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Lets say I have an equation like: $$\frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$$

Our teacher said you have to verify the equality by simplifying the left hand side or the right hand side without moving(doing operations to both sides) anything from one side of the equation to the other. I have a problem with this because I proposed that we can verify the equality by multiplying the denominator by 1 and moving it to the other side. Then you can proceed to verify the equation by substituting trig identities. My argument is that if correct algebraic manipulation is done, then the equality is the same. It won't suddenly become false or true.

My method: **the addition to the classical method is that you can multiply the equation by both sides while preserving the equality which is specifically not multiplying by 0 or applying operators such as derivatives or squaring etc as these do not preserve equalities.

$$\sin^2x+\cos^2x = \cos^2x·\sec^2x$$ $$1=\frac{\cos^2x·1}{\cos^2x}$$ $$1=1$$ equality is true

It requires ingenuity to be able to do the right manipulations...

Is my teacher justified in requiring that nothing is moved between the sides? is there a convincing argument for my side or am I wrong? I would be grateful if someone can enlighten me; just why is it wrong to do algebraic manipulation before verifying an equality? Keep in mind that this method is meant to be extended to the more complicated problems. I just gave a simple example to illustrate it.

For reiteration and clarity, my question is why is my method wrong? What evidence is available to back up any certain answer?

Bandoo
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  • Don't think about it as "moving". Think about it as performing an operation to both sides of the equality. Here, "moving the denominator to the other side" would be "multiplying both sides of the equality by the denominator". – wjmolina Mar 17 '19 at 16:40
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    I'd advise listening to your teacher. – Angina Seng Mar 17 '19 at 16:41
  • Yes that is what I meant, i just used what the teacher exactly said – Bandoo Mar 17 '19 at 16:41
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    Why should I listen? Could you elaborate why my method is wrong for verifying an equation? – Bandoo Mar 17 '19 at 16:42
  • Your method is correct if the steps are reversible. If you multiply both sides by a nonzero expression, for example, then the original and transformed equations are equivalent. I've always thought the reason they tell you not to consider both sides of the equation is that in real life, you're trying to simplify the left-hand side with yet knowing the right-hand side. – saulspatz Mar 17 '19 at 16:57
  • @saulspatz: you might want to look at the final example in my answer; I don't think that "nonzero expression" is unambiguous enough in what you wrote. :) In my example, $\sin x$ is certainly a nonzero expression. I'm not inclined to get more specific, because I think that it might help OP to actually look at the subtleties of that example. – John Hughes Mar 17 '19 at 17:49
  • In the absence of an explicit signal (like ⇐ or ⇔) to read your sequence of equations in the reverse direction, it is natural to interpret your presentation as "A implies B, which implies C, which implies 1=1", which is certainly not a proof of A. I wrote more here and here. – ryang Jun 26 '23 at 02:20

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Consider the equation $1 = -1$. Allow me to prove this equation by performing the same operation on both sides, namely squaring. Hence, $1^2 = (-1)^2$ or $1 =1 $. Therefore, the first equation must be true. Does this seem like sound logic? The key observation here is that a false statement can be turned into a true statement if you abuse properties that are true for true statements. "Do the same thing" to equal values does indeed preserve their equality, but "doing the same thing" to unequal values can make them "equal".

You may find this related queation enlightening on on the logic consequences of how you proceed with problems such as this.

