A trigonometry text like Sullivan's Algebra & Trigonometry often has a prohibition like this (Sec. 7.3):

WARNING: Be careful not to handle identities to be established as if they were conditional equations. You cannot establish an identity by such methods as adding the same expression to each side and obtaining a true statement. This practice is not allowed, because the original statement is precisely the one that you are trying to establish. You do not know until it has been established that is, in fact, true.

I'll look at the first such example in Sullivan. Establish the identity: $\csc \theta \cdot \tan \theta = \sec \theta$.

The given solution is: $\csc \theta \cdot \tan \theta = \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} = \sec \theta$.

But let's consider using the prohibited operations that generate equivalent equations. We may write: $\csc \theta \cdot \tan \theta = \sec \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \sin \theta = \sin \theta$. [Multiplying both sides by $\sin \theta \cdot \cos \theta$ in the last step.] Now, personally, this does persuade me that the original equation is an identity, because it is equivalent to an obviously-true equation (excepting any values not in the domain of one side, which are identical to the official answer above).

If we want to be completely rigorous, then I think we could just reverse the sequence of statements above. Starting with the reflexive property of equality: $\sin \theta = \sin \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \csc \theta \cdot \tan \theta = \sec \theta$.

The official book proof is slightly shorter, and it can be written down directly without the reversal, but it seems mostly like a stylistic point -- fighting with students to abandon their equivalent-equation techniques is a great burden, and right now I'm not entirely seeing the necessity for it. Does anyone permit this alternative as an acceptable proof in their trigonometry class? If not, what is the compelling reason for Sullivan's dire warning?

Edit: This question presumes that any applied operations do correctly generate equivalent equations (noting gaps in the domain as required). The question is whether Sullivan's "cannot establish an identity by such methods as adding" statement is strictly true.


4 Answers 4


Well, if their "equivalences" are in fact equivalences, then everything is fine.

The problem seems to be mostly a psychological one: When students want to verify an equivalence, they check the left-to-right implication, but they often seem to take the right-to-left one for granted. As a result, they perform operations like multiplying both sides of an equation with some value (that is not guaranteed to be non-zero!), and they put, quite routinely, an equivalence sign between both formulas – which is wrong. Unfortunately, when they start with the property to be proved, the left-to-right implication is pointless, and the right-to-left implication is the important one.

If I want to demonstrate a proof on the blackboard and start with the property to be proved, I'd rather use a right-to-left implication sign than an equivalence sign; just because it makes students focus on the important one of the two directions.

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    $\begingroup$ I'm not sure that I agree that the left-to-right implication is pointless and should be left out. It seems like a tiny piece of writing to gain maximal truth content/generality. I guess I got in the habit of including that working through Casella/Berger's Statistical Inference. $\endgroup$ Jul 11, 2017 at 21:24
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    $\begingroup$ @DanielR.Collins: You can have situations where the RTL implication is preserved but the LTR is not. For example, you could have $a^2 = b^2 \impliedby a = -b$. Obviously the reverse implication is (potentially) false in that case. $\endgroup$
    – Kevin
    Jul 12, 2017 at 5:06
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    $\begingroup$ @Kevin: That's not the point here. Uwe is suggesting dropping the equivalence sign even when it would be a true statement. $\endgroup$ Jul 12, 2017 at 5:22
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    $\begingroup$ I agree with Uwe on that. Students are often drilled to put an equivalence sign between all equations and some of them don't even think about it anymore, i.e. they also don't check if they really can put one. So putting only one direction might indeed make them look up and think a little bit about why this particular direction is true. It might then be a nice exercise for the students to show that every $\Leftarrow$ can indeed be replaced by a $\Leftrightarrow$, but I agree with Uwe on the point that you should first focus on the important part, the one you want to teach and make clear. $\endgroup$
    – Dirk
    Jul 12, 2017 at 8:51

When you solve an algebraic equation like $2x - 4 = 0$, what you're really doing is finding assignments for the variables assuming the given equation is true. Doing the same thing to both sides of an equation is meant to take a true equation and turn it into another true equation. That is, it can be used to form correct implications, like $2x - 4 = 0 \implies 2x = 4$ and $2x = 4 \implies x = 2$. Those implications can be chained to get $2x - 4 = 0 \implies x = 2$, which is the "answer". This implication is all that the problem is asking you to show, that assuming $2x - 4 = 0$ is true, $x = 2$ is also true.

When you're trying to prove an identity, you don't assume that the identity is true, and an implication doesn't serve as an answer. Applying the same sort of steps used above to $\csc \theta \cdot \tan \theta = \sec \theta$ gets you $\csc \theta \cdot \tan \theta = \sec \theta \implies \sin \theta = \sin \theta$. But you need to show $\csc \theta \cdot \tan \theta = \sec \theta$ by itself. You can't use the implication you just derived because it goes in the wrong direction (and like Uwe pointed out in their answer, this direction of implication isn't used in the problem at all, and could be distracting to include). You instead need $\sin \theta = \sin \theta \implies \csc \theta \cdot \tan \theta = \sec \theta$; then, since you already know $\sin \theta = \sin \theta$, you'd get just $\csc \theta \cdot \tan \theta = \sec \theta$.

In this case, the particular algebraic steps you used can be inverted to get this implication. The biggest problem is that some steps aren't invertible, which rnrstopstraffic mentioned in their answer. Another problem, IMO, is that students wouldn't understand the correct logical steps of their answer if they wrote it like solving an equation. They probably wouldn't recognize that the correct proof uses the inverses of the operations they wrote down. It'd be clearer to them if they wrote down such inverse steps explicitly if they were proving it this way.

