三角形ABC中,4sin(3pi-A)sin^2(A/2+pi/4)-cos(pi-2A)=√3+1,求A
三角形ABC中sin(2Pi-A)=-根号2cos(3Pi/2+B)根号3cos(2Pi-A)=根号2sin(Pi/2+
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
cosa=2/3 a是第四象限角求sin(a-2Pi)+sin(-a-3Pi)cos(a-3Pi)/cos(Pi-a)-
化简间sin(a-2pi)cos(a+pi)tan(a-99pi)cos(pi-a)sin(3pi-a)sin(-a-p
若cos(pi/6-a)=1/2,sin(a+pi/3)=?cos(2pi/3+2a)=?注:pi指圆周率
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),
已知a∈(pi,3pi/2),cosa=-1/3,求sin(a-2pi/3)的值
tanθ=a,求[sin(pi/4+θ)/sin(pi/2-θ)]*tan2θ
tanθ=a,求[sin(pi/4+θ)/sin(pi/2-θ)]*tan2θ,
已知a是第一象限角,且cosa=3/5,那么[1+√2cos(2a-pi/4)]/sin(a+pi/2)=