Trigonometry  Identities and proof
posted by Sam on .
Show that cot((x+y)/2) =  (sin x  sin y)/(cos x  cos y) for all values of x and y for which both sides are defined.
I tried manipulating both sides in terms of trig identities but I don't really have a solution....help would be appreciated, thanks.

Using the sumtoproduct formulas,
sinxsiny = 2cos((x+y)/2)sin((xy)/2)
cosxcosy = 2sin((x+y)/2)sin((xy)/2)
now just divide to get
cos((x+y)/2) / sin((x+y)/2) = cot((x+y)/2) 
Two of the standard conversion formulas are:
sinA  sinB = 2sin((AB)/2) cos( (A+B)/2)
and
cosA  cosB =  2sin( (AB)/2) sin( (A+B)/2)
see:
near bottom of page under:
SumtoProduct Formulas
http://www.sosmath.com/trig/Trig5/trig5/trig5.html
RS
= (sinx  siny)/(cosx  cosy)
=  2sin((XY)/2) cos( (X+Y)/2)/ 2sin( (XY)/2) sin( (X+Y)/2)
= cos((X+Y)/2) / sin(X+Y)/2)
= cot((X+Y)/2)
= LS