The book provides that warning because it wants to point out that solving it like an equation wouldn't be using correct logic (because it's a textbook, a place where such a detail should be mentioned). The statement is also stated generally to be applicable for proofs in general (like those found in upper-level college math courses). If you're more concerned about students just doing the algebra correctly, then you don't need to mark your students off for doing it like this, but I wouldn't solve it like an equation for examples you show during class.

Personally, I prefer how the book solved it, by just simplifying the expressions on one (or both) sides of the identity. It's simpler, and the expression on each side of the equal sign remains equivalent to the ones in the original identity, so it's clearer what was actually shown: $\frac{1}{\cos \theta} = \frac{1}{\cos \theta}$ proves that $\csc \theta \cdot \tan \theta = \sec \theta$ because the respective left sides are equal to each other, and same for the right sides. You didn't need any equivalent equations at all, just equivalent expressions in the same equation.

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    $\begingroup$ I think this answer makes a good point by contrasting "solving an equation" with "establishing an identity." $\endgroup$
    – user52817
    Jul 12, 2017 at 12:59
  • $\begingroup$ This answer is based on the false premise "When you solve an algebraic equation like $2x - 4 = 0, \ldots\ldots 2x - 4 = 0 \implies x = 2 \ldots\ldots$ This implication is all that the problem is asking you to show, that assuming $2x - 4 = 0$ is true, $x = 2$ is also true." $\endgroup$
    – ryang
    Nov 27, 2021 at 17:47
  • $\begingroup$ On the contrary! If this claim were true, then extraneous solutions wouldn't exist, it wouldn't be logical to filter out solutions, and $\sqrt x=2{\implies}x^2=4{\implies}(x+2)(x-2)=0{\implies}x=\pm2$ would indicate that $\sqrt x=2$ has solution set $\{-2,2\}.$ Don't just take my word for it. $\endgroup$
    – ryang
    Nov 27, 2021 at 18:00

For nearly any identity that we will run across in a standard trig class, I believe that the method that you described (the rigorous one) would be sufficient.

I think the problem with the logic is that not all modifications that we do to equations are invertible, so one would need to take great care in some situations. Consider the incorrect identity $\sin(\theta)=\sqrt{1-\cos^2(\theta)}$ which we could "prove" by the following: $$\begin{align}\sin(\theta)&=\sqrt{1-\cos^2(\theta)}\\ \iff\sin^2(\theta)&=1-\cos^2(\theta)\\ \iff \sin^2(\theta)+\cos^2(\theta)&=1\\ \iff 1&=1\end{align}$$

You can of course see where this is wrong and make appropriate adjustments and notes, but our students would not.

In general though, I think that the idea that your presenting is useful for finding the correct thought process much the same way that we find the appropriate $\varepsilon$ in analysis by working backward. It can be used as part of the "figuring out" process, but in my opinion there is still a value for teaching a more rigorous approach (logically speaking).


I believe that the point of having students prove trigonometric identities is to give them practice rewriting expressions involving trigonometric equations.

The actual problems they are preparing to solve involve, for example, two people working the same problem independently and then comparing answers.

For example: If I got $\csc{x}\cdot\cot{x}$ and you got $\cos{x}$, did we get the same answer?

I think that this is the motivation for the kind of quote offered in the question.

In addition, the method in the quote teaches students to think in terms of chains of equal expressions rather than chains of equivalent equations. I see some value in that as well.

That said, I also have not much difficulty with the idea of teaching students to 'prove' an identity by solving the equation. If the solution set includes all numbers in the domains of both sides of the original equation, then the equation if an identity.

That seems more difficult to teach and likely to be more difficult for students to understand than the approach in the text quoted above.

Edit: I am not ignoring anyone. The rules do not permit me to comment until I have more reputation points.

I doubt that every textbook author takes the same view I do. I would like to see more open questions in textbooks. In trigonometry , there should be more: "Are these two expressions equivalent when both are defined?" questions than I have seen in the books I have been assigned.

  • $\begingroup$ Re: "the point of having students prove trigonometric identities is to give them practice rewriting expressions" is a good theory (+1), but if that was truly the goal then I think it would be better served by other, more direct exercises. For example, Sullivan already has exercises of form (Sec 7.3): "Simplify each trigonometric expression by following the indicated direction... Rewrite in terms of sine and cosine: $\tan{\theta} \cdot \cot{\theta}$." Expanding on those, I think, would better clarify that goal. $\endgroup$ Sep 3, 2017 at 21:55
  • $\begingroup$ And on that point I'll note that a source like OpenStax Algebra and Trigonometry does a much better job: e.g., in Sec 9.1 it has more exercises (13) of the form "simplify the first trigonometric expression by writing the simplified form in terms of the second expression", and fewer exercises (5) of the form "verify the identity". $\endgroup$ Sep 3, 2017 at 22:03
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    $\begingroup$ I think this might be my favorite answer so far, and has helped to clarify my thinking on the issue; thank you. $\endgroup$ Sep 3, 2017 at 22:05
  • $\begingroup$ As I compare books today, re: your edit "there should be more: 'Are these two expressions equivalent when both are defined?'", I note favorably that OpenStax has exercise blocks of exactly that skill, whereas Sullivan does not. $\endgroup$ Sep 3, 2017 at 23:35

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