如何把 2sin(π/12)sin(nπ/6)化成 cos(2n-1)π/12 - cos(2n+1)π/12请尽量详细点,因为我连基本公式都几乎忘掉了,补补那些年忘记的三角函数!

lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=? sin(n*π/2)*sin(n*π/3)*sin(n*π/4)*...*sin(n*π/n-1) 求化简成一个关于n的表达式, 令n趋近于无穷大,且n存在,求sin(π/n)+sin(2π/n)+sin(3π/n)+...+sin(π)=?. 计算,sin(π/12)-sin(5π/12)+2sin(π/8)sin(3π/8) 当n趋于无穷时,求[sin(π/n)/(n+1)+sin(2π/n)/(n+1/2)+.sinπ/(n+1/n)]的极限 sin(π/n)的收敛性 n大于等于1,θ不是π的倍数,求证sinθ+sin3θ+...+sin((2n-1)θ)=sin^2(nθ)/sinθ [2sin(π/12)sin(nπ/6)]/2sin(π/12)={cos[(2n-1)π/12]-cos[(2n+1)π/12]}/2sin(π/12)我想知道这是怎么变成的? 如何把 2sin(π/12)sin(nπ/6)化成 cos(2n-1)π/12 - cos(2n+1)π/12请尽量详细点,因为我连基本公式都几乎忘掉了,补补那些年忘记的三角函数! 化简sin×[a+(2n+1)π]+2sin×[a-(2n+1)π]/sin(a-2nπ）coS(2nπ-a) (n属于Z) lim[sinπ/(√n＾2＋1)+sinπ/(√n^2+2)+．．．＋sinπ/√n^2+n),n—>无穷 sin(nπ+π/2n)条件收敛 级数(1/n) × sin(πn/2)的敛散性 级数(1/n) × sin(πn/2)的敛散性 求∑sin(n^2+1)π/n条件收敛 an=sin(nπ/2),n→无限,极限. 化简sin(a+nπ)+sin(a+nπ)/sin(a+nπ)cos(a-nπ)(n∈z) {sin(θ+nπ)+sin(θ-nπ)}/sin(nπ+θ)cos(θ-nπ) n∈Z 化简