Proving trigonometric identities can be a major challenge for students, as it is often very different from anything they have previously done. Often they confuse this concept with solving an equation. But also, they may be give overly rigorous standards to comply with. Here, I will look at several discussions we have had about different styles of proof (or verification), and how to change a proof initially obtained in an unacceptable style, to one that will please the teacher.

## A general method

I’ll go through these in chronological order, starting with a broad question from 1999:

Proving Trigonometric Identities Can you teach me how to prove trig identities? I don't know where to start or what to do exactly.

Doctor Rob gave an appropriate introduction to the topic, giving an overview of proof styles, and then listing a general procedure. First, though, he points out that there is usually more than one way to do it, and that it takes experience to develop the skill:

There are usually many ways to prove trigonometric identities. Some are short and very elegant. Others are longer and more tedious, but any proof should do. The shortest proofs involve *pattern recognition*. In the equation you are trying to prove, you detect a pattern that appears in one of the standard identities you know, and that allows you to make a substitution which simplifies the identity a lot. There is a knack to this which is a bit hard to teach, but it comes with practice. See the following web page for a list of standard identities: http://mathforum.org/dr.math/faq/formulas/faq.trig.html There is a systematic way to produce a proof, which may be fairly long, but still valid. I will describe it below.

First, you can work on each side, separately but simultaneously (more on this later!):

A proof of an identity is often constructed by starting with one expression on the lefthand side of the equation and another on the righthand side which you want to prove as equal. Youmake substitutions on both sidesuntil they are both reduced to the same identical expression. Then, since the last equation is identically true, you canreverse the stepsto conclude that the first equation that you were trying to prove is also true.

The reversal is necessary so that you are starting with a known fact and working toward the new, rather than starting with something you don’t know to be true.

Second, the work can instead be arranged as a single chain of transformations, from one side to the other:

Alternatively, you can rearrange your work to start with the lefthand side of the first equation, work down thechain of equal expressionson the lefthand sides until you reach the bottom, then work up the chain of equal expressions on the righthand side until you are back at the top.

He gives an example, using the first format and then the second:

Here is an example: cos^4(x) - sin^4(x) = cos(2*x) [cos^2(x) - sin^2(x)]*[cos^2(x) + sin^2(x)] = 2*cos^2(x) - 1 Here wefactored the lefthand sideas the difference of two squares, and weused the cosine double-angle formula on the righthand side. [cos^2(x) - sin^2(x)]*1 = 2*cos^2(x) - 1 Here weused the identity cos^2(x) + sin^2(x) = 1. cos^2(x) - [1 - cos^2(x)] = 2*cos^2(x) - 1 Here weused 1 - cos^2(x) = sin^2(x). 2*cos^2(x) - 1 = 2*cos^2(x) - 1 Here we justexpanded and combined like terms. Now we have an obviously true equation at the bottom. To prove the original identity, we can justreverse the steps, (1.) 2*cos^2(x) - 1 = 2*cos^2(x) - 1 (2.) cos^2(x) - [1 - cos^2(x)] = 2*cos^2(x) - 1 (3.) cos^2(x) - sin^2(x) = 2*cos^2(x) - 1 (4.) [cos^2(x) - sin^2(x)]*[cos^2(x) + sin^2(x)] = 2*cos^2(x) - 1 (5.) cos^4(x) - sin^4(x) = cos(2*x) and supply the reasons.Alternatively, we can use this form: cos^4(x) - sin^4(x) = [cos^2(x)] - sin^2(x)]*[cos^2(x) + sin^2(x)] = cos^2(x) - sin^2(x) = cos^2(x) - [1 - cos^2(x)] = 2*cos^2(x) - 1 = cos(2*x) by working down the left and up the right of our original derivation, and supply the reasons.

Finally, he lists specific steps you can take (not all of which are required):

Here is a systematic way to produce a trigonometry proof: 1. Convert all cosecants, secants, cotangents, and tangents to expressions involvingonly sines and cosines. The first group of five identities on the above web page will allow you to do this easily. That will give you a "simpler" equation to prove. 2. Look at all the angles in the sines and cosines in the new equation you are trying to prove. If any is a sum or difference of angles, use a sine or cosinesum-of-angles formula, as in the 9th and 10th groups of identities from the Dr. Math FAQ web page. That will give you a "simpler" equation to prove. 3. Look at all the angles in the sines and cosines in the new equation you are trying to prove. If any is a multiple of some angle, use a sine or cosinemultiple-angle formula, as in the 11th through 16th groups of identities from the web page. Likewise for half-angles and the 12th group of identities. That will give you a "simpler" equation to prove. 4.Expand all the expressionsin sight, combine like terms, and simplify. That will give you a "simpler" equation to prove. 5. Replace all occurrences of thesquare or higher power of a cosineusing the identity cos^2(u) = 1 - sin^2(u). Expand, combine, and simplify again. That will give you a "simpler" equation to prove. 6.Factornumerator and denominator if possible andcancel common factorsif any. That will give you a "simpler" equation to prove. 7. At this point, you should have an equation that is obviously true, of the form E = E, where E is some expression involving sines to powers and possibly some cosines to the first power. 8. From this sequence of equations,construct a proofof the original equation using one of the two methods described above: either working from thebottom up, orelse working from thetop left down, over, and up, to the top right. I did warn you that this was tedious! It will work, however, in almost all cases, and with minor variations in essentially all cases. Be sure to do the algebra correctly when you are doing "expand, combine, simplify, and factor" operations!

