化简:(1)sin[α+(2n+1)π]•2sin[α−(2n+1)π]sin(α−2nπ)cos(2nπ−α)(n∈Z
已知cos(π+α)=1/2,计算sin(2π-α) sin[(2n+1)π+α]+sin[α-(2n+1)π]/sin
设f(x)=cos^(nπ+x).sin^(nπ-x)/cos^[(2n+1)π-x](n∈z)求f(π/6)的值
紧急:求 lim n*sin(π(n^2+2)^0.5)*(-1)^n,n趋向无穷大;
级数收敛性之sin(1/n)>(2/π)×(1/n)
级数(1/n) × sin(πn/2)的敛散性
n→无穷大 sin^n(2nπ/3n+1)的极限怎么求解
求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷
计算极限lim(n→∞){1+ sin[π√(2+4*n^2)]}^n
已知向量m=(1,cosα),n=(-1,sinα-1/2),且m//n,α∈[0,π],则sinαcosα=
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/
数列an=n^2((cos(nπ/3))^2-(sin(nπ/3))^2)
证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1)