Trig Proof: Two Methods

Date: 03/25/98 at 23:00:34
From: Seven of Nine Tertiary Adjunct of Unimatrix Zero One
Subject: Trigonometry

Dear Dr. Math,

I am a grade 11 student living in Toronto. I've been given an equation 
and have to prove that everything to the left of the equation is equal 
to what's on the right of the equation. I just keep going in circles 
and never seem to get anywhere. Could you explain it to me please?

This is the question:  (the ------- is the fraction line).

Prove:

      1 + sin x + cos x       1 + sin x
     -------------------  =  -----------
      1 - sin x + cos x         cos x


Thank you very much.

-J.W.

Date: 03/26/98 at 19:45:47
From: Doctor Wilkinson
Subject: Re: trigonometry

This is equivalent to showing that:

     (cos x) (1 + sin x + cos x) = (1 + sin x) (1 - sin x + cos x)

The left side is:

     cos x + (sin x)(cos x) + cos^2 x

and the right side is:

     (1 - sin^2 x) + cos x + (sin x) (cos x)

but:  

     sin^2 x + cos^2 x = 1, so

     1 - sin^2 x = cos^2 x

and from this it is easy to see that the two sides are equal.
 
-Doctor Wilkinson,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   

Date: 03/26/98 at 20:43:37
From: Seven of Nine Tertiary Adjunct of Unimatrix Zero One
Subject: trigonometry proof

Dear Dr. Math,

My teacher told my class that in proving, she does not want us to 
cross-multiply. She wants us to treat the lefthand side of the 
equation by itself and work with just the left side to derive the 
righthand side.  Is there any way to do that?

Thank you again.

-J.W.

Date: 03/27/98 at 20:49:26
From: Doctor Pete
Subject: Re: trigonometry proof

Hi,

Well, if you only have:

     1 + Sin[x] + Cos[x]
     -------------------
     1 - Sin[x] + Cos[x]

and are told to simplify, here's what to try: 

First, I performed long division on it, so I got:

               2 Sin[x]
     1 + -------------------
         1 - Sin[x] + Cos[x]

But then I didn't know what to do with the remainder.  So I started 
over, and recognized that you could group the terms like this:

     (1+Cos[x]) + Sin[x]
     -------------------
     (1+Cos[x]) - Sin[x].

So the numerator is of a form like A+B, and the denominator is like 
A-B. We multiply the numerator and denominator by A+B, so that the 
denominator becomes a difference of squares, A^2 - B^2.  We get:

     ((1+Cos[x]) + Sin[x])^2   
     ----------------------- 
     (1+Cos[x])^2 - Sin[x]^2       

              (1+Cos[x])^2 + 2(1+Cos[x])Sin[x] + Sin[x]^2
            = -------------------------------------------
                 1 + 2 Cos[x] + Cos[x]^2 - Sin[x]^2


which is kind of messy, until you notice the trig identity:

     1 - Sin[x]^2 = Cos[x]^2

and complete expanding the numerator.  This gives:

     1 + 2 Cos[x] + Cos[x]^2 + 2 Sin[x] + 2 Sin[x]Cos[x] + Sin[x]^2
     --------------------------------------------------------------
                        2 Cos[x] + 2 Cos[x]^2

            2 + 2(Cos[x] + Sin[x] + Sin[x]Cos[x])
          = -------------------------------------
                    2 Cos[x](1 + Cos[x])

and when we cancel out the 2's, we get:

     1 + Cos[x] + Sin[x] + Sin[x]Cos[x]
     ---------------------------------- .
            Cos[x](1 + Cos[x])

But notice the numerator can be factored into (1+Sin[x])(1+Cos[x]), 
and so we can cancel out a common factor of 1+Cos[x] to give the final 
result:

     1 + Sin[x]
     ---------- .
       Cos[x]

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