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I taught three basic trigonometric identities. One of my 9th grade students asked: How can we say $$\sec^2x-\tan^2x=1$$ is an identity since when we plug in $x=\frac{\pi}{2}$ the identity fails which creates an ambiguity for the definition of an identity in mathematics?

How should I try to explain him without entering into calculus part?

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    $\begingroup$ What is the definition of an identity in your book or lectures? $\endgroup$ – Daniel R. Collins Jul 5 '17 at 13:48
  • $\begingroup$ Does he also say that secants don't exist (because of the pi/2 problem?) If he is OK with the general concept of a secant, knowing that there are times it doesn't exist, than what is the big issue with the identity? $\endgroup$ – guest Jul 6 '17 at 1:11
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An identity always holds for some values of the free variables. Sometimes the allowed values are all real numbers, sometimes something else. In this case the identity holds whenever $\sec x$ and $\tan x$ are defined (they are defined in the same set). In a certain limiting sense the identity is also true at $x=\pi/2$, but you probably don't want to go into that with your student. The point is: An identity should always come with information about when it is supposed to hold. It's not just an equation.

Even with the more common identity $\sin^2x+\cos^2x=1$ you can start asking questions: This holds for geometrical reasons for $x\in(0,\pi/2)$, but does it hold for all $x\in\mathbb R$? Ok, does it hold for all $x\in\mathbb C$ as well? Well, what if $x$ is a matrix (you can still define the functions via the matrix exponential)? The answers are all yes, but certainly only the real case is meant when it is first encountered. It is not clear how far an identity should extend unless you explicitly tell what you mean by it. You cannot always trust that your students will intuitively see what values the variables should take.

For comparison, the Pythagorean theorem does not state just $a^2+b^2=c^2$. It states this equation and tells what the three variables should be for this to hold. The equation only becomes meaningful once there is a triangle with a right angle and the lengths of the sides are named appropriately — it is all this information together that makes the theorem.

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    $\begingroup$ Thanks that was pretty clear $\endgroup$ – Ekaveera Kumar Sharma Nov 27 '15 at 14:06
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    $\begingroup$ This answer is halfway there, but it obscures what might be a subtlety to most students. An identity like the above is really a statement of equality between functions. Although the right hand side looks like a number rather than a function, "1" really is referring to an identity function and thus you also need to clarify the domain for it. $\endgroup$ – Chan-Ho Suh Nov 27 '15 at 18:40
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    $\begingroup$ In fact, failure to put the domain of an identity in is usually the root cause of these silly "1 = 0 proofs" we see high schoolers engage in. $\endgroup$ – corsiKa Nov 27 '15 at 20:00
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    $\begingroup$ @Chan-HoSuh, I would not regard this as an equality of functions. Instead, I see it as an equality of expressions. The two expressions are equal for certain values of the variable. You can of course view it as an equality of functions, but I feel that adds a layer of unnecessary structure. This is a matter of taste, but it is certainly not a must to relate it to functions. $\endgroup$ – Joonas Ilmavirta Nov 27 '15 at 20:29
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    $\begingroup$ @Chan-HoSuh: there is an equality of functions, but it would rather be written $\sec^2-\tan^2=1$ (and then your remark about domain of $1$ applies; note that here $1$ would be called a constant function rather than an identity function). $\endgroup$ – Benoît Kloeckner Nov 27 '15 at 20:47
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If $x$ is meant to be a real variable, then your student is right; $\sec^2 x - \tan^2 x = 1$ is not* an identity; it suffers from the same sort of problems that $\frac{x}{x} = 1$ does.

It fails to be an identity in two different ways, depending on fine details of what one actually means: either the left hand side is undefined because $x$ is not restricted to the domain of $\sec$ and $\tan$, or the identity fails because the domain of the left hand side differs from the domain of the right hand side.

(the former is syntax/semantics based on total functions, the latter on partial functions)

To make a correct statement, limit $x$ appropriately: e.g. $\sec^2 x - \tan^2 x = 1$ whenever $x \neq (2n+1) \frac{\pi}{2}$.

As another answer suggested, the version $1 + \tan^2 x = \sec^2 x$ has the advantage of actually being an identity in the partial function syntax/semantics when $x$ is a real variable, since both sides have the same domain. Think of it as saying "if one side is defined, then so is the other and they're both equal".

*: Technically, there are other settings where this would be an identity; e.g. if the two sides don't refer to functions, but instead to equivalence classes of functions where two functions are equivalent if they differ on a set of measure zero, or if every operation is to be followed by "... then continuously extend the result"

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    $\begingroup$ If you go that way, ignoring that there is a missing part that might be implied but should be there at least implicitly, I will argue that $\cos^2x + \sin^2x =1$ fails to be an identity: it makes no sense unless one defines $x$ one way or another. $\endgroup$ – Benoît Kloeckner Nov 27 '15 at 20:52
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    $\begingroup$ @Benoit: While true, keep in mind that implicit contexts like $x$ being of real number type are very broadly applied. If you make the assumptions explicit all the time, you just make the exceptions less memorable (as well as increase cognitive load). Yes, it's good to know that they are making background assumptions, but they generally belong in the background. If you really want to push that line of thought to the extreme, note that $\cos$ and $\sin$ are things whose meanings might vary in different contexts too. (for that matter, the same goes for exponentiation, $+$, $2$, $1$, and even $=$) $\endgroup$ – user797 Nov 27 '15 at 22:40
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One answer is to say that, properly, the identity is $\sec^2x=1+\tan^2x$, where the sides fail to be defined at the same values.

