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Proving Identities Rigorously


Date: 01/23/2002 at 18:35:19
From: Robert Chen
Subject: Proving Identities

This would probably apply to many of the proofs about identities, but 
since I was looking at identities, that'll be the topic.

One submission in the Dr. Math archives was asking about 
(1-tan A)/sec A + (sec A/tanA) = (1+tan A)/(sec A tan A). 
One of the Doctor Maths answered it by saying to multiply both sides 
by sec A tan A.

I do these two-sided operations frequently in math, and after I did
VERY badly on my identities test, my teacher wrote "You can only work 
with one side at a time" beside my work. I was wondering why you work 
with both sides. I've been taught that the "proper" way is to work 
with one side only


Date: 01/25/2002 at 00:02:39
From: Doctor Pete
Subject: Re: Proving Identities

Hi Robert,

You bring up a very good point, and you are correct: the proper way to 
prove such identities is to begin on one side and algebraically 
transform it into the form shown on the other side. Working on both 
sides is technically incorrect because in doing so we are assuming 
that the equality to be proven is already true.

However, in the case of many identities, it is allowable to work on 
both sides, in the sense that it is not mathematically incorrect - if 
the identity is true, then the result of working on both sides will 
eventually result in an equality that is "obviously" or more easily 
seen to be true. That is, such a method won't lead to a wrong decision 
on the truth or falseness of the identity in question. But in so far 
as mathematical rigor, it is insufficient because of the reason 
mentioned at the end of my first paragraph. Often, however, working on 
both sides can help us form a rigorous proof of an identity. I will 
illustrate this with the example identity you provided.

Identity to be proven:

     (1-Tan[a])/Sec[a] + Sec[a]/Tan[a] = (1+Tan[a])/(Sec[a]Tan[a])

If we work on both sides, the first step is to multiply both sides by 
Sec[a]Tan[a]:

     (1-Tan[a])Tan[a] + Sec[a]Sec[a] = 1+Tan[a].

Multiplying through, we find

     Tan[a] - Tan[a]^2 + Sec[a]^2 = 1+Tan[a],
or
     -Tan[a]^2 + Sec[a]^2 = 1,

which is equivalent to the identity 1+Tan[a]^2 = Sec[a]^2, so we are 
done. This is not a rigorous proof, but it leads to one, in which we 
see the proper method is as follows:

     (1-Tan[a])/Sec[a] + Sec[a]/Tan[a]
   = ((1-Tan[a])Tan[a])/(Sec[a]Tan[a]) + Sec[a]^2/(Sec[a]Tan[a])
   = (Tan[a] - Tan[a]^2 + Sec[a]^2)/(Sec[a]Tan[a])
   = (Tan[a] - Tan[a]^2 + 1 + Tan[a]^2)/(Sec[a]Tan[a])
   = (Tan[a] + 1)/(Sec[a]Tan[a]).

Therefore we have taken the left-hand side and transformed it into the 
right-hand side using algebraic manipulation and other known 
trigonometric identities. But it should not be too hard to see that 
the basic "flow" of the proof is essentially the same procedure as the 
"improper" method, just rearranged a bit.

Another, more sophisticated example of this principle is the proof of 
the following fact, known as the Arithmetic-Geometric Mean Inequality, 
for two numbers a, b >= 0:

     (a+b)/2 >= Sqrt[ab].

A common mistake is to use the two-sided method, like this:

     a + b >= 2 Sqrt[ab]     (multiply both sides by 2)
     (a+b)^2 >= 4ab          (square both sides)
     a^2 + 2ab + b^2 >= 4ab  (expand the right side)
     a^2 - 2ab + b^2 >= 0    (subtract 4ab from both sides)
     (a - b)^2 >= 0          (factor the right side)

and the last inequality is obviously true, since the square of any 
real number is never negative. But this is bad form, especially 
because we are dealing with an inequality. Therefore, the correct way 
is to begin with

     0 <= (a - b)^2
        = a^2 - 2ab + b^2
        = a^2 + 2ab + b^2 - 4ab
        = (a + b)^2 - 4ab.

Hence ((a+b)^2)/4 - ab >= 0, since if a number is greater than or 
equal to 0, that number divided by 4 is also greater than or equal 
to 0. Adding ab to both sides (we are allowed to do this because we 
already know that the inequality we began with is true),

     ((a+b)^2)/4 >= ab,

and since a, b > 0, we may take the square root of both sides without 
affecting the inequality: thus (a+b)/2 >= Sqrt[ab], as was to be 
shown.

In short, to be completely rigorous in our proof of trigonometric 
identities, we mustn't assume the identity, or in general, whatever 
needs to be proven, is true in the first place: it is much better to 
work on one side, trying to obtain the other, in a single chain of 
reasoning, and in other cases, working from already known identities 
(in which we may work on both sides) to obtain the desired result.

- Doctor Pete, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Trigonometry
High School Trigonometry

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