin[(n+1/2)π+π/4]+cos[(n-1/2)π+π/4]+tan[(n+1)π+π/4].求f(2011)的值
题目详情
已知f(n)=sin[(n+1/2)π+π/4]+cos[(n-1/2)π+π/4]+tan[(n+1)π+π/4].求f(2011)的值
▼优质解答
答案和解析
f(n)=sin(nπ+3/4π)+cos(nπ-1/4π)+tan(π/4)
f(2011)=sin(π+3/4π)+cos(3/4π)+1
=-sin(π/4)+cos(3/4π)+1
=1-sqrt(2)
sqrt:代表根号
f(2011)=sin(π+3/4π)+cos(3/4π)+1
=-sin(π/4)+cos(3/4π)+1
=1-sqrt(2)
sqrt:代表根号
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