sin pi/12-根号3cos pi/12=
三角形ABC中sin(2Pi-A)=-根号2cos(3Pi/2+B)根号3cos(2Pi-A)=根号2sin(Pi/2+
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
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),
sin(pi/12)=?
cos(a-pi/3)=12/13,pi/3
函数f(x)=sin^2(x+pi/12)+cos^2(x-pi/12)的最大值
化简sin(x+PI/3)+2sin(x-PI/3)-根号3*cos(2pi/3-1)
化简:sin(2pi-a)sin(pi+a)cos(-pi-a)/sin(3pi-a)cos(pi-a)
tana=2求sin(pi-a)cos(2pi-a)sin(-a+3pi/2)/tan(-a-pi)sin(-pi-a)
f(a)=sin(pi-a)cos(2pi-a)tan(-a+3pi/2)/cos(-pi-a) 求 f(-31pi/3
cos(-pi/3)=?