The answer i got from Yahoo'er FirstName is:
Let a be the angle for which cos(a)=-1/3
tan((1/2)arccos(-1/3))=tan(a/2)= sin(a)/(1+cos(a))=
sqr(1-cos^2(a))/(1+cos(a))
Since cos(a)=-1/3
sqr(1-cos^2(a))/(1+cos(a))= sqr(1-1/9)/(1-1/3)= sqr(8/9)/(2/3)= sqr(2)

but i dont quiet get how tan(a/2)= sin(a)/(1+cos(a))
come from and y is cos=-1/3 instead of arccos -1/3

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