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If you input the trig identity: $$\cot (x)+\tan(x)=\csc(x)\sec(x)$$ Into WolframAlpha, it gives the following proof:

Expand into basic trigonometric parts: $$\frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{\cos(x)} \stackrel{?}{=} \frac{1}{\sin(x)\cos(x)}$$ Put over a common denominator:

$$\frac{\cos^2(x)+\sin^2(x)}{\cos(x)\sin(x)} \stackrel{?}{=} \frac{1}{\sin(x)\cos(x)}$$

Use the Pythagorean identity $\cos^2(x)+\sin^2(x)=1$:

$$\frac{1}{\sin(x)\cos(x)} \stackrel{?}{=} \frac{1}{\sin(x)\cos(x)}$$

And finally simplify into

$$1\stackrel{?}{=} 1$$

The left and right side are identical, so the identity has been verified.

However, I take some issue with this. All this is doing is manipulating a statement that we don't know the veracity of into a true statement. And I've learned that any false statement can prove any true statement, so if this identity was wrong you could also reduce it to a true statement.

Obviously, this proof can be easily adapted into a proof by simply manipulating one side into the other, but:

Is this proof correct on its own? And can the steps WolframAlpha takes be justified, or is it completely wrong?

Asaf Karagila
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Nico A
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    Can the downvoter explain their reasoning? – Nico A Dec 07 '17 at 15:44
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    You are correct that this sort of proof is wrong - you really need to go the other direction. In this case, you can (assuming $\cos x\neq 0$ and $\sin x\neq 0$. If either is zero, then both sides of the original equation are undefined.) – Thomas Andrews Dec 07 '17 at 15:44
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    You can read the proof as saying that each identity holds if and only if the following one holds. Since the last identity is indeed true, the result follows. Whether additional short comments need to be added to verify that indeed each new potential identity is equivalent to the preceding one probably depends on the precise phrasing used by Alpha. – Andrés E. Caicedo Dec 07 '17 at 15:44
  • @ThomasAndrews disagree. It is really an $\iff$ for each line. Also +1, I think this was a well thought out question. – Andres Mejia Dec 07 '17 at 15:44
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    Well, I suppose how the proof is phrased, I guess, but unless the writer says "if and only if." This kind of proof is risky because it moves in one direction. (It is not true that if $1=1$ then the previous equation is true, for example, unless $\cos x\sin x\neq 0$. You'd spot that error if you wrote the proof in the reverse.) @AndresMejia And there are many cases of "fake proofs" that look like this where reverse steps don't work, so it is a valid complaint for this proof, as well. – Thomas Andrews Dec 07 '17 at 15:51
  • That $\cos(x)\sinx\neq 0$ is a hypothesis for the theorem though. I agree, not the best way to write it... – Andres Mejia Dec 07 '17 at 15:54
  • Sorry, that hypothesis is not stated. It is implied, perhaps, just as the $\iff$ is implied, but the OP is clearly at a learning stage for proofs, and the concern OP expresses is valid. It is true that this proof can be made correct, but it is good the OP is also suspicious of this form of proof. – Thomas Andrews Dec 07 '17 at 16:01
  • "All this is doing is manipulating a statement that we don't know the veracity of into a true statement" How else do you expect a proof to be done? – Rolazaro Azeveires Dec 07 '17 at 21:52
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    @RolazaroAzeveires By showing a statement to be equivalent to a true statement, or manipulating a true statement into our statement of unknown veracity. But starting with a statement that we don't know the truth value of, it could be false, and a false statement implies any true statement. – Nico A Dec 07 '17 at 22:50
  • This is sometimes referred to as "synthetic proof". You go from the statement you want to prove back to the given hypothesis. Of course, that is invalid- unless every step is reversible. In fact, it is not uncommon to analyze a proof that way- starting from the conclusion and going back to the hypothesis. Then writing down in the reverse, from hypothesis to conclusion. Again, that is only possible if every step is reversible. If it is clear that every step is reversible then you don't really need to explicitly show the "direct" proof. – user247327 Dec 08 '17 at 13:36
  • "showing a statement to be equivalent to a true statement" is what is done in the example. no? You start with an expression, which you do not know to be true or not. You change it in ways that keep it true, whenever it was true and keep it false, whenever it was false. If you reach a obviously true proposition then all the previous ones were also always true. Because the manipulation has not changed the truthfulness (or falsity). Note that the last step in your example is not such a change. 1=1 is always true, but your expressions have no meaning for sin(x)=0 or cos(x)=0 – Rolazaro Azeveires Dec 09 '17 at 03:22

