试证明：cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0

试证明：cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0 证明：cos( π/2n+1)+cos( 3π/2n+1)+cos( 5π/2n+1)+cos(7 π/2n+1)+...+cos[(2n-1) π/2n+1)]=1/2 证明cos【(4n+1)π/4+x】-cos[(4n-1)π/4-x]=0(n属于z） cos(π/17)*cos(2π/17)*cos(4π/17)*cos(8π/17) 证明cos(π/(2n+1))cos(2π/(2n+1))cos(3π/(2n+1))……cos(nπ/(2n+1))=1/2^n 证明∫cos^nxsin^nxdx＝(1/2^n)∫＜0,π/2＞cos^nxdx n为整数 化简cos[(4n+1)π/4+x]+cos[(4n-1)π/4+x] cos(π/32)*cos(π/16)*cos(π/8)= 证明 Σcos((k/n)π)=0;k=0,1,2,...2n-1 为什么cos(n-1)π=cos(n+1)πcos(n-1)π=cos(n-1+2)π=cos(n+1)π 对吗? cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4 cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4 COS(π/17)COS(2π/17)COS(4π/17)COS(8π/17)要过程 化简cos{[(4n+1)π/4]+α}+cos{[(4n-1)π/4]-α},（n∈Z） 请教,化简： cos(4/4n+1 π +a)+cos(4/4n-1 π -a) n E Z cos(4/4n+1 π +a)+cos(4/4n-1 π -a) n E Z 高二推理证明证明等式2cosπ/4=√2,2cosπ/8=√(2+√2),2cosπ/16=√(2+√(2+√2))…并从中归纳出一般结论 求SINπ/64 COSπ/64 COSπ/32 COSπ/16 COSπ/8