化简sin(nπ+2π/3)*cos(nπ+4π/3)

数列an=n^2((cos(nπ/3))^2-(sin(nπ/3))^2) 计算sin(2nπ/3+π/6)+cos(2nπ/3+π/6) 已知向量M=（2sinπ/4,cosπ/2）n=（cosπ/4,根号3）函数f（X）=m·n 已知角α终边上一点P(-4a,3a)(a不等于0),求[cos(nπ+a)+sin(nπ-a)]/[cos(nπ-a)- 化简：sin(nπ+a)/cos(nπ-a)（n属于Z） 已知向量m=(cosθ,-sinθ),n=(根号2+sinθ,cosθ),θ∈(π,3π/2),且cos(θ/2+π/8 求极限(sin(2/n)+cos(3/n))^(-n) n→无穷大 sin^n(2nπ/3n+1)的极限怎么求解 已知数列｛an｝的通项公式为an=2n(cos^2nπ/3-sin^2nπ/3),求a1+a2+…+a100 已知向量M=(sin(x+π/4),-根号3cos(x+π/4)) 向量N=(sin(x+π/4),cos(x-π/4) 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 已知f(n)=sin[(n+1\2)π+π\4]+cos[(n-1\2)+π\4]+tan[(n+1)π+π\4],求f