I’m sure that this question has been asked in a million high school math offices, so apologies for the rudimentary nature of the question.
I’m teaching Precalculus for the first time. And I’m about to teach proving trig identities, like:
I understand that the standard ways to prove trig identities is:
(a) pick one side of the equation, and keep morphing it until it matches the second side of the equation
(b) individually modify both side of the equations independently until they equal the same thing.
I always learned that what you cannot do is start mixing both sides of the equations. So, for the equation above, you can’t cross multiply to get:
and keep on simplifying both sides to show they are the same and the equality is true.
The reasons I’ve heard this is not allowed:
1. Because I said so.
2. You can only cross multiply if you know the equality is true. But that’s precisely what you’re trying to prove. You are assuming the statement is true to prove the statement is true.
However, both explanations are unsatisfying to me. The first one is for obvious reasons. My objection with the second one is that it seems to always work for these problems. Although I know it is logically unsound, I can’t quite pinpoint why with a concrete example to demonstrate it..
My questions are the following:
What do you do to explain to your kids why you can only work the sides of the equality independently? Does it convince them?
Does anyone have a good example involving trigonometric identities that illustrates that bad things happen when you don’t solve the sides independently, but start mixing them together? Like proving something that isn’t true actually is true… or proving something true that actually isn’t true?
Thanks for any help. I feel a little foolish, like I’m missing something obvious. Like I should know this. But hey, if I knew everything, I wouldn’t need all y’all.
Continuous Everywhere but Differentiable Nowhere 
Great question! I don’t have an answer, though… I’ve always used #2 (of course), but I agree with you that it would be very helpful to the students if I could show them that it doesn’t always work out.
To me, if just doesn’t seem logical to assume they’re equal when that’s what you’ve trying to prove. But if every step works backward and forward, and your assumption gets you to something simple and right instead of a contradiction, then .. yeah, maybe it is a reasonable way to proceed. So what we’d need for our counter-example would be something that uses a step that doesn’t work backward and forward. Hmm…
Of course part of the problem is that the books say “prove this”, instead of “prove or disprove”.
I must admit that I have never tried to justify the approach. However one thought is that an identity must be true for all values of x, and if, for example, 1-cot(x) is ever zero (which it is), it isn’t valid to multiply by it and then continue. It is even more clear that we can’t divide by anything that can take the value zero.
I’ll continue to think about this.
Judith
I don’t believe this is the case, I did just learn this in class 5 minutes ago and thats why i asked the internet but thats besides the point. I don’t believe the fact that it can be 0 would make a difference since you can have an equation where 1-cot(x) is on the bottom of the starting equation. This just means that the equation is undefined and obviously the it is not an identify(unless maybe the other side is also an identity not sure what happens with that). So I don’t think that what you suggested is true. I do however have a thoery which i will post in a little.
For a true trig identity, the “mixing” of sides will always work. To show the problem with the mixing of sides, you’d have to create an identity that isn’t true (by inspection, or substituting values) and then “prove” that it is an identity by mixing the sides somehow. So you’d end up with something that is obviously false, but the mixing of sides “proves” that it’s true.
I can’t come up with a good example right at the moment, though! Sorry.
Here’s how I explain it (kind of, I usually end up spending a fair amount of time on this issue over the course of the unit)…
It will always work, that’s because you are starting with a true statement. As you’ve already stated the problem is that you have been asked to PROVE it works, and instead you’ve built a circular argument showing that when you solve the equation it gives a solution of all real numbers. The problem is you don’t KNOW it is a valid equation; you are trying to PROVE it is a valid equation. I liken it back to proofs from Geometry class. If I want to prove that 2 lines are parallel, I can’t use the Corresponding Angle Postulate because that assumes the lines are parallel, even though the angles are congruent because the teacher was nice enough to give us a problem where the lines actually are parallel.
However, assuming it is true can give insights to the proof. One suggestion I make for constructing the proof is go ahead and treat it like an equation during your “rough draft” phase to see how things go. That can give hits on how to do the proof (especially for things like decomposing fractions, etc). But for your final answer, because you were asked to write a PROOF, you have to follow the laws of logic and make sure that your argument isn’t circular.