Brian61354270
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  • In this case, you are squaring both quantities which is not doing the same operation to both sides of the equation because on one side it is multiplying 1 and on the other its multiplying by negative one. I am still confused – Bandoo Mar 17 '19 at 17:00
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    The operation of squaring is raising the base to the exponent of 2, the same operation on both sides – user650025 Mar 17 '19 at 17:12
  • No it is not an operation... https://math.stackexchange.com/questions/568780/why-cant-you-square-both-sides-of-an-equation To save you some of the reading, if $x=1$ then the only solution is $x=1$ but for $x^2=1$ then you introduce new solutions. – Bandoo Mar 17 '19 at 17:20
  • @Bandoo would you mind defining what an "operation" is? The interpretation that I, and presumably user650025, have is any mapping between real numbers. One you have a valid equality, it is perfectly valid to apply a mapping to both sides of an equation, since, by definition, two equal input values will be "mappped" to the same output value. The issues that you correctly raise is that not all mappings are injective, meaning that for some mappings, such as squaring, two input values can be "mapped" to the same output value. This can result in a false equality becoming true, as shown in my answer – Brian61354270 Mar 18 '19 at 14:50
  • @Bandoo In my answer, I purposely omitted logic symbols since they seemed mostly irrelevant to the essence of the argument. You will note that since squaring real numbers is not an injective mapping, the transition between $1 = -1$ and $1^2 = (-1)^2$ is only a logical implication ($\implies$), not an equivalence ($\iff$). Hence, we can say that "if $1=-1$, then $1^ 2= (-1)^2$", but not the other way around. This is what is meant by using only "reversible" steps when proving an equality. – Brian61354270 Mar 18 '19 at 15:02
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I think your equality should be $\displaystyle \frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$.

When I read your work, I naturally interpret it as

$\displaystyle \frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1 \implies \sin^2x+\cos^2x=\cos^2x\sec^2x \implies 1=\frac{\cos^2x\cdot1}{\cos^2x} \implies 1=1$.

It does not make sense. I already know that $1=1$, but it is not because of $\displaystyle \frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$. $\displaystyle \frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$ is what you want to show and you should not start from it.

The correct sequence should be

$\displaystyle 1=1 \implies1=\frac{\cos^2x\cdot1}{\cos^2x} \implies \sin^2x+\cos^2x=\cos^2x\sec^2x \implies\frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$.

Let me 'prove' that $ \sin^2x+ \cos^2 x=2$.

$ \sin^2x+ \cos^2 x=2 \implies (0)(\sin^2x+ \cos^2 x)=(0)(2) \implies 0=0$

equality is true.

Can you see the danger of the approach?

