a*sin(pi/4 - x) - b*cos(pi/4 - x) = a*sin(pi/4 + x) - b*cos(
a*sin(pi/4 - x) - b*cos(pi/4 - x) = a*sin(pi/4 + x) - b*cos(
a,b∈(3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)=12/13,则cos(a+pi/4)=
已知a,b,属于(3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)=12/13,则cos(a+pi
已知函数f(x)=cos(2x-pi/3)+2sin(x-pi)*sin(x+pi/4)
x=0:pi/100:6*pi;b=(cos(x)+i*sin(x))*(x);
sin x + cos x = √2 * sin (x + Pi/4)
已知cos(x)=sqrt(10)/10,x属于(0,pi/2),求sin(pi/4+2x)的值
sin(PI()X)+cos(pi()x)的周期
三角形ABC中sin(2Pi-A)=-根号2cos(3Pi/2+B)根号3cos(2Pi-A)=根号2sin(Pi/2+
是否存在a属于(-pi/2,pi/2),b属于(0,pi),使等式sin(3Pi-a)=根号2cos(pi/2)-b),
为什么y=2sin(3x-3/4 *pi) 可以写成y=-cos(3x+3/4 *pi)
x=4*sin(2*pi*0.01*t).*sin(2*pi*3*t)+2*cos(pi*t*t/4);在matlab中