I guess basically what I am saying is that we only work with one side at a time because those are the constraints that logic puts on us.
We have the same issue with geometry proofs (if students are actually thinking about them!). When we prove something, we tend to assume that theorems aren’t biconditional – and therefore we can’t just use the converse of a statement whenever we want. The problem is that most theorems kids see, particularly those involving algebra, are biconditional. If your students have experience with indirect proofs, this is one context to talk about why you can’t go in whatever order you want. I remember one of my colleagues setting up a problem about determining if an angle is bisected, where each piece of the angle had an associated quadratic expression. If you start by using the given straight angle in the diagram, you find two values of x, one of which creates a bisected angle & one of which doesn’t. If you start by assuming that the angle is bisected, you can show that the angle must be straight. The problem with this second method is that you never realize that there’s another valid value of x – which doesn’t create a bisected angle. Sadly, I don’t have access to the particular problem right now, but maybe that gives you something to think about.
I don’t think you can get easy errors, because all the operations kids are expected to do are tautologies everywhere each expression is well-defined (e.g., all values of x so that denominators aren’t zero). It’s hard for students, because they could arrive at a valid proof if they did the problem their way, then just wrote all their steps in reverse order starting from the last*.
Thus, both my suggestions are bad:
For demonstration purposes only, you could try a 1=0 proof, like http://www.math.hmc.edu/funfacts/ffiles/10001.1-8.shtml , starting with $$x/y = 1/1$$.
Or maybe, try an inequality? $${-1 \over -4} > {-2 \over -3} \\ \mbox{ but } (-1)(-3) \not> (-2)(-4) $$
* In fact, if after every line of their proof, students wrote $$ \Leftrightarrow$$ instead of the usual implied $$\Leftarrow$$, I think mathematicians would be happy.
Oops, wrong direction in my last line: if after every line of their proof, students wrote $$ \Leftrightarrow$$ instead of the usual implied $$\Rightarrow$$, I think mathematicians would be happy.
I think the answer is “this isn’t really a proof”. You are not trying to prove the equality–that’s given. You are trying to show that you can manipulate the trig identities on one side of the equations to morph them into the identities on the other side of the equation. The student is demonstrating that they understand how to use the trig identities appropriately.
In real life, the only time I’ve used trig identities is to simplify a complex formula, to make the complex formula more “user friendly”. I didn’t know, a priori what the new formula would be.
My thinking involves the age old trick of proof by contradiction. If the equation is initially set up as being false, then if we work the problem as though it is true (mixing both sides) it should end up not being valid at some point… But I suppose they won’t know its not valid unless they make some identity substitutions to get different things on each side. I’m not sure you could start with a false equation and wind up with something that works if you mix the sides. I’d like to see that counterexample, but just on initial thoughts I’m not sure it exists. I may have to mess around with it.
In some there is absolutely nothing wrong with cross multiplying to prove an identity since A/B=C/D if and only if AD = BC for real numbers A, B, C, and D provided B and D are not equal to 0. Since it is implied in the original claim that the identity is only true in the case that the denominators are not 0 (although this really should be explicit ie in the example you gave the claim is that the two sides are equal for all values of x except when x is pi/4 plus an integer multiple of pi) the proof is valid.
That being said, there are two reasons we generally don’t encourage students to do that. One is, as Katherine and others noted, we are taking advantage of a biconditional statement. Since later students will be asked to prove things using some statements that only work in one direction, this creates a really bad habit. In this sense to see a false proof, start with a false identity and multiply both sides by 0, or start with sin(x)=sin(-x) and square both sides. But then the matter is just being careful about what you are allowed to do to both sides. That being said, it is a poor habit that can be costly.
More importantly, though, is that the point of having students “prove identities” is not about proof. The point is to give them practice in “simplifying” expressions involving trigonometric functions, but students are not experienced in what is “simple” which is not a well-defined term and can change depending on context. So we provide the goal. Thought of this way, the whole point, is to work with one side. (Edward Barbeau’s book on Mathematical Fallacies, Flaws, and Flimflam has a nice discussion on this from Richard Askey , which I’m borrowing from extensively) Askey credits Mark Saul with using the idea of giving students two lists of expressions and asking students to pair them, and show that they are equal.
