y=a^sin(2x-pi/4)（a>0,且a≠1）是否为周期函数?是,求周期,不是,说明理由.求单调增区间,最大值

已知x属于[0,pi],则函数y=2sin(pi/6-2x)为增函数的区间是a.[0,pi/3] b.[5pi/6,pi] y=sin(2x+a)+(√3)cos(2x+a)为奇函数,且在[0,pi/4]上是减函数的a的一个值为? MATLAB编程 求大侠改错a=-pi:pi/50:pi;b=pi:pi/50:4*pi;x=1:0.1:8;subplot(1,3,1),plot(a,a.*cos(a))xlabel('x'),ylabel('y'),title('y=xcosx')subplot(1,3,2),plot(b,b.*tan(1/b).*sin(b.^3))gtext('y=xtanxsinx '),axis([3 13 0 100])subplot(1,3,3),plot(x,ex 已知f(x)连续可导,证明g((x,y),(a,b))亦连续.已知f(x)在(-pi/2,pi/2)上连续可导,定义g(x,y)在集合E:=(-pi/2,pi/2)*(-pi/2,pi/2),g(x,y)=[f(x)-f(y)]/[sin(x)-sin(y)],证明g(x,y)在E上连续. sin(PI/2+a)=-4/5 a属于(Pi,3pi/2)则cos(Pi/3-a)为 y=3*sin(3*x-pi/2) 和 y= -3*sin(3*x+pi/2) 一样么?f(x)=Asin(ωx+θ)的图像与x轴相交的两个邻点坐标为(pi/6,0),(pi/2,0),且过(0,-3),则f(x)为?A.y=3*sin(3*x+pi/2)B.y=3*sin(3*x-pi/2)C.y= -3*sin(3*x+pi/2)本人用函数作图器得y=3*sin sin(pi/4+2a)sin(pi/3-2a)=1/4,a属于(pi/4,pi/2),求2sin^2a+tana-cota-1 急!matlab,求大神解答x=solve('9530*(sin(pi*x)^3)/pi+14300*x*(1-sin(2*pi*x)/(2*pi*x))/(1.25-3*x)*(sin(pi*x)+sin(pi*(1.25-2*x))/pi-983.04=0','x')想求解这个含三角函数的一元三次方程,但显示Error using ==> solve is not a valid expr a*sin(pi/4 - x) - b*cos(pi/4 - x) = a*sin(pi/4 + x) - b*cos(pi/4 + x) 怎么得到（a + b）* sinx = 0 将函数y=sin(2x-pi/3)的图形按向量a平移后所得的图像关于点(-pi/12,0)中心对称 matlab surf画三维图的问题n=4;c=3e8;f=30e6;lamda=c/f;beta=2*pi/lamda;d=0.5*lamda;alpha=0;a=linspace(0,2*pi);b=linspace(0,pi);psi=beta*d*cos(a)+alpha;Fa=abs(sin(n*psi/2)./(n*sin(psi/2)));z=Fa.*cos(a)；x=(Fa.*sin(a)).*cos(b);y=(Fa.*sin(a)).*sin( sin(pi/4+2a)sin(pi/4-2a)=1/4,a(pi/4,pi/3),q:2sin2 a+tana-cota-1 cos(pi/4-a)=12/13,pi/4-a是第一象限角,则sin(pi/2-2a)/sin(pi/4+a)的值是 cosa=2/3 a是第四象限角求sin(a-2Pi)+sin(-a-3Pi)cos(a-3Pi)/cos(Pi-a)-cos(-Pi-a)cos(a-4Pi) t1=input('t1=') w=50/3*pi a=2/3*pi for t=0:0.0001:t1 b=2*w*(pi-a)*t/pi c=cos(2/3*pi+b) d=sin(2/3*pit1=input('t1=')w=50/3*pia=2/3*pi for t=0:0.0001:t1b=2*w*(pi-a)*t/pic=cos(2/3*pi+b)d=sin(2/3*pi)h=cos(2/3*pi)[x,y,z]=solve('x^2+y^2+z^2=100^2+170^2','x^ tanθ=a,求[sin(pi/4+θ)/sin(pi/2-θ)]*tan2θ tanθ=a,求[sin(pi/4+θ)/sin(pi/2-θ)]*tan2θ, 英语翻译n = 12;a = .2; %小筒的直径c = .6; %灯泡的直径t1 = pi/4 :pi/n :5*pi/4; %沿管参数t2 = 5*pi/4 :pi/n :9*pi/4; %角管周长u = pi/2 :pi/n :5*pi/2;[X,Z1] = meshgrid(t1,u);[Y,Z2] = meshgrid(t2,u);%手柄len = sqrt(sin(X).^2 + cos(