化简根号下[1-cos(a-pi)]\2
化简根号下[1-cos(a-pi)]\2
sin a +cos a = (根号下2)sin(a+pi/4)如何化简?
三角形ABC中sin(2Pi-A)=-根号2cos(3Pi/2+B)根号3cos(2Pi-A)=根号2sin(Pi/2+
化简三角函数[sin(α+pi)*cos(pi+α)*cos(α+2pi)]/[tan(pi+α)*cos^3(-α-p
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),
化简根号下(1)1-cos^2 240 (2)tana分之根号下1-cos^2 a(a为第二象限角)
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)
若cos(pi/6-a)=1/2,sin(a+pi/3)=?cos(2pi/3+2a)=?注:pi指圆周率
sina=5/4,那么sin(a+4/pi)-二/根号2•cos(pi-a)=
化简根号下[(1-cos x)/(1+cos x)] + 根号下[(1+cos x)/(1-cos x)]
设tan2a=-2根号2,2a在180度与360度之间,求[2cos^2(a\2)-sina-1]/根号2(a+pi/4