Proof Writing – Using Both Sides of an Equation to Prove Equality

proof-writingsoft-question

For example:

$\color{red}{\text{Show that}}$$$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$

In high school my maths teacher told me

To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side.

So starting from the LHS: $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{4\cos^2(x)-2}{\cos^2(x)}=4-2\sec^2(x)$$ $\large\fbox{}$

At University, my Maths Analysis teacher tells me

To prove a statement is true, you must not use what you are trying to prove.

So using the same example as before:

LHS = $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{2\Big(2\cos^2(x)-\left[\sin^2(x)+\cos^2(x)\right]\Big)}{\cos^2(x)}=\frac{2(\cos^2(x)-\sin^2(x))}{\cos^2(x)}=\bbox[yellow]{2-2\tan^2(x)}$$

RHS =$$4-2\sec^2(x)=4-2(1+\tan^2(x))=\bbox[yellow]{2-2\tan^2(x)}$$

So I have shown that the two sides of the equality in $\color{red}{\rm{red}}$ are equal to the same highlighted expression. But is this a sufficient proof?

Since I used both sides of the equality (which is effectively; using what I was trying to prove) to show that $$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$

One of the reasons why I am asking this question is because I have a bounty question which is suffering from the exact same issue that this post is about.

EDIT:

Comments and answers below seem to indicate that you can use both sides to prove equality. So does this mean that my high school maths teacher was wrong?

$$\bbox[#AFF]{\text{Suppose we have an identity instead of an equality:}}$$ $$\bbox[#AFF]{\text{Is it possible to manipulate both sides of an identity to prove that the identity holds?}}$$

Thank you.

Best Answer

There's no conflict between your high school teacher's advice

To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side.

and your professor's

To prove a statement is true, you must not use what you are trying to prove.

As in Siddarth Venu's answer, if you prove $a = c$ and $b = c$ ("working from both sides"), then $a = c = b$ by transitivity of equality. This conforms to both your teacher's and professor's advice.

Both your high school teacher and university professor are steering you away from "two-column proofs" of the type: \begin{align*} -1 &= 1 &&\text{To be shown;} \\ (-1)^{2} &= (1)^{2} && \text{Square both sides;} \\ 1 &= 1 && \text{True statement. Therefore $-1 = 1$.} \end{align*} Here, you assume what you want to prove, deduce a true statement, and assert that the original assumption was true. This is bad logic for at least two glaring reasons:

  1. If you assume $-1 = 1$, there's no need to prove $-1 = 1$.

  2. Logically, if $P$ denotes the statement "$-1 = 1$" and $Q$ denotes "$1 = 1$", the preceding argument shows "$P$ implies $Q$ and $Q$ is true", which does not eliminate the possibility "$P$ is false".

What you can do logically is start ("provisionally", on scratch paper) with the statement $P$ you're trying to prove and perform logically reversible operations on both sides until you reach a true statement $Q$. A proof can then be constructed by starting from $Q$ and working backward until you reach $P$. Often times, the backward argument can be formulated as a sequence of equalities, conforming to your teacher's advice. (Note that in the initial phase of seeking a proof, you aren't bound by anything: You can make inspired guesses, additional assumptions, and the like. Only when you write up a final proof must you be careful to assume no more than is given, and to make logically-valid deductions.)

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