设f(x)=sin(πx/2+π/3)+cos(πx/2+π/6),x属于z,求f(1)+f(2)+······+f(2007)的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 15:29:36
![设f(x)=sin(πx/2+π/3)+cos(πx/2+π/6),x属于z,求f(1)+f(2)+······+f(2007)的值](/uploads/image/z/1628019-27-9.jpg?t=%E8%AE%BEf%28x%29%3Dsin%28%CF%80x%2F2%2B%CF%80%2F3%29%2Bcos%28%CF%80x%2F2%2B%CF%80%2F6%29%2Cx%E5%B1%9E%E4%BA%8Ez%2C%E6%B1%82f%281%29%2Bf%EF%BC%882%EF%BC%89%2B%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%2Bf%EF%BC%882007%EF%BC%89%E7%9A%84%E5%80%BC)
设f(x)=sin(πx/2+π/3)+cos(πx/2+π/6),x属于z,求f(1)+f(2)+······+f(2007)的值
设f(x)=sin(πx/2+π/3)+cos(πx/2+π/6),x属于z,求f(1)+f(2)+······+f(2007)的值
设f(x)=sin(πx/2+π/3)+cos(πx/2+π/6),x属于z,求f(1)+f(2)+······+f(2007)的值
f(x)=sin(πx/2+π/3)+cos(πx/2+π/6)
=sin(πx/2+π/3)+sin[π/2-(πx/2+π/6)]
=sin(πx/2+π/3)+sin(π/3-πx/2)
=2cos(πx/2)sin(π/3)
=√3cos(πx/2)
又,cos(π/2)+cos(2π/2)+cos(3π/2)+cos(4π/2)=0
所以,f(1)+f(2)+f(3)+f(4)=0
因为,cosx的周期为2π
所以,连续4个f(x)的和为0
由,2007=4×501+3
所以,f(1)+f(2)+······+f(2007)=f(2005)+f(2006)+f(2007)=f(1)+f(2)+f(3)
而,f(1)+f(2)+f(3)=√3[cos(π/2)+cos(2π/2)+cos(3π/2)]=-√3
所以,f(1)+f(2)+······+f(2007)=-√3
设函数 f(x)=sin(2x+y),(-π
设函数f(x)=sin(2x+φ)(-π
设函数f(x)=sin(2x+φ)(-π
设函数f x=SIN(2X+φ)(-π
设函数f(x)=sin(2x+φ)(-π
设函数f(x)=sin(2x+φ)(-π
设函数f(x)=sin(2x+φ)(-π
设函数f(x)=sin(2x+φ)(-π
设函数f(x)=sin(2x+ φ)(-π
设函数f(x)=sin(2x+φ)(-π
设函数f(x)=sin(2x+ φ)(-π
设函数f(x)=sin(2x+φ)(-π
设f(x)=sinπx/3,求f(1)+f(2)+f(3)+...+f(2011)的值RT
设函数f(x)=sinπ/6(x),则f(1)+f(2)+f(3)+…f(2008)=?
设函数f(x)=sinπ/6(x),则f(1)+f(2)+f(3)+…f(2008)=?
设f(x)={sinπx(x
设f(x)={sinπx(x
设 f(x)= {sinπx(x