## Can you work on both sides?

The next question, from 2002, questions a previous answer (which I can’t find – possibly it was really on another site) in which we worked on both sides of an identity in a different way than Doctor Rob did. He rewrote *each* side, keeping each side’s value unchanged, but never used the common solving technique of doing something to *both* sides (changing their values, but in the same way).

Proving Identities Rigorously One submission in the Dr. Math archives was asking about (1-tan A)/sec A + (sec A/tan A) = (1+tan A)/(sec A tan A). One of the Doctor Maths answered it by saying to multiply both sides by sec A tan A. I do these two-sided operations frequently in math, and after I did VERY badly on my identities test, my teacher wrote"You can only work with one side at a time"beside my work.I was wondering why you work with both sides.I've been taught that the "proper" way is to work with one side only.

Doctor Pete went through that proof as described, showing how it was actually valid despite not following the usual rules. First, an introduction:

You bring up a very good point, and you are correct: the proper way to prove such identities is tobegin on one side and algebraically transform it into the form shown on the other side. Working on both sides is technically incorrect because in doing so we areassuming that the equality to be proven is already true. However, in the case of many identities, it is allowable to work on both sides, in the sense that it is not mathematically incorrect - if the identity is true, then the result of working on both sides will eventually result in an equality that is "obviously" or more easily seen to be true. That is, such a method won't lead to a wrong decision on the truth or falseness of the identity in question. But in so far as mathematical rigor, it is insufficient because of the reason mentioned at the end of my first paragraph.Often, however, working on both sides can help us form a rigorous proof of an identity. I will illustrate this with the example identity you provided.

In other words, working as if you are solving an equation can be, at the least, an exploratory activity.

Identity to be proven: (1-Tan[a])/Sec[a] + Sec[a]/Tan[a] = (1+Tan[a])/(Sec[a]Tan[a]) If we work on both sides, the first step is tomultiply both sidesby Sec[a]Tan[a]: (1-Tan[a])Tan[a] + Sec[a]Sec[a] = 1+Tan[a]. Multiplying through, we find Tan[a] - Tan[a]^2 + Sec[a]^2 = 1+Tan[a], or -Tan[a]^2 + Sec[a]^2 = 1, which is equivalent to the identity 1+Tan[a]^2 = Sec[a]^2, so we are done.This is not a rigorous proof, but it leads to one, in which we see the proper method is as follows: (1-Tan[a])/Sec[a] + Sec[a]/Tan[a] = ((1-Tan[a])Tan[a])/(Sec[a]Tan[a]) + Sec[a]^2/(Sec[a]Tan[a]) = (Tan[a] - Tan[a]^2 + Sec[a]^2)/(Sec[a]Tan[a]) = (Tan[a] - Tan[a]^2 + 1 + Tan[a]^2)/(Sec[a]Tan[a]) = (Tan[a] + 1)/(Sec[a]Tan[a]). Therefore we have taken the left-hand side and transformed it into the right-hand side using algebraic manipulation and other known trigonometric identities. But it should not be too hard to see thatthe basic "flow" of the proof is essentially the same procedure as the "improper" method, just rearranged a bit.

The exploration determined what identities are of use; then multiplying by the denominator was replaced by using a common denominator, leading to very similar work in the end.

Doctor Pete then gave another example that is not trigonometric, which I will omit.

## Working from both ends

Another question from 2002 just asked for help with a particular problem, which led me to discuss a general strategy that amounts to Doctor Rob’s alternative:

Bridging Trig Identities Hi...please help. I have been staring at this question for ages and I'm getting nowhere: Prove tan 2A x sec A = 2sin A x sec 2A My teacher told me that when proving these identitiesyou have to choose one sidewhich is what I have been doing. I started off by opening up the identity tan2A = 2tanA / (1 - tan*2 A) and then that didn't get anywhere so Ichanged sidesand opened up sec2A but that didn't help either.

I didn’t actually help with the details, but just with the one thing Susannah was probably struggling with: the fact that **neither side was simple**, so that it seems necessary to start on both sides. What I do here is not the same as Doctor Pete’s exploratory “solving”, as I am only simplifying each side separately:

Often when you are first working out a proof, it helps to work on both sides at once; then you can rewrite what you have done in a more proper way. This is sort of likeconstructing a bridge by building both ends toward the middle, but then driving across it in one direction. In this case, as in many, it helps a lot to write everything in terms of sine and cosine: tan(2A) * sec(A) =? sec(A) * sec(2A) sin(2A) 1 1 1 ------- * ------ =? ------ * ------- cos(2A) cos(A) cos(A) cos(2A) (Notice how I indicated that the "=" is not yet known to be true; that's my way to keep myself honest when I work this way!) Now you can apply a double-angle formula and do some canceling, and you will be able to make both sides look the same. Then you will have a real equation.

