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( π/2n+1)+cos( 3π/2n+1)+cos( 5π/2n+1)+cos(7 π/2n+1)+...+cos[(2n-1) π/2n+1)]=1/2 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(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 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? 证明 Σcos((k/n)π)=0;k=0,1,2,...2n-1 证明cos 2x cos x=1/2(cos x+cos 3x) 用数列极限的ε-N定义证明证明lim 1/n*cos 2n=0cos 2n不在分母上。 (cos a)^2 + (cos b)^2 =1证明 sin(n+1)A+2sin(n)A+sin(n-1)A/cos(n-1)A-cos(n+1)A怎么证明等于cot(A/2), lim (n趋于无穷)π/n(cos^2 π/n + .+ cos^2 (n-1)π/n )=? 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 不等式证明1已知α、β、γ∈(0,π/2),且tanα+tanβ+tanγ＝3,求证：1/(cosαcosβ)+1/（cosβcosγ)+1/(cosγcosα)≥6. 1+cos(2θ)+cos(4θ)+cos(6θ)=4cosθcos(2θ)cos(3θ)证明相等. 如何证明2cosA*cos(kA)-cos[(k-1)A]=cos[(k+1)A] limn^2(1-cosπ/n),n趋向无穷大 证明cos【(4n+1)π/4+x】-cos[(4n-1)π/4-x]=0(n属于z）