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    $\begingroup$ -1 if I had the rep. "Properly"? There's nothing improper about writing it with only "1" on one side. I guess writing it that way does highlight confusion about what functional identities mean, but that's hardly a recommendation to write it the way you suggest. $\endgroup$ – Chan-Ho Suh Nov 27 '15 at 18:24
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    $\begingroup$ @Chan-HoSuh: That depends on your definitions. If you define an identity to require equality at all points, one way is an identity and one way isn't (assuming one considers both sides being undefined to count as equality). That actually seems to be the commonly used definition, which makes this form an identity and the OP's original version close to, but not quite, an identity. I'm not sure that's a good idea---perhaps it's better to highlight the issue of domain rather than cleverly hiding it in a definition---but it might be easier than insisting that every identity has a domain. $\endgroup$ – Henry Towsner Nov 27 '15 at 19:59
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For typical inputs the equality holds. Thus, whatever further qualifications or discussion may be interesting or necessary, we're "most of the way there". To deny that it's "an identity" is to pretend to disqualify the fact on some technical grounds that do not address significant issue.

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    $\begingroup$ What I get from this is that in contrast to the (usual?) direction/order where one starts with hypotheses and sees what can be derived, here one first has a conclusion in mind to aim for (the equation, the $\int f'g = fg - \int fg'$ formula, etc.) that feels in some sense that it 'ought' to hold and then imposes conditions to 'make it happen' as in the quote by Hadamard about intuition or something, and that it is sometimes convenient to be agnostic about the conditions as opposed to specifying them precisely because ... $\endgroup$ – Vandermonde Nov 28 '15 at 7:29
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    $\begingroup$ (i) said conditions might be easily broadened so as to defeat any objections, making the original conditions kind of pointless (e.g., even if $x/x=1$ doesn't technically qualify as an identity of real numbers, it could well have been taken as an assertion about $\mathbb{R}(x)$, and then it is genuine. Or particular functions may not be classically differentiable or Riemann integrable yet satisfy integration by parts in a more general sense.), and $\endgroup$ – Vandermonde Nov 28 '15 at 7:30
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    $\begingroup$ (ii) trying to articulate the conditions for the very strongest statement possible so you get an if-and-only-if (and refusing to settle) can result in really fussy conditions (well-behaved + Cauchy-Riemann-satisfying = analytic, now what is the least I can get away with for 'well-behaved'?). $\endgroup$ – Vandermonde Nov 28 '15 at 7:30
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    $\begingroup$ @Vandermonde, yes, your narrative here is how I feel about most of these things, seconding Hadamard, etc. From at least the viewpoint of "taste", I'm not a fan of fussy conditions, myself, and my observations over some decades strongly indicate that the general population already (often correctly) suspects math people of making up nearly-pointless, technical objections to otherwise reasonable ideas. Indeed, some "testing" amounts to bait-and-switch or "gotchas", which is not good, and I work hard to counter that unfortunate thread. $\endgroup$ – paul garrett Nov 28 '15 at 15:46
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    $\begingroup$ Oh, one more thing I remember which might tangentially relate: I was once reading about quantum mechanics and the text considered wavefunctions that in the position representation were Dirac deltas. I was OK with this since they at that point had done only legitimate operations with them (which could all in principle be justified and which I interpreted as just saving a few words; no squaring such things or anything smarmy like that) and I could treat them as a localised distribution anyway and save the limit for later $\endgroup$ – Vandermonde Dec 6 '15 at 5:36
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$π/2$ and $−π/2$ are not in the domain. I'd take a step back and offer examples from algebra where, when observing an equation, the first discussion is "what is the domain?"

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    $\begingroup$ Personally I wouldn't expect 'domain' to have been defined by that age. $\endgroup$ – Jessica B Nov 27 '15 at 15:13
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    $\begingroup$ Domain and range are part of first year algebra, trig wouldn't ever come before this in any curriculum. $\endgroup$ – JoeTaxpayer Nov 27 '15 at 15:37
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    $\begingroup$ @JoeTaxpayer: "...in any curriculum"? Beg to differ. In the Australian Curriculum, trig ratios are Year 9 (~15 years old), but specific considerations like "function", "domain" and "range" are Year 10 at the earliest. $\endgroup$ – Tim Pederick Nov 27 '15 at 16:57
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    $\begingroup$ We did trig aged 13, but domain and range as terms didn't appear until I reached university. The concepts appeared a couple of years earlier, about age 17. $\endgroup$ – Jessica B Nov 27 '15 at 18:35
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    $\begingroup$ I stand corrected. What I should have written was "In my opinion, and my experience, algebra, and the concept of domain and range precedes trigonometry." I've seen the sequence of learning for the schools in my state, and of course that doesn't reflect the world, let alone other parts of my own country. $\endgroup$ – JoeTaxpayer Nov 27 '15 at 18:39
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Start with $\sin^2(x) + \cos^2(x) = 1$, which is defined for $0\leq x\leq 2\pi$.