The rest of the exploratory work should be easy if she knows her identities. (By the way, you can see that I am using several of Doctor Rob’s steps.)

Nowhow do we walk across the bridgeand make it a proper proof? You will have written a sequence of equivalent equations of the form a =? b c =? d e = e (this one doesn't need the "?"!)As long as you have been working on each side separately, you can now say this: a b = = c d = = e = e That is, a = c = e = d = b and you have proved that a = b! Can you see now how we were working from both ends? We built the span on the left (a=c) and on the right (b=d) first, then the middle spans (c=e, d=e) and found we had connected the two shores. Then we just had to put it all in the right order and we were done.

## When a “bad” proof is valid

Finally, let’s look more deeply into a case like Doctor Pete’s sort of “exploratory solving”, and see how in fact, if explained properly, it can be a valid proof even without a rewrite. This question is from 2006:

Proving Trig Identities vs. Solving Trig Equations I need to prove that: sin x/(1 - cos x) - (1 + cos x)/sin x = 0. I am told that I need to show that the left side of the [equation] is equal to zero. I thought that I had done this so now I am really confused. Here's what I did: First: Use the additive property to add (1 + cos x)/sin x to both sides of the equation to give sin x/(1 - cos x) = (1 + cos x)/sin x. Next: Multiply both sides of the equation by sin x and also by (1 - cos x) which will give sin^2x = (1 + cos x)(1 - cos x) or 1 - cos^2x so we have sin^2x = 1 - cos^2x. Then: Use the additive property and add cos^2x to both sides, giving sin^2x + cos^2x = 1 This is the Pythagorean identity. It holds for all real values of x. This proves that the original equation is also true for all values of x except for where either sin x = 0 or where 1 - cos x = 0. This means that the equation is true except for where x = n(pi), where n is any integer. Sin(x) = 0 when x = n(pi) Cos(x) = 1 when x = 2n(pi) Therefore x =/= n(pi)

Angel has done just what you would do to solve an equation; the final result was a true equation. The teacher rejected this, because it starts with what could be a false equation, and derives from it something true. This is not how a proof works. But there’s more to the story:

I think you are saying that you submitted this proof, and were told it was wrong, without explanation. Is that right?It is basically valid, but is not in the form that many teachers require, namely working only on one side of the equation at a time to transform one side into the other. You should have been told that, rather than only being told that it is wrong, so that you would know how to correct it. What you have done DOES constitute a proof, IF you state it very carefully. You have done what you would do to SOLVE an equation, which is a somewhat different process.What you need to do here is to explicitly state that you are transforming the equation into an EQUIVALENT equation--one that is true when the original equation is true, and vice versa.This is not strictly true, since multiplying an equation by an expression containing the variable does not necessarily result in an equivalent equation, since that expression might be zero. You haven't ignored this problem, as many students would do, but have pointed out that you would have multiplied by zero only if the variable is a multiple of pi. The one thing you didn't mention is that in those cases, the equation you are trying to prove can't be evaluated in the first place, since you would be dividing by zero; so you have shown that it is true over its entire domain.With that one little exception, I think you've got a fine proof.The trouble, as I said, is thatmany teachers don't accept this-- because, I think, they want to avoid the subtleties involved in this sort of thinking (essentiallyprotecting their students from the errors that could creep in if they were not as careful as you) by only allowing them to take the "safe" path, which consists of a single chain of equal expressions. You can usually turn an "unsafe" proof such as yours into this kind of proof by sort of turning it on its side, doing to the left side only, the operations equivalent to what you did with the whole equation.

So although Angel’s work is excellent, it has to be redone for use in the typical class:

Here's how: Your first step just moved one fraction to the other side; we have to keep the whole left side together, so this can just be ignored. The second step is to multiply by the LCD. To do the same thing without changing the value of the expression, you can instead rewrite the fractions with a common denominator (the LCD), and then add them. That is, rewrite them both with denominator (sin x)(1 - cos x). Your third step just moved part back to the other side; the equivalent is just to simplify the combined fraction you now have. Then you can use the Pythagorean identity to simplify it further, replacing sin^2x + cos^2x with 1, and you will be done.

This is very much like Doctor Pete’s rewrite.

For another case in which two methods are used, see:

Trig Proof: Two Methods Prove: 1 + sin x + cos x 1 + sin x ------------------- = ----------- 1 - sin x + cos x cos x

An almost identical identity is proved here:

Proof of a Trig Identity 1 + sin x + cos x 1 + cos x ------------------- = ----------- 1 + sin x - cos x sin x

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