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Warning: new definition for norm
Warning: new definition for trace
##### Problem

Show that if u and v are vectors then
|v| u + |u| v

bisects the angle between u and v.

First notice that we only need to show that (u+v)
bisects the angle between u and v when |u|=|v|=1.
The general case will follow from here, for if the result is
true for vectors of length 1 then,

(u/|u|) + (v/|v|)

will bisect the angle between u/|u| and v/|v| which
is the same as the angle between u and v (we are only changing
their lengths). Also, if a vector b bisects an angle then so does
any scalar multiple of it. Thus, multiplying by (|u|*|v|)
we obtain the original result.
OK but we still need to show that u+v does the trick.
That is not difficult to see. We loose no generality if
we choose a coordinate system where:

i=u, and j on the plane containing u and v

> u := vector([1,0]); v := vector([cos(t),sin(t)]);
u := [ 1, 0 ]
v := [ cos(t), sin(t) ]
#The vector v forms an angle of "t" with u,
> angle(u,u+v);
1 + cos(t)
arccos(----------------------------)
2 2 1/2
((1 + cos(t)) + sin(t) )
# so this is the angle between u and u+v... is this t/2 ?
First, simplify with trig,
> ang := simplify(",trig);
1 + cos(t)
ang := arccos(-----------------)
1/2
(2 + 2 cos(t))
#we are getting there... but not yet. Now let's use
the formula for cos(t) in terms of cos(t/2).
Let's make s=t/2 and simplify.
ang := subs(t=2*s,ang);
1 + cos(2 s)
ang := arccos(-------------------)
1/2
(2 + 2 cos(2 s))
> ang := arccos(expand(cos(ang)));
1/2 2
4 cos(s)
ang := arccos(1/2 ------------)
2 1/2
(cos(s) )
> simplify(");
arccos(csgn(cos(s)) cos(s))
#The inside of arccos is nothing but |cos(s)| and since
we are assuming that 0 < s < Pi/2, the above is just s.

Q.E.D.
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