Then divide through by $\cos^2(x)$ to get $\tan^2(x) + 1 = \sec^2(x)$, which is not defined when either of the denominators $= 0$.

Rearrange to get $\sec^2(x) - \tan^2(x) = 1$, as per the question.

The reason it is an identity is because it derives from an identity but is undefined when trying to divide by zero.

Also, I always taught an identity with an = sign with 3 bars ($\equiv$). This may not be common in other parts of the world, as suggested by some of the answers above, which are a bit stuck on the meaning and purpose of the 'ordinary' = sign: I suggest don't use it.

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    $\begingroup$ Many thanks to @Joonas Ilmavirta for the formatting help :) $\endgroup$ – KiwiSteve Dec 27 '15 at 8:40
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This question simply turns on the definition of the word "identity". Michael E2's answer is currently the best one, because he actually references a definition. Any college algebra or precalculus book that I can touch has such a definition, and every one that I see is effectively equivalent. In addition to Michael E2's examples, I can add:

Ratti & McWatters, Precalculus: A right triangle approach, 2E (Sec 1.1) [2011]:

An equation that is satisfied by every real number in the domain of the variable is called an identity.

Sullivan, Algebra & Trigonometry, 7E (Sec 1.1) [1987-2005]:

An equation that is satisfied for every value of the variable for which both sides are defined is called an identity.

Rietz & Crathorne, Introductory College Algebra, Revised (Ch. II, Art. 18) [1923-1933]:

The two members of an identity are equal for all values of the symbols for which the expressions are defined.

I include dates and the latter example to highlight that the customary definition of an "identity" has been stable and consistent for at least the last century.

The OP should refer the student to the definition in use in the course's textbook and/or lecture notes (which hopefully matches all of the preceding examples). In particular, an identity only needs values to match when both sides are defined; so it is not foiled by looking at values outside the domain of the expressions. This answers the student's question.

For a similar example that one should be preparing students to look very carefully at their definitions, consider the fact that the definition of a continuous function similarly looks only at points in the domain (so that, e.g., $1/x$ qualifies as a continuous function).

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    $\begingroup$ The definition that you quoted from Sullivan has the curious consequence that, if we're considering only real numbers, $\sqrt{-x}=\sqrt x$ is an identity. The only value of the variable for which both sides are defined is zero, and there the equation is satisfied. For the same reason, $\sqrt{-x}=7\sqrt x$ is also an identity by this definition (but of course $\sqrt x=7\sqrt x$ is not an identity). There seems to be a possibility of serious confusion here. $\endgroup$ – Andreas Blass Jul 5 '17 at 21:09
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    $\begingroup$ @AndreasBlass: That's an interesting point. $\endgroup$ – Daniel R. Collins Jul 6 '17 at 1:49
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For a right triangle with angle $x$, hypotenuse $H$, adjacent side $A$ and opposite side $O$:

$sec(x) = H/A$

$tan(x) = O/A$

$sec^2(x) - tan^2(x) = 1$

$H^2/A^2 - O^2/A^2 = 1$

$H^2 = A^2 + O^2$, which is just the Pythagorean Theorem

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    $\begingroup$ I don't think the OP really wants a proof of the identity. The problem seems to be about making sense of it or explaining it to students because of issues when both trigonometric functions blow up. Your proof does not make sense when $A=0$, so the problem at $x=\pi/2$ remains. $\endgroup$ – Joonas Ilmavirta Nov 27 '15 at 12:13
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    $\begingroup$ @JoonasIlmavirta Show that for the equation to make sense, $A$ must be nonzero, thus, $x$ must not be $\pi/2$ or $-\pi/2$ (or other angles coterminal with them). $\endgroup$ – Joel Reyes Noche Nov 27 '15 at 12:40
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Look at the "identity". On the right-hand-side, you apparently have the number 1 and on the left hand side, you have the difference of two functions (which is also a function). A number is not a function, so the stated identity would seem to be wrong. Undoubtedly the student is trying to make sense of this by thinking of "=" as some sort of process where it really means (or so s/he thinks) "plug in numbers on the left-hand-side and the result should be the same as the numerical value on the right". This is wrong.

The identity is in fact asserting equality of the two sides. You just have to be careful about defining what the sides are. The right hand side, which is represented by the symbol "1", is really the constant function giving 1 on a specified domain. It would be better to represent it as something like $1_D$ where $D$ is defined to be the domain that matches the domain of the left-hand-side.

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    $\begingroup$ I think you've slightly misunderstood the question. The key point is about the degeneracy at $\pi/2$. $\endgroup$ – Jessica B Nov 27 '15 at 21:52

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