化简sin[(4n-1)π/2-a]+cos[(4n+1)π/2-a]

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] 化简sin(a+nπ)+sin(a+nπ)/sin(a+nπ)cos(a-nπ)(n∈z) 化简sin{[(4n-1)/4]π-a}+cos{{(4n+1)/4}π-a} sin{[(4n-1)/4]π-a}·cos{{(4n+1)/4}π-a} 化简 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, 在三角形ABC中,角A.B.C所对的边分别为a,b,c,已知向量m=(a,3b-c),n=(cosA,cosC),满足m平行n（1）求cosA的大小（2）求sin^2B+C/2-2sin(A-π/4)sin(A+π/4)的值 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] 已知n∈Z化简sin[(4n-1/4)π-a]+cos[(4n+1/4)π-a] 化简SIN((4N-1)/4 π-A)+COS((4N+1) π-A) N属于Z 化简sin(a+nπ)+sin(a-nπ)/sin(a+nπ)(cosa-nπ)要步骤, matlab解方程组方程组1：(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2-n*sin(c/2+d)+e*cos(f-b))^2-(h+n*cos(c/2+d)-e*sin(f-b))^2=0;方程组2：(m/2-n*sin(c/2)+e*cos(f))^2+(h+n*cos(c/2)-e*sin(f))^2-(m/2+e*cos(a+f)-n*sin(c/2-d))^2-(h 若f（n）=sin(¼nπ+a),求证f(n).f(n+4)+f(n+2).f(n+6)=-1 lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=? 【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） 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), 已知A+b+C=π 求证 cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2)