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),

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), 化简sin×[a+(2n+1)π]+2sin×[a-(2n+1)π]/sin(a-2nπ）coS(2nπ-a) (n属于Z) 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) 求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2] lim n→∞ sin pi√(n^2+a^2) (a不等于0)自己想出来了sin(pi√(n^2+a^2)=(-1)^n sin(pi√(n^2+a^2)-npi)=(-1)^n*sin(pia^2/(√(n^2+a^2)+n))(-1)^n是有界函数 lim n→∞sin(pia^2/(√(n^2+a^2)+n))＝0所以sin(pi√(n^2+a^2)=0SNOWHORSE70121 两道三角函数题 求证sin(3a/3)+3sin(3a/3^2)+.+3^(n-2)sin(3a/3^n-1)=0.25*(3^nsin(a/3^n)-sina)同上.求证：sin(3a/3)+3sin(3a/3^2)+.+3^(n-2)sin(3a/3^n-1)=0.25*(3^nsin(a/3^n)-sina)求证：tan(a)+2tan(2a)+2^(n-1)tan(2^n-1)=cot(a)-2^(n)cot(2^(n) 化简 【sin(a+nπ)+sin(a-nπ)】/【sin(a+nπ)cos(a-nπ)】 化简sin[(4n-1)π/2-a]+cos[(4n+1)π/2-a] 化简sin(a+nπ)+sin(a-nπ)/sin(a+nπ)(cosa-nπ)要步骤, 化简:sin(nπ+a)cos(nπ-a)/cos[(n+1)π-a] (sin2a*sin4a*sin8a.sin(2^n)a)/(2sina*2sin2a*2sin4a.2sin(2^n-1)a)=sin(2^n)a/(2^n)sina这个式子左边为什么等于右边,是怎么来的, 求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷 lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=? 求极限（sin^2)n/n+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 若f(n)=sin(n派/4+a)求证f(n)*f(n+4)+f(n+2)*f(n+6)=-1 若f（n）=sin(¼nπ+a),求证f(n).f(n+4)+f(n+2).f(n+6)=-1