cos(pi/4-a)=1/7,sin(3pi/4+b)=11/14
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
a*sin(pi/4 - x) - b*cos(pi/4 - x) = a*sin(pi/4 + x) - b*cos(
第一题 cos(a+b)=4/5.cos(a-b)= -4/5 .a+b∈[7pi/4,2pi].a-b∈[3pi/4,
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),
三角形ABC中sin(2Pi-A)=-根号2cos(3Pi/2+B)根号3cos(2Pi-A)=根号2sin(Pi/2+
tana=2求sin(pi-a)cos(2pi-a)sin(-a+3pi/2)/tan(-a-pi)sin(-pi-a)
若cos(pi/6-a)=1/2,sin(a+pi/3)=?cos(2pi/3+2a)=?注:pi指圆周率
已知函数f(x)=cos(2x-pi/3)+2sin(x-pi)*sin(x+pi/4)
sin(pi/6+a)=1/3 则cos(pi/3-a)是多少
T=[-1 1] q=cos(pi/8*(cos(pi*t/T)).^3-3pi/8*cos(pi*t/T))+pi/4
f(a)=sin(pi-a)cos(2pi-a)tan(-a+3pi/2)/cos(-pi-a) 求 f(-31pi/3