Trig Proof - Newton's Formula

Date: 04/09/98 at 23:37:14
From: Jay
Subject: Newton's Formula

The problem I've been having trouble with is:

     (a - b)    sin [(1/2)(A - B)] 
     ------- = -------------------
        c        cos [(1/2)(C)]

I've tried desperately to solve this problem analytically but it
just turned into a mess. Is it possible that you (Dr. Math) could tell 
me an easier way to prove this formula (i.e. graphically)? If so, 
please let me know.

Jay

Date: 04/10/98 at 13:15:23
From: Doctor Anthony
Subject: Re: Newton's Formula

Start with the Sine formula:

     sin(A)   sin(B)   sin(C)
     ------ = ------ = ------
       a        b        c

From the note below, we find that:

     sin(A)-sin(B)    sin(C)
     ------------- = -------     
          a-b           c

Thus:

     2cos((A+B)/2)*sin((A-B)/2)       sin(C)
     --------------------------   =  -------         (Equation 1)
              a - b                     c


Since (A+B+C)/2 = 90:

     (A+B)/2 = 90 - C/2

So:

     cos((A+B)/2) = cos(90 - C/2) =  sin(C/2)        (Equation 2)

Substituting Equation B into Equation A, we find that:

     2sin(C/2)*sin((A-B)/2)   2sin(C/2)cos(C/2)
     ---------------------- = ----------------
            a - b                    c


      sin((A-B)/2)     cos(C/2)
      ------------  =  --------
         a - b            c


     sin((A-B)/2)     a - b
     ------------  =  -----       
       cos(C/2)         c
          
as required.

NOTE
----

If: 

      a       c
     ---  =  ----  = k      
      b       d

then:

      a = bk    and    c = dk


So:

      a-c     bk - dk     k(b-d)         a     c
     ----- = --------- = -------- = k = --- = ---
      b-d       b-d        b-d           b     d

-Doctor Anthony,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/