Law of Cosines in Informal Trigonometric Proof

Date: 12/22/2004 at 22:01:20
From: andy
Subject: determining if triangle ABC is either isosceles or right

I'm in trig in high school and came across this problem involving 
cosine:

If cosA/cosB = b/a, then show that triangle ABC is either isosceles 
or right.

I drew a right triangle with the right angle at C.  Then I labeled
sides a,b,c opposite the repective angle letters; thus, side c is the
hypotenuse.  Now, using this right triangle, I come up with:

  cosA = b/c
  cosB = a/c

Since they say cosA/cosB, I substituted cosA and cosB with b/c and a/c
to get (b/c)/(a/c) which simplifies to b/a.  

It seems like that proves that cosA/cosB = b/a using a right triangle.
 However, I feel like I haven't answered the question completely and
I'm not sure what to do with the isosceles triangle.  Can you tell me
what I'm missing?

Date: 12/23/2004 at 04:05:59
From: Doctor Floor
Subject: Re: determining if triangle ABC is either isosceles or right

Hi Andy,

Thanks for your question. We should use the Law of Cosines:

  a^2 = b^2 + c^2 - 2bc cosA

or, after some rewriting.

  cosA = (-a^2 + b^2 + c^2)/(2bc)

Similarly

  cosB = (a^2 - b^2 + c^2)/(2ac)

This gives

  cosA   a (-a^2 + b^2 + c^2)
  ---- = --------------------
  cosB   b (a^2 - b^2 + c^2)

Now cosA/cosB = b/a is equivalent to

  b   a (-a^2 + b^2 + c^2)
  - = --------------------
  a   b (a^2 - b^2 + c^2)

and cross-multiplying we get

 a^2(-a^2 + b^2 + c^2) = b^2(a^2 - b^2 + c^2).

Now, can you finish by further solving this equation?  If you have
more questions, just write back.

Best regards,

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

Date: 12/23/2004 at 16:12:32
From: andy
Subject: determining if triangle ABC is either isosceles or right

Dear Dr. Floor,

Thank you for the help so far.  I've followed through every step on 
my own piece of paper and it seems like if I prove that a = b, then 
this equation works for an isosceles triangle, but I'm not really 
sure.  

OK, here's my work from the point you left off which was

  a^2(-a^2 + b^2 + c^2) = b^2(a^2 - b^2 + c^2)

  -a^4 + a^2b^2 + a^2c^2 = a^2b^2 - b^4 + c^2b^2 (distributed)

  -a^4 + a^2c^2 = -b^4 + c^2b^2 (cancel a^2b^2 on both sides)

  b^4 - c^2b^2 + a^2c^2 - a^4 = 0 (bring all to one side)..

However, first I'm not sure exactly what I'm trying to actually solve, 
and second, how, when I have solved it, I will prove that this
triangle is isosceles or right.  I hope you can help me a little bit
more.  Thanks a lot!

Date: 12/27/2005 at 03:08:02
From: Doctor Floor
Subject: Re: determining if triangle ABC is either isosceles or right

Hi Andy,

Your work so far looks good.  You have now found an equation that is
quadratic with regard to b^2 (try replacing b^2 with x if you're not
sure what I mean):

    b^4  -  c^2b^2  + a^2c^2 - a^4 = 0
 (b^2)^2 - c^2(b^2) + a^2c^2 - a^4 = 0
  (x)^2  -  c^2(x)  + a^2c^2 - a^4 = 0

This you can solve for b^2 with the quadratic formula, where in the
formula a = 1, b = -(c^2), and c = a^2c^2 - a^4.  Note that those are
not the same a,b,c you are working with from your triangle, those are
the a,b,c that are traditionally used in the quadratic formula,
defined in terms of the a,b,c from the triangle.  That gives you 

  b^2 = .... or b^2 = ....  (because of the +/- in the formula)

Hopefully you will find that b^2 = a^2 (isosceles triangle) or 
b^2 = c^2 - a^2 (right triangle).  Good luck!

Best regards,

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