Trigonometric Proofs

Date: 09/13/2004 at 03:57:39
From: Ippy
Subject: Trigonometric Proofs

How do I prove that (sinA - sinB)/(cos A + cosB) = tan(A/2 - B/2)?

How can I obtain: 1) sinA - sinB = 2sin(A/2 - B/2)cos(A/2 + B/2)
       and        2) cosA + cosB = 2cos(A/2 + B/2)cos(A/2 - B/2)

using standard trigonometric proofs?

Date: 09/13/2004 at 04:33:36
From: Doctor Luis
Subject: Re: Trigonometric Proofs

Hi Ippy,

You can come up with (1) and (2) from the sum and difference 
identities by using a clever substitution:  

  A = u + v, B = u - v

Then,

    sin(A) - sin(B)
  = sin(u + v) - sin(u - v)
  = [sin(u)cos(v) + cos(u)sin(v)] - [sin(u)cos(v) - cos(u)sin(v)]
  =  sin(u)cos(v) + cos(u)sin(v) - sin(u)cos(v) + cos(u)sin(v)
  = 2 cos(u)sin(v)

So now we need to know u and v in terms of A and B. Fortunately,
that's easy

  A + B = (u + v) + (u - v) = 2u  ==>  u = (A + B)/2
  A - B = (u + v) - (u - v) = 2v  ==>  v = (A - B)/2

To summarize,

  sin(A) - sin(B) = 2 cos((A + B)/2) sin((A - B)/2)

which is exactly (1).

Using this method, you should be able to prove (2).  Once you have 
these two results, you can prove your original identity.

Please write back if you have further questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/