tan(5+a)=-2,且cos>0,则sin(-pi+a)=?/
tan(5+a)=-2,且cos>0,则sin(-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
化简间sin(a-2pi)cos(a+pi)tan(a-99pi)cos(pi-a)sin(3pi-a)sin(-a-p
已知cosa=3/5,且a属于(3pi/2,2pi),则cos(a-pi/3)=
sin^2(-a)cos(2pi+a)tan(-a-pi)化简
sin^3(-a)cos(2pi+a)tan(-a-pi)化简
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
tanθ=a,求[sin(pi/4+θ)/sin(pi/2-θ)]*tan2θ
tanθ=a,求[sin(pi/4+θ)/sin(pi/2-θ)]*tan2θ,
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),