Wow, I can’t believe so many people commented. Thank you. I pretty much agree with most of what I’ve read and you’ve helped me think about this in different ways.
To be clear, I spend the first day having students graph both sides of the equation to decide whether they believe they are equivalent or not. On the second day (on Monday), we’re going to start saying “hey, if they look equivalent on their graphs, let’s see if we can prove definitively that they are equivalent.” (this is atcha @sue!)
@sam, I think I’m in agreement.
@kendall, that’s how I explained it to my class…
@Katherine, I’ve never though about through the “biconditional” lens… thanks!
@timteachesmath: “It’s hard for students, because they could arrive at a valid proof if they did the problem their way, then just wrote all their steps in reverse order starting from the last.” CHECK.
@lisa: I think it is a proof… Albeit a very simple one… at least the way I present it to them. I have them first graph both sides to see if they think it is true, and if their graph suggests it, I say “prove it.”
@groupact: I agree with both of your points. And it’s not about the “proof” aspect for me, mainly. But I want to be prepared if any of my more questioning/precocious students asks the question I’m sure I asked… WHY CAN’T WE PROVE IT BY MIXING BOTH SIDES? On a side note, I was a counselor at a math camp when I was in college, and that summer it was held at U of T, and Ed Barbeau was one of the instructors! I am excited to check out this book of his. Thanks for letting me know about it!
I say, don’t let them do it, but let that guide what they’re doing. If they want to “multiply both sides by (sinx – cosx)” then have them multiply by (sinx – cosx)/(sinx-cosx). Distribute across the numerators and leave the denominators alone. If it’s a path worth pursuing, things will “cancel out” later anyways or the denominators will match and you can just compare numerators.
We’re doing rational functions right now and I’m trying to hit hard that (x^2 – x)/(x-1) is not the same as x (because of the domain) and I think that’s the same issue here.
Personally, as a former mathematician, I would encourage cross-multiplying for these identities. There is no point to hobbling them in their proof techniques for artificial reasons. If you want them to manipulate trig identities for some other reason (like changing an integration of cos^2 x into an integration of cos 2x), then have them practice that manipulation directly, rather than giving them a false idea of what proofs are about.
I think Katherine (and others) nailed it here: it’s OK as long as the steps are reversible. Multiplying both sides by 0 will lead to a true statement, for example, but doesn’t prove anything about the original statement. Squaring both sides leads to similar but less severe problems; it proves the original things have the same absolute value but not that they’re equal.
I often had my students do the proof starting with the equation and ending with 1 = 1 or whatnot, and then simply number their steps in reverse order. Sometimes I’d ask them to write a short reason for each step as they worked back up the page to make sure that they saw that the reasoning had to flow up in that way. But of course we all discover and experiment and prove things working the “illegal” way (or we are careful to check that our steps are all of the if and only if variety)
The problem with cross-multiplying lies in the implied domain of the identities. Consider the identity
sin(t) / cos(t) = tan (t)
Notice that this equation is not true when t = pi/2, since “undefined” does not necessarily equal “undefined”. Is it really an identity, then? To say no might be considered nitpicking, since it’s an identity everywhere both sides of the equation are defined.
But now multiply both sides by cos(t).
sin(t) = tan(t) * cos(t)
Is this an identity? It’s still not true when t = pi/2, but now one side of the equation is defined at t = pi/2, and the other isn’t.
And when we multiply tan(t) and cos(t), we get the following result:
sin(t) = sin(t)
This is clearly an identity. Is it equivalent to the original identity? Well, this equation is true when t = pi /2, while the original was not. So they aren’t the same.
Does this mean you shouldn’t be allowed the cross the equal sign? I don’t know. But it is a relevant objection.
I think Josh hit on the explanation that can make the most sense to kids. It’s completely fine so long as you make sure the steps are reversible, but doing so can be a big pain. For example, multiplication is not necessarily OK because if you’re multiplying by something that might equal zero then you can’t reverse it. Same with squaring both sides. In your particular example, it’s OK: the equality doesn’t hold (it’s not even defined) if sin(x) = cos(x) or if cot(x) = 1, so your cross-multiplication is reversible because you’re never multiplying by zero (you aren’t even considering those cases). Thus, when reversing, you don’t divide by zero.
