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
热心网友
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+genhao22。(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。
热心网友
1的答案是1+根号2
热心网友
f(1)=sin(π／4)=(2^0.5)/2f(2)=sin(2π／4)=1f(3)=sin(3π／4)=(2^0.5)/2f(4)=sin(4π／4)=0f(5)=sin(5π／4)=-(2^0.5)/2f(6)=sin(6π／4)=-1f(7)=sin(7π／4)=-(2^0.5)/2f(8)=sin(8π／4)=0f(9)=sin(9π／4)=(2^0.5)/2......................f(1)+f(2)+.....+f(7)+f(8)=0100/8=12...4f(1)+f(2)+.....+f(99)+f(100)=(2^0.5)/2)+1+(2^0.5)/2+0=1+2^0.5
热心网友
f(1)=sin(π／4)=(2^0.5)/2f(2)=sin(2π／4)=1f(3)=sin(3π／4)=(2^0.5)/2f(4)=sin(4π／4)=0f(5)=sin(5π／4)=-(2^0.5)/2f(6)=sin(6π／4)=-1f(7)=sin(7π／4)=-(2^0.5)/2f(8)=sin(8π／4)=0f(9)=sin(9π／4)=(2^0.5)/2......................f(1)+f(2)+.....+f(7)+f(8)=0100/8=12...4f(1)+f(2)+.....+f(99)+f(100)=(2^0.5)/2)+1+(2^0.5)/2+0=1+2^0.5