sin(x-pi)cos(x-pi)=0.25,sin2x

sin(x-pi)cos(x-pi)=0.25,sin2x Matlab u(x,t)=sin(5*pi*x)cos(5*pi*t)+2sin(7*pi*x)cos(7*pi*t) 其中0 matlab rotate程序求大神修改t=0:pi/1000:3.5*pi;k=fix(2*t/pi)+1;x=0.5*sqrt(2)*cos(0.5*k*pi+0.25*pi)+k.*cos(t);y=0.5*sqrt(2)*sin(0.5*k*pi+0.25*pi)+k.*sin(t);h=plot(x,y,'r');m=pi:pi/20:4.5*pi;n=fix(2*m/pi)-1;x0=2*cos(1.25*pi);y0=2*sin(1.25*pi);x2= sin(PI()X)+cos(pi()x)的周期 化简sin(2x-pi)sin[x+(3/2)pi]/cos[x+(5/2)pi][1-sin(pi+x)] 已知cos(x)=sqrt(10)/10,x属于(0,pi/2),求sin(pi/4+2x)的值 函数f(x)=sin^2(x+pi/12)+cos^2(x-pi/12)的最大值 f(x)=sin(Pi/4-x)+cos(Pi/4+x)的周期 f(2x)=sin (x+3*pi/4)+cos (x-pi/2), 2(sin^4 x + cos^4 x) =1 [ - pi ≤ x ≤ pi ] f(x)h(x)=√2/2sin(x+pi/4)cos(x+5pi/4) =-√2/2sin(x+pi/4)cos(x+pi/4)√2/2sin(x+pi/4)cos(x+5pi/4) =-√2/2sin(x+pi/4)cos(x+pi/4)这个是怎么化出来的? sin（pi/2+x）=?如题哈：（1）sin（pi/2+x）=cos(x) ------> （2）cos(pi/2+x)=-sin(x) ----------> 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 函数y=sin(pi/2+x)cos(pi/6+x)的最大值 sin(2x+pi/3)=cos(2x-pi/6) 成立吗 y=sin(2x+Pi/3)+cos(Pi/6-2x)周期为 用MATLAB画图,例>> x=0:pi/15:2*pi; >> y1=sin(x)； >> y2=cos(x)； >> plot(x,y1,x,y2)这个 x=0:pi/15:2*pi f（x)=2sin(pi*x+pi/6),若f(a/2pi)=1/2,求cos (2*pi/3-a)的值