Squaring could give you a good, non-obvious example of a place where you could prove a false equality. The students could “prove” that cos(x) = cos(x + pi) if you square both sides first and then do some complicated manipulations. You could obscure the fundamental trick by doing some manipulations to both sides first, so that it’s not obvious that one side is the negative of the other. :)
So, maybe it’s a breath of fresh air to know that this actually is, on some level, OK — just so long as you’re careful!
I like groupact’s example (slightly modified)
-sin(x)=sin(x)
(square both sides)
(-sin(x))^2=(sin(x))^2
sin^2(x)=sin^2(x)
So did we just prove -sin(x)=sin(x)? Of course not, because it’s not true. But we ended up with a true statement… so a true statement as a result of manipulating both sides doesn’t mean what you started with was true.
I guess I’m just repeating what’s been said here several times about being reversible, but I think the above example would help students tremendously because it is simple, fast, and to the point. Thanks to everyone for helping me understand this better!
Perfect. I also was trying to come up with an example involving squaring, because I knew part of this dealt with reversibility (in addition to the multiplying/dividing by 0 issue), but I kept on coming up blank.
But this – so obvious, so simple!
As for Patrick’s discussion about implied domain, I think Hung-Hsi Wu describes the issue quite well ( see pages 12-14) He writes:
My thinking on this is similar to groupact.
If you want some variety, do the same problem twice. First start on the left side, then the right. If you want even more variety, alternate sides after each step – make a change to the left, then the right, then the left, etc.
I like to use more familiar equations first. Can you prove the identity: 2(x – 3)^2 + 5 = 2x^2 – 12x + 23? Can you prove it by only working on the right side of the equation?
I think it’s terrible to teach proofs this way (going from what you want, to something true, even if the steps are reversible). Contrast it with epsilon-delta proofs, where that type of thinking must be stamped out.
The way I teach epsilon-delta proofs could apply equally well to trig identities. I show them two separate things: my “scratch paper” (in a separate box) and my “actual proof”. In the scratch paper, I start with what I want and work backwords. Then in the proof, I start with whatever tautology and work forwards. The result is more logically coherent and actually agrees with the way mathematicians actually prove things in journals. Plus you can joke about how if the students show their proof to someone without showing their scratch paper, they’ll look like amazing geniuses.
When we use an equals sign (=) with trigonometric identities, we really mean that both sides agree except on a set of measure zero (specifically a finite union of lattices). Even in your first example,
, the RHS is undefined when 
unless one redefines the RHS using a limiting procedure. This is similar to birational equivalence in algebraic varieties…or more simply to canceling when simplifying rational expressions. Why not use the wealth of available symbols to indicate what you mean? For instance, a real identity (true at all points) could use “identical to” (≡; U+2261). Equal except on a periodic set (as above) could be “equivalent to” (≍; U+224D) or “pretty darn close to” (≅; U+2245 or ≈;U+2248).
As for the proving
or 
issue, I would argue that it doesn’t matter. To emphasize that students must be sure that 
, i.e. that their implication is going in the right direction, why not make this a useful lemma named after the student who first wanted to try it? Try to prove it and figure out what conditions would be sufficient. Of course, all this should be done in the context of the expanded meaning of the equals sign (=) or one of the alternate symbols from the previous paragraph.
To underscore the need for a different notion of equivalence, try adding
or a similar expression to either side of a so-called trig identity and look at the domains of each side.
Late tot he party here and I am not sure I have much to add, but I do want to make a vocabulary point here. When we multiply each side by some quantity what we are relying on is the multiplication property of equality. Maybe this is too picky on my part, but I want my cherubs to really think about this name. This property tells us something we are allowed to do to equations. The theory behind trig identities is that we don’t yet know that it is an equation. (I also make a big deal about the vocabulary of conditional versus identity equations here)
Sam – I love the fact that you start with graphs of both sides.
