简单的高一数学题,答对有奖!谢谢!
1.已知f(n)=sin nπ/4 (n∈N+) 求f(1)+f(2)……+f(100)的值 2.根据任意三角函数的定义证明: (1+secα-tanα)/(1+secα+tanα)=(1+sinα)/ cosα
1。 f(n+4)=sin[(n+4)π/4] = -sinnπ/4 (n∈N+) ==> f(1)+f(2)……+f(100) = f(1)+f(2)+f(3)+f(4)+{[f(5)+f(9)]+[f(6)+f(10)]+。。。。+[f(96)+f(100)]} = f(1)+f(2)+f(3)+f(4) = 1+genhao2 2。
(1+secα-tanα)/(1+secα+tanα)=(1+1/cosα-sinα/ccosα)/(1+1/cosα+sinα/ccosα) = (1+cosα-sinα)/(1+cosα-sinα) = [2*(cosα/2)^2 - 2*sinα/2*cosα/2]/[2*(cosα/2)^2 +2*sinα/2*cosα/2] = (cosα/2 - sinα/2)/(cosα/2 + sinα/2) = (cosα/2 - sinα/2)(cosα/2 + sinα/2)/(cosα/2 + sinα/2)^2 = (cosα)/(1+2*cosα/2 *sinα/2) = cosα/(1+sinα) = (1-sinα)/cosα 原题有误。
1的答案是1+根号2
f(1)=sin(π/4)=(2^0.5)/2 f(2)=sin(2π/4)=1 f(3)=sin(3π/4)=(2^0.5)/2 f(4)=sin(4π/4)=0 f(5)=sin(5π/4)=-(2^0.5)/2 f(6)=sin(6π/4)=-1 f(7)=sin(7π/4)=-(2^0.5)/2 f(8)=sin(8π/4)=0 f(9)=sin(9π/4)=(2^0.5)/2 ...................... f(1)+f(2)+.....+f(7)+f(8)=0 100/8=12...4 f(1)+f(2)+.....+f(99)+f(100)=(2^0.5)/2)+1+(2^0.5)/2+0=1+2^0.5
f(1)=sin(π/4)=(2^0.5)/2 f(2)=sin(2π/4)=1 f(3)=sin(3π/4)=(2^0.5)/2 f(4)=sin(4π/4)=0 f(5)=sin(5π/4)=-(2^0.5)/2 f(6)=sin(6π/4)=-1 f(7)=sin(7π/4)=-(2^0.5)/2 f(8)=sin(8π/4)=0 f(9)=sin(9π/4)=(2^0.5)/2 ...................... f(1)+f(2)+.....+f(7)+f(8)=0 100/8=12...4 f(1)+f(2)+.....+f(99)+f(100)=(2^0.5)/2)+1+(2^0.5)/2+0=1+2^0.5
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