如何把 2sin(π/12)sin(nπ/6)化成 cos(2n-1)π/12 - cos(2n+1)π/12

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 已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n 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) sin(n兀+4兀/3),cos(2n兀+兀/6),sin(2n兀+兀/3),cos((2n+1)兀-兀/6),sin( 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 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满足a1=1,a2=2,a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)],n=1.2. 求极限(sin(2/n)+cos(3/n))^(-n) 已知向量m=(cosθ,-sinθ),n=(根号2+sinθ,cosθ),θ∈(π,3π/2),且cos(θ/2+π/8