2 Answers2

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It is good that you are wary of proving identities this way. Indeed, I could "prove" $0=1$ by saying

\begin{align*} 0 &\stackrel{?}{=}1\\ 0\cdot 0 &\stackrel{?}{=} 0 \cdot 1\\ 0 &=0. \end{align*}

The important point is that every step WolframAlpha did is reversible, while the step I took (multiplying by $0$) was not. That is what allows the proof from WolframAlpha to be rearranged into a proof that starts with one side of the identity and ends at the other:

\begin{align*} \cot(x)+\tan(x) &= \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{\cos(x)}\\ &= \frac{\cos^2(x)}{\sin(x)\cos(x)} + \frac{\sin^2(x)}{\sin(x)\cos(x)}\\ &= \frac{\sin^2(x)+\cos^2(x)}{\sin(x)\cos(x)}\\ &=\frac{1}{\sin(x)\cos(x)}\\ &=\csc(x)\sec(x). \end{align*}

So no, the WolframAlpha proof is not wrong, but it neglects to emphasize the important fact that every step is reversible. I am not a fan of that sort of proof, as it gives students the idea that they can prove an identity by manipulating both sides in any way they like to arrive at a true statement.

kccu
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    +1, I think that they should edit the page just for pedagogical reasons. – Andres Mejia Dec 07 '17 at 15:50
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    +1: it's absurd and counter-intuitive to write the chain of equalities the way WolframAlpha does. – Rob Arthan Dec 07 '17 at 19:16
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    @RobArthan I would argue that it's absurd, but not counter-intuitive. If trying to prove such an identity, it's perfectly reasonable to begin from the statement you wish to prove and, taking care that all steps are reversible, try to manipulate it into something you know is true. – Carl Schildkraut Dec 07 '17 at 21:29
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    I would say that it is neither absurd nor counter-intuitive. That's exactly how you find proofs for such things. – tomasz Dec 07 '17 at 21:39
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    I agree, neither absurd nor counter-intuitive, just sloppy. Mathematicians do sloppy things all the time when trying to come up with proofs, but they must be cleaned up and made rigorous eventually. – kccu Dec 07 '17 at 22:01
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    @AndresMejia: I don't think this page is a human generated wiki. It's a dynamic step-by-step "proof" done automatically by WolframAlpha. – Eric Duminil Dec 08 '17 at 08:38
  • @tomasz: please read carefully what I wrote. It would be absurd and counter-intuitive for a human being to write what turns out to be a chain of equalities in the form that WolframAlpha did. If your proof has the form $a = d$ because $a = b = c = d$ where each equality holds via algebraic reasoning, it's not helpful to present it it as a list of deductive transformations on the target equation $a = d$ (which then has to be backed up by an argument that the deductive steps are reversible). – Rob Arthan Dec 09 '17 at 22:52
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You can make it rigorous by going in the reverse direction. $1=1 \implies \frac{1}{\cos x \sin x}=\frac{1}{\cos x \sin x} \implies \dots$.

But this is a silly looking "proof" and it is really clunky. The point is that this is not $X \implies 1=1$, but rather the proof given ensures that $X \iff 1=1$ by following equality (which is symmetric) in either direction for the proof.

Personally the way you suggest is much preferred , $$\frac{1}{\cos x \sin x}=\frac{\cos^2x+\sin^2x}{\cos x\sin x}=\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}=\cot x+\tan x.$$

Giuseppe
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Andres Mejia
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