## ProblemShow that if u and v are vectors thenbisects 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, 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: |

> 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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Carlos Rodriguez <carlos@math.albany.edu> Last modified: Wed Oct 23 11:39:01 EDT 1996