Jim Doherty
In response to Jim’s last comment, I don’t really agree: we don’t know that it’s an identity, but it has an equals sign so it is an equation. What we’re trying to do is establish that the solution set is (almost) all real numbers. So we can certainly manipulate it like any other equation. The only issue is that we might introduce extraneous solutions, just like we do in solving square-root equations when we square both sides. There, rather than keep careful track of things, the usual procedure is to only do things that *increase* the number of solutions, and then check them when we’re done and discard the ones that didn’t solve our original equation. With proposed trig identities, this procedure won’t work, because we would have infinitely many solutions to check, so we need to be more careful with our steps.
Hi Sam: This is from my colleague, Kevin Weis:
It’s a lovely explanation. read it all.
I prove that -1=1.
I start with the statement that -1=1. Then I square both sides. This gives 1=1. Proof complete.
The “thing that goes wrong” is that squaring both sides isn’t reversible. And so I can’t follow the logic of the “proof” backwards to actually prove that -1=1.
For an example involving cross multiplying: In order for it not to “work” we would need the cross multiplying process to be not reversible like squaring both sides is. This could happen if the thing we multiplied by was, say, equal to 0. Most of the time cross multiplying is reversible which is why you are having problems finding an example that “doesn’t work”.
So how about this:
Prove that: (tan^2(x)-sec^2(x)+1)/(cot^2(x)-csc^2(x)+1)=sin(x)/cos(x)
This is not true. The left hand side of the equation is “equal” to 0/0 everywhere and thus undefined. However, when we cross multiply and use some Pythagorean identities we end up with 0=0, which is obviously true.
The truth of the matter is, we are allowed to work with both sides of the equation we are trying to prove as long as the things we do to both sides are reversible. For example if I am asked to prove the identity sin(x)=cos(x) (which is not true), I could multiply both sides by 2 and instead prove that 2sin(x)=2cos(x). I would have to explain in my proof why proving 2sin(x)=2cos(x) suffices to prove the original identity but it does suffice. I find that students do have difficulty understanding this (especially in the 11th grade) and so I usually “outlaw” it. That said, when I have some strong students in the class that are able to follow the logic of this, I do allow them to be more creative.
Hope that helps.
Kevin Weis
This. Here. This is what I need. It helped me and will make it understandable to kids who push (if they are any). Thanks!
The 0/0 technique is brilliant. Thanks!
@Josh
I am processing your idea, I guess the difference is that I have always treated the equals sign as more of a simple divider rather than an equals sign. In fact, I usually write a question mark over it. Then, if we can simplify each side so that they are identical I conclude that it is not ‘just’ an equation, but one that is true for all values in the domain. Does this make some sense, or am I misleading my cherubs?
@mrdardy I guess it depends on how differently you want to teach “proving identities” and “solving equations”. I think your approach is valid if you’re approaching them as really different questions: “Is this true for all x in the domain?” vs “For what x is this true?” But you can see that these questions are also in some sense the same; you could ask “For what x?” and if it turns out to be the complete domain then you’ve answered the other question as well. So I prefer treating all equations as being fundamentally the same thing, with only the questions we ask about them being different, but I can see that you don’t need to approach it that way.
Yay! I’m glad it helped. Kevin teaches many honors-level classes, and has been “battle tested” with very precocious and inquisitive kids.
I think it’s important for us to remember that often in calculus and algebra we solve equations by a solving an equation that is NOT equivalent to the original, and then “checking our answers.”
Consider: sin(x)+cos(x) = 1, or sqrt(4-x) = 2x. These can yield extraneous solutions. We square both sides, solve, check answers, throw some out. When kids “call us out” about why this is OK, they are justified to demand a satisfying answer. The idea that we give them grief about working on both sides of a possible identity rings as inauthentic, when we somehow go against it later on.
Hi. Groupact above reminded me of this problem: : Is it possible to find functions f(x) and g(x) such that f(x) ≠ g(x) for infinitely many values of x, but yet f(x) = g(x) is an identity? (I think this is a trick question.)
Let f(x) = 0 for x rational, and f(x) = 1 for x irrational.
Let g(x) = 0 for x rational, and g(x) undefined for x irrational.
