化简三角函数:sin[(2n-1/2)π-α]+cos[(2n+1/2)π+α],n∈N
设f(x)=cos^(nπ+x).sin^(nπ-x)/cos^[(2n+1)π-x](n∈z)求f(π/6)的值
已知cos(π+α)=1/2,计算sin(2π-α) sin[(2n+1)π+α]+sin[α-(2n+1)π]/sin
求极限(sin(2/n)+cos(3/n))^(-n)
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/
已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n
紧急:求 lim n*sin(π(n^2+2)^0.5)*(-1)^n,n趋向无穷大;
n→无穷大 sin^n(2nπ/3n+1)的极限怎么求解
数列an=n^2((cos(nπ/3))^2-(sin(nπ/3))^2)
2^n/n*(n+1)
求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
已知f(n)=sin[(n+1/2)π+π/4]+cos[(n-1/2)π+π/4]+tan[(n+1)π+π/4].求