作业帮已知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
来源：学生作业帮助网 编辑：作业帮 时间：2024/12/02 07:53:13
![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](/uploads/image/z/12628507-67-7.jpg?t=%E5%B7%B2%E7%9F%A5f%28n%29%3Dsin%5B%28n%2B1%2F2%29%CF%80%2B%CF%80%2F4%5D%2Bcos%5B%28n-1%2F2%29%CF%80%2B%CF%80%2F4%5D%2Btan%5B%28n%2B1%29%CF%80%2B%CF%80%2F4%5D.%E6%B1%82f%282011%29%E7%9A%84%E5%80%BC)
x){}K44m35
47ho7N/颈$BEmlJ0204|>i "}_`gCKW379ߠ 2M 6U iVf Q ԆEPDb$md W<[iv<;ܔS
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
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
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
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