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Proving Trig Identities vs. Solving Trig Equations

Date: 10/08/2006 at 13:25:51
From: Angel
Subject: Trigonometry: Prove that a proof = 0

I need to prove that: sinx/(1 - cosx) - (1 + cosx)/sinx = 0.  

I am told that I need to show that the left side of the proof is equal 
to zero.  I thought that I had done this so now I am really confused.  
Here's what I did:

First:  Use the additive property to add (1 + cosx)/sinx to both sides 
of the equation to give sinx/(1 - cosx) = (1 + cosx)/sinx.

Next:  Multiply both sides of the equation by sinx and also by (1 - 
cosx) which will give sin^2x = (1 + cosx)(1 - cosx) or 1 - cos^2x so
we have sin^2x = 1 - cos^2x.

Then:  Use the additive property and add cos^2x to both sides, giving 
sin^2x  + cos^2x = 1

This is the Pythagorean identity.  It holds for all real values of x.  
This proves that the original equation is also true for all values of 
x except for where either sinx = 0 or where 1 - cosx = 0.  This means 
that the equation is true except for where x = n(pi), where n is any 

Sin(x) = 0 when x = n(pi)

Cos(x) = 1 when x = 2n(pi)

Therefore x =/= n(pi)

Date: 10/08/2006 at 22:55:02
From: Doctor Peterson
Subject: Re: Trigonometry: Prove that a proof = 0

Hi, Angel.

I think you are saying that you submitted this proof, and were told it
was wrong, without explanation.  Is that right?

It is basically valid, but is not in the form that many teachers 
require, namely working only on one side of the equation at a time to
transform one side into the other.  You should have been told that,
rather than only being told that it is wrong, so that you would know
how to correct it.

What you have done DOES consitute a proof, IF you state it very
carefully.  You have done what you would do to SOLVE an equation,
which is a somewhat different process.  What you need to do here is to
explicitly state that you are transforming the equation into an
EQUIVALENT equation--one that is true when the original equation is
true, and vice versa.  This is not strictly true, since multiplying an
equation by an expression containing the variable does not necessarily
result in an equivalent equation, since that expression might be zero.
You haven't ignored this problem, as many students would do, but have
pointed out that you would have multiplied by zero only if the 
variable is a multiple of pi.  The one thing you didn't mention is
that in those cases, the equation you are trying to prove can't be
evaluated in the first place, since you would be dividing by zero; so
you have shown that it is true over its entire domain.  With that one
little exception, I think you've got a fine proof.

The trouble, as I said, is that many teachers don't accept this--
because, I think, they want to avoid the subtleties involved in this
sort of thinking (essentially protecting their students from the 
errors that could creep in if they were not as careful as you) by only
allowing them to take the "safe" path, which consists of a single 
chain of equal expressions.  You can usually turn an "unsafe" proof
such as yours into this kind of proof by sort of turning it on its
side, doing to the left side only, the operations equivalent to what
you did with the whole equation.

Here's how:

Your first step just moved one fraction to the other side; we have to
keep the whole left side together, so this can just be ignored.

The second step is to multiply by the LCD.  To do the same thing
without changing the value of the expression, you can instead rewrite
the fractions with a common denominator (the LCD), and then add them.
That is, rewrite them both with denominator (sinx)(1 - cosx).

Your third step just moved part back to the other side; the equivalent
is just to simplify the combined fraction you now have.  Then you can
use the Pythagorean identity to simplify it further, replacing sin^2x
 + cos^2x with 1, and you will be done.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 
Associated Topics:
High School Trigonometry

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