高一数学正弦函数的周期性f(n)=sin(πn/6 +π/6),求f(1)+f(2)+...+f(100)
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高一数学正弦函数的周期性f(n)=sin(πn/6 +π/6),求f(1)+f(2)+...+f(100)
高一数学正弦函数的周期性
f(n)=sin(πn/6 +π/6),求f(1)+f(2)+...+f(100)
高一数学正弦函数的周期性f(n)=sin(πn/6 +π/6),求f(1)+f(2)+...+f(100)
分析:
f(1)=sin(π/6 +π/6)=sin(2π/6)
f(2)=sin(2π/6 +π/6)=sin(3π/6)
.
f(12)=sin(6π/6 +π/6)=sin(13π/6)
f(13)=sin(6π/6 +π/6)=sin(14π/6)=sin(14π/6-2π) =sin(2π/6)
由sinx周期为2π,2π/(π/6)=12所以f(n)周期为12
100/12=8,余4
所以原式=8*(f(1)+f(2)+...f(12))+f(1)+f(2)+f(3)+f(4)=?【结果略】
f(1)+f(2)+...f(12)具体值自己算一下.
得数应该是0
f(1) = sin(2π/6)
f(13) = sin(14π/6) = sin(2π + 2π/6) = sin(2π/6) = f(1)
所以f(n)以12为一个周期,∴f(1) + ... + f(100) = 8*[f(1) + f(2) + ... + f(12)] + f(97) + f(98) + f(99) + f(100)
= ...
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f(1) = sin(2π/6)
f(13) = sin(14π/6) = sin(2π + 2π/6) = sin(2π/6) = f(1)
所以f(n)以12为一个周期,∴f(1) + ... + f(100) = 8*[f(1) + f(2) + ... + f(12)] + f(97) + f(98) + f(99) + f(100)
= 8 * [sin(2π/6) + sin(3π/6) + sin(4π/6) + .. + sin(13π/6)] + f(1) + f(2) + f(3) + f(4)
= 8 * 0 + sin(2π/6) + sin(3π/6) + sin(4π/6) + sin(5π/6)
= √3/2 + 1 + √3/2 + 1/2
=√3 + 3/2
收起
2分之根号3,也就是sin60度
之前加一个f(0)=1/2。则f(0)加到f(11)是0,f(12)加到f(12+11)=0,以此类推,可得从f(0)加到f(95)等于0,则最后的结果是f(96)+f(97)+f(98)+f(99)+f(100)=f(0)+f(1)+f(2)+f(3)+f(4)=2+根号3,再减去最初多加的f(0)则最后的结果是
根号3 + 3/2...
全部展开
之前加一个f(0)=1/2。则f(0)加到f(11)是0,f(12)加到f(12+11)=0,以此类推,可得从f(0)加到f(95)等于0,则最后的结果是f(96)+f(97)+f(98)+f(99)+f(100)=f(0)+f(1)+f(2)+f(3)+f(4)=2+根号3,再减去最初多加的f(0)则最后的结果是
根号3 + 3/2
收起