[2sin(π/12)sin(nπ/6)]/2sin(π/12)={cos[(2n-1)π/12]-cos[(2n+1)
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
数列an=n^2((cos(nπ/3))^2-(sin(nπ/3))^2)
已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n
sin(n兀+4兀/3),cos(2n兀+兀/6),sin(2n兀+兀/3),cos((2n+1)兀-兀/6),sin(
计算sin(2nπ/3+π/6)+cos(2nπ/3+π/6)
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/
sinπ/24*cosπ/24*sinπ/12=—— 答案是这么写的:1/2(2sinπ/24*cosπ/24)cosπ
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
数列an满足a1=1,a2=2,a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)],n=1.2.
已知向量m=(cosθ,-sinθ),n=(根号2+sinθ,cosθ),θ∈(π,3π/2),且cos(θ/2+π/8