Jerry Tuttle
onlinecollegemathteacher.blogspot.com
What a wonderful conversation! (Not that I managed to read everything … )
My view is that when it comes to working with equations — whether solving them, or establishing an identity, or using them to show how an expression was evaluated — the words surrounding the equations are just as important as the equations themselves. The words provide the reasoning and logic that connect the equations. With that perspective, it will be perfectly reasonable to cross-multiply (i.e., multiply both sides of the equation by the product of the two denominators) as long as the words make clear what the reasoning is. Here’s one way to do it:
sin(x)/(sin(x) – cos(x)) = 1/(1 – cot(x)) [Equation 1]
sin(x)(1 – cot(x)) = 1(sin(x) – cos(x)) [Equation 2]
sin(x) – sin(x)cot(x) = sin(x) – cos(x) [Equation 3]
sin(x) – cos(x) = sin(x) – cos(x) [Equation 4]
We want to prove that Equation 1 is true for all values of x for which the two expressions are both defined. So now let x be such a value. Then Equation 1 is true if and only if Equation 2 is true because Equation 2 is obtained from Equation 1 by multiplying both sides by something that is not 0 (namely the product of the two denominators to the left and right of the equal sign). Equation 2 is true if and only if Equation 3 is true, which is true if and only if Equation 4 is true because the expressions in Equations 3 and 4 are equivalent to the expressions in Equation 2, having been obtained by applying properties of arithmetic to the expressions, and by definition of cotangent. Since Equation 4 is true, we have established the identity in Equation 1.
Even though equation 4 is true for more values than Equation 1, this doesn’t matter because we restricted the values of x under consideration from the get go. Now I know that nobody really wants to write all those words. But the point is that unless one has those words in mind — or some others that make a logical argument — it’s probably not at all clear what one is doing when manipulating the equations.
I’m going to post something about this on the Mathematics Teaching Community, https://mathematicsteachingcommunity.math.uga.edu which is a forum for all of us who teach math at any level. I think we need to connect and communicate across all levels of math teaching — and learn with and from each other! I invite all of you to join the community and post there too — including links to blog posts.
I apologize if I’m being redundant, as I only scanned the first 15 or so comments, but here’s my take:
We aren’t asking the kids to prove or disprove. We are starting with something that we know to be true. This is a fault in our problem generation, but that doesn’t mean that we should approach the identity with any less caution than we might approach an identity that is not true. In my opinion, the reason that we always have students prove true identities is that they are novices in the area and we have just given them 15 or so new tools to use. Many students already spend a great deal of time chasing their tails while proving identities and we want to foster a solid understanding of the ways in which we can manipulate trigonometric equations to get what we want. Thus, providing a true identity gives the students a much better idea as to whether they are making “good” moves.
Just because we have sacrificed “prove or disprove” to help students become more comfortable with the identities does not mean that we should also abandon the caution we would normally take if we didn’t already know the identity was true. I believe crossing the assumed equal sign fosters bad habits in the students (and a lack of understanding the difference between manipulation of an equation vs. the manipulation of an expression).
Ryan, I totally agree with you, and I like this point about their confusion between how to handle equations versus how to handle expressions. So we come back to Sam’s original question – how do we explain to the kids what’s wrong with them doing it that way.
But I think that is the explanation – We don’t cross the equal sign because we have only conveniently chosen identities that are true and we want to prepare ourselves for any case where they are not. Granted, I don’t believe most students will ever encounter a trig verification that is not true, but that doen’t seem like a compelling reason to do away with the idea that the equality is not guaranteed.
In order to aide my students in this, I often write a little question mark over my equal sign in order to indicate that we aren’t sure it is a valid equal sign (and this serves as a visual aide that they should not cross it).
Allow me to also go on a tangent here – I came from a very small highschool that didn’t even offer a Calculus class and went on to major in mathematics at the U of I. I mention this because I often felt like I was blindsided by what was happening “behind the curtain” in many types of problems. I was taught mostly algorithms and tricks, and it has become clear to me that my teachers were more interested in the path of least resistance than the path of enlightenment. I often find myself wishing that it had been the other way (and this is one of those cases, as I struggled a great deal at first in my formal proofs class)
I’m not sure I understand your explanation Ryan. Do you also have students use an equal sign with question mark that you can’t cross when solving equations? When solving an equation we also don’t know whether the equality is valid. In fact, the goal when solving an equation is to try to find values for which the equality is valid. The only difference with verifying an identity is that you’re expecting to show that the solution set is a “large” set of numbers.
