化简sin×[a+(2n+1)π]+2sin×[a-(2n+1)π]/sin(a-2nπ）coS(2nπ-a) (n属于Z)

化简sin×[a+(2n+1)π]+2sin×[a-(2n+1)π]/sin(a-2nπ）coS(2nπ-a) (n属于Z) 化简sin[(4n-1)π/2-a]+cos[(4n+1)π/2-a] lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/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/2), 懂得进.证明,sina+sin(a+b)+sin(a+2b)+...sin(a+nb)=sin(a+ab/2)sin[(n+1)b/2]/sin(b/2) 化简sin(a+nπ)+sin(a+nπ)/sin(a+nπ)cos(a-nπ)(n∈z) 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, 化简 【sin(a+nπ)+sin(a-nπ)】/【sin(a+nπ)cos(a-nπ)】 化简:sin(nπ+a)cos(nπ-a)/cos[(n+1)π-a] 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 化简sin(a+nπ)+sin(a-nπ)/sin(a+nπ)(cosa-nπ)要步骤, 若f（n）=sin(¼nπ+a),求证f(n).f(n+4)+f(n+2).f(n+6)=-1 C语言编程找错误:计算 sin(x)=x - x3/3!+ x5/5!- x7/7!+ ……直到最后一项的绝对值小于10-6.#includemath.hmain(){float sin,x,a,b,t;int s,n;scanf(%f,&x);sin=0,a=x;n=1,b=1;t=a/b;do{sin=sin+s*t;a=a*x*x;b=b*(n+1)*(n+2);s=-s;t=a/b;n=n 求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2] 【1】化简：sin(a-5π）/cos(3π-a）×cos(π/2-a）/sin(a-3π）×cos(8π-a）/sin(-a-4π）【2】已知f（x)=sin(nπ-x）cos(nx+x)/cos[(n+1)π-x]×tan(x-nπ）cot（nπ/2+x)（n∈Z）,求f（7/6π） 化简:sin(-a)cos(2π+a)sin(-a-π) 求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷 级数(1/n) × sin(πn/2)的敛散性