Proof of a Trig Identity

Date: 03/13/98 at 20:03:02
From: Chris Hebert
Subject: Proving Trigonomic Identies

How can I prove that

      1 + sinX + cosX      1 + cosX
    ------------------- = -----------   
      1 + sinX - cosX        sinX

I have tried solving this problem using

     Basic Trigonometric Identities,
     Pythagorean Identities,
     Double-Angle Formulas, and
     Half-Angle Formulas.

This problem has stumped the entire Pre-Calculus class, including my 
teacher! Any help would be appreciated. Even if this problem stumps 
you, will you please E-mail and tell me so, or let me know any 
progress you have made? (I do know that this problem is solvable 
though, because it was published on a work sheet).

Date: 03/16/98 at 17:14:13
From: Doctor Fred
Subject: Re: Proving Trigonomic Identies

Proof:
Often, multiplying by conjugates is helpful in trig identities.

     1 + sinX + cosX           (1 + sinX + cosX)^2
     ---------------- =  -----------------------------------
     1 + sinX - cosX      (1 + sinX - cosX)(1 + sinX + cosX)

               1 + 2sinX + 2cosX + 2sinXcosX + (sinX)^2 + (cosX)^2
            = -----------------------------------------------------
                         1 + 2sinX + (sinX)^2 - (cosX)^2

Now, notice almost everything has a coefficient of 2. Substituting 1 
for (sinX)^2 + (cosX)^2 in the numerator and (sinX)^2 for 1 - (cosX)^2 
in the denominator gives

                       2 + 2sinX + 2cosX + 2sinXcosX
                    = -------------------------------
                             2sinX + 2(sinX)^2

Factor by grouping in the numerator, and factor out sinX in the 
denominator

                       (1 + sinX)(1 + cosX)
                    = ----------------------
                         (1 + sinX)sinX

Dividing numerator and denominator by 1+sinX gives

                       1 + cosX
                    = -----------
                         sinX

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