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### 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

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/
```
Associated Topics:
High School Trigonometry

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