Again it is worth noting that cross-multiplying, or adding the same thing to both sides of the equation is a logically valid way of proving an identity. In general one way to prove a statement P is to show that P is true if and only if Q is true, and then show that Q is true. This is also valid to try to determine whether the statement P is true. We show that is true if and only if Q is true and then try to determine whether the statement Q is true. For example, suppose we asked students to “verify” the (false) identity cos(x + π)=cos(x) + cos(π). One could rewrite the left side as -cos(x) and the right side as cos(x)-1 to get -cos(x)=cos(x)-1. Next add cos(x)+ 1 to both sides (crossing the equal sign!) to get 1=2cos(x), divide both sides by 2 (again crossing) to get one-half=cos(x). I can then see that this is not valid for most values of x. In fact, it is only valid for plus or minus one-third π plus integer multiples of 2π.
I still maintain that the reason we ASK students not to verify identities in that manner is because (1) we are trying to get them to learn how to manipulate expressions involving trig functions, and (2) it’s too easy for students to neglect the importance of the IF AND ONLY IF aspects of the solution and to think incorrectly that in other situations may prove a statement by assuming it is true and implying another true statement from that.
When we are solving equations, we are on a different mission, if you will, which is to show that there exists a value that makes the statement true. To me, this is very different from seeking to show that the statement is true for all values. So no, I do not have them use a question mark over an equal sign for solving. I realize that this is a false construct for the students.
I don’t think we’re disagreeing about whether the moves are logically valid. I think the point here is that the students haven’t reached a place where they are able to identify when steps are logically invalid. Part of this is a lack of exposure to formal logic, although I’m sure there are other factors at play here as well. This is not to say that there aren’t any students capable of understanding this at a highschool level, but rather that in general instruction, the safer assumption (at least from my experience) is that they don’t understand. Since the logical validity (and formal logic in general) is not the goal of a PreCalculus class, we provide constructs that allow us to avoid the stickier details of what’s happening. Some of these constructs are as simple as “Always check your answers” and others are a little more intricate like “Don’t cross the equal sign”.
That said, I think we are providing the exact same reasoning for our instruction – experience with manipulating trig functions and avoiding pitfalls of a logical system they don’t fully understand.
It seems to me that some people are saying that “not crossing the equal sign” is the equivalent of electricians working with one hand in their pocket, to keep from doing something dangerous that would allow large currents to flow through their hearts and kill them. Others seem to be saying that it is like typing with one hand—possible, but a silly thing to do unless absolutely necessary.
Personally, I lean more towards the latter view. Teaching a rule that really does not provide much safety and is very limiting, rather than teaching how to manipulate identities safely, does not seem to lead to a desirable end: students who can confidently and correctly use, manipulate, and prove trig identities. I think that proving identities should be viewed as being the same problem as solving equations: the proof of an identity is determining for what values it is true (and often the “identities” are true for almost all, rather than all values, an important distinction that is lost in the blind “don’t cross the equals sign” rule).
Ive Been Trying To Prove Identities but i Cant get the correct answers so is there anyone who can help me please?
You can cross multiply or mix the two sides of a trig identity.
The only reason not to do it is that you’d get marked wrong by a teacher. The two arguments cited by textbooks make no sense, sorry to tell you that.
As long as you don’t multiply both sides by zero or add infinity to both sides, you will be ok. Also one has to use reversible steps.
For example, prove
(1-sinx)/cosx=cosx/(1-sinx)
If this is true, it can only be true for x’s not equal to {(2k+1)Pi/2}
Therefore Cosx is not 0 and sinx is not 1.
So you can multiply both sides by cosx(1-sinx) (so cross multiply); this is an allowed reversible step, since we are multiplying by a nonzero expression.
The equivalent expression is cos^2(x)=1-sin^2(x), which is obviously true, so therefore Te first identity is also true, but only on the set R-{(2k+1)Pi/2}
Another example: You are not allowed to multiply both sides by 1-sin^2(x)-cos^2(x) for instance, since this would be multiplying both sides by zero.
So the rule is there to prevent students to make mistakes, but this is a bad reason.
Also, many other countries do not use this “rule”. Real math book should explain everything and Pearson doesn’t.
You are right to be uneasy about this.