cos(x+1)½-cosx½→-2sin{[(x+1)½+x½]/2}*sin{[(x+1)½-x½]}如何转化

cos(x+1)½-cosx½→-2sin{[(x+1)½+x½]/2}*sin{[(x+1)½-x½]}如何转化
cos(x+1)½-cosx½→-2sin{[(x+1)½+x½]/2}*sin{[(x+1)½-x½]}如何转化

cos(x+1)½-cosx½→-2sin{[(x+1)½+x½]/2}*sin{[(x+1)½-x½]}如何转化
a=(x+1)½, b=x½

cos(x+1)½=cosa=1-2sin^2(a/2)
cosx½=cosb=1-2sin^2(b/2)
cos(x+1)½-cosx½=1-2sin^2(a/2)-[1-2sin^2(b/2)]
=2sin^2(b/2)-2sin^2(a/2)
=2[sin(b/2)+sin(a/2)][sin(b/2)-sin(a/2)]
=2{sin(x½/2)+sin[(x+1)½/2]}*[sin(x½/2)-sin[(x+1)½/2]}
=-2{sin(x½/2)+sin[(x+1)½/2]}*{[sin[(x+1)½/2]-sin(x½/2)}