化简:sqr[1/2-1/2*sqr(1/2+1/2*cos2a)] a属于(3π/2,2π)

问题描述:

化简:sqr[1/2-1/2*sqr(1/2+1/2*cos2a)] a属于(3π/2,2π)

∵a属于(3π/2,2π) ,a/2属于(3π/4,π)
∴ cosa>0,sin(a/2)>0
sqr[1/2-1/2*sqr(1/2+1/2*cos2a)]
=sqr[1/2-1/2*sqr(cosa)^2]
=sqr[1/2-1/2*(cosa)]
=sqr[sin(a/2)^2]
=sin(a/2)