sin{[(4n-1)/4]π-a}·cos{{(4n+1)/4}π-a} 化简

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{[(4n-1)/4]π-a}+cos{{(4n+1)/4}π-a} 化简sin[(4n-1)π/2-a]+cos[(4n+1)π/2-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 若f（n）=sin(¼nπ+a),求证f(n).f(n+4)+f(n+2).f(n+6)=-1 与cos13π/3的值相同的是 ：A.nπ/3 B.sinπ/4 C.sinπ/6 已知A+b+C=π 求证 cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2) 化简sin×[a+(2n+1)π]+2sin×[a-(2n+1)π]/sin(a-2nπ）coS(2nπ-a) (n属于Z) matlab 错误Error using zy (line 4) Not enough input arguments.function f=zy(p,n,m,t)int n m t;syms a c;if(p>0|p3|2*t*sqrt(1/(n*tan(pi/n)))c/sqrt(a*c)&2*m*sqrt(2/(n*sin(2*pi/n)))==c/sqrt(a*c)|m*t 化简( cos(4n-1/4π+x)·sin(4n+1/4π-x)(n∈Z）化简( cos(4n-1/4π+x)·sin(4n+1/4π-x)(n∈Z） 已知△三个内角A、B、C,向量m=(4,1),n=(sin²A/2,cos2A),且m·n=1/2求角A大小4sin²A/2 +cos2A=1/2.4cos²A-8cosA+5=0我哪里算错了? 证明cosA＋cosB＋cosC=1+4sin(A/2)sin(B/2)sin(C/2)证明：cosA＋cosB＋cosC=1+4sin(A/2)sin(B/2)sin(C/2)尽量详细一点选做cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2) cos(180-B-C)+cosB+cosC=1+2sin(A/2)[2sin(B/2)sin(C/2)] cos(180-B-C)+cosB+cosC 如何用for循环编写最好是c,c#语言n=1 时y=sin(1)*x1n=2 y=sin(1/2)*x1+sin(2/2)*x2n=3 y=sin(1/3)*x1+sin(2/3)*x2+sin(3/3)*x3;n=4 y=sin(1/4)*x1+sin(2/4)*x2+sin(3/4)*x3+sin(4/4)*x4;...n=n y=sin(1/n)*x1+sin(2/n)*x2+.sin(n/n)*xn;x1-xn的值是分 二分法的问题#include #include #include #define MIN 0.001double f (double a,double b){double c;c = (sin(a) - a*a/4) * (sin(b) - b*b/4);printf ( c is :%lf ,c);return c;}double erfen( double a ,double b){double n;n = a - b;printf ( n is :%lf 设向量m=（sinα,cosα-（1/2）y）,n=（-2,sinα）,若m//n,则y的最大值为A.2 B.1 C.0 D.4