CY Aries
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  • You don't know 1 = 1...the question is to VERIFY the equation so it could turn out to be false. Your method assumes 1 = 1 with all certainty. – Bandoo Mar 17 '19 at 17:02
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    To prove or to verify something is to deduce something you want to know from something you already know. The statement to be verified is something you want to know (whether it is true or not) and should appear as a conclusion but not the starting point. And of course, we know that $1=1$. – CY Aries Mar 17 '19 at 17:07
  • The way this is taught in all textbooks and other media always ends at 1 = 1 because it only simplifies one side at a time(with the underlying goal of making both sides equal to each other). In the aforementioned 'classical' method, no operations are done to both sides of the equation just like when you solve equations, instead, substitutions from trigonometric identities. You started your proof with 1 = 1; whereas the starting step should be what the question asked:$\frac{sin^2+cos^2x}{cos^2sec^2}=1$ What I offered is a variation to the classical method. – Bandoo Mar 17 '19 at 17:08
  • Actually I won't start from $1=1$. I can start from $1=\frac{\cos^2x\cdot 1}{\cos^2x}$. – CY Aries Mar 17 '19 at 17:13
  • Yes that is the classical method and you are going to follow as any textbook or teacher would do. However, I still do not understand your argument for why method is incorrect. – Bandoo Mar 17 '19 at 17:15
  • As saulspatz pointed out, your approach is fine if the steps are reversible. And your steps are actually reversible. That's why I write your work reversely and obtain a correct proof. In those textbooks that verifying statements by works ending with $1=1$, the steps should be reversible. I won't say this approach (or your work) is totally wrong, as the steps are reversible. If one of your steps is not reversible, you will be in trouble. As said in Brian's answer. – CY Aries Mar 17 '19 at 17:19
  • Multiplication by 0 is the only exception. I should edit that into the main post. – Bandoo Mar 17 '19 at 17:25
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    Multiplying both sides by $0$ is not reversible. Squaring both sides are also not reversible. That's why these steps cause problems. If you restrict your steps to direct application of identities, adding or subtracting some quantities on both sides, multiplying or dividing some nonzero quantities on both sides, you will not obtain an absurd conclusion. That's why I said that the approach is not totally wrong. But you can't only restrict yourself to these simple operations if you are dealing with more complicated problems. The best way is to write your work in a logical sequence. – CY Aries Mar 17 '19 at 17:34
  • What is wrong with such restriction? For more complicated problems,I can multiply $cosx$ by both sides if x represents a specific real quantity – Bandoo Mar 17 '19 at 17:41
  • you will be in trouble as $\cos \frac{\pi}{2}=0$. – CY Aries Mar 17 '19 at 18:01
  • Yes. But that does not matter unless you are dividing by cosine, for which the domain is restricted. This is exactly the same issue brought up by @john hughes – Bandoo Mar 17 '19 at 18:06
  • You can't multiply both sides by $0$. "Multiplication by 0 is the only exception." – CY Aries Mar 17 '19 at 18:07
  • There is no exception being made here. It is just being mindful of the domain restrictions on any equation. x/sin(x) is always restricted so that x cannot be 0. Obviously you have to take into account that when trying to verify an identity – Bandoo Mar 17 '19 at 18:11
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    If you are careful enough to avoid all the problems, then you will not be in trouble. You work is actually $\frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1 \iff \sin^2x+\cos^2x=\cos^2x\sec^2x \iff 1=\frac{\cos^2x\cdot1}{\cos^2x} \iff 1=1$. but we tend to read the work line by line and intepret it as $\frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1 \implies \sin^2x+\cos^2x=\cos^2x\sec^2x \implies 1=\frac{\cos^2x\cdot1}{\cos^2x} \implies 1=1$. It is a question of bad style rather than a logical mistake, if you are carefully enough. – CY Aries Mar 17 '19 at 18:23
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As you observe, multiplying by zero is the only unreversable step. (Well...that's not quite true, but close enough.)

The problem with applying this in the context of proving trig identities is that it's sometimes tough to know that you are multiplying by zero.

For instance, you might multiply both sides by $2\sin^4 x + \sin^2 (2x) + 2 \cos^4 x - 1$. Are you multiplying by zero or not? As it turns out, you're multiplying by zero in this case ... but you don't know that until you've proved another trig identity, namely that $$ 2\sin^4 x + \sin^2 (2x) + 2 \cos^4 x - 1 = 0. $$ You can see how this can lead you down a rathole pretty fast.

So your logic (with appropriate restructuring) is fine: if you apply an invertible operations to both sides of your supposed equality, and the resulting equality is true, then so was your supposed equality.

But in practice, it's almost useless, because especially in the context of trig identities, it's tough to know whether a given function is injective or not. In particular, multiplication by some trigonometric expression may turn out to be multiplication by zero, and then it's "game over."

Let me give one last example for you to consider: $$ \frac{\sin (2x)}{\sin x} = 2\cos x $$ If we multiply both sides by $\sin x$, we get $$ \sin (2x) = 2 \sin x \cos x $$ which is true. But the original statement is false. Why? Because at $x = 0$, the left hand side is undefined, while the right-hand side is $2 \cos 0 = 2$.

Notice that in this example, I did an apparently reversible operation (multiplying by $\sin x$, which is certainly not the 0 function!) and derived a true statement, but the original statement was not true.

When you can carefully explain this example, and how you might avoid situations like this one, you can probably safely start ignoring your teacher's advice.

John Hughes
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  • The first equation needed the domain to be restricted because you cannot divide by 0. sin(2x) = 2sin x cos x is only true if :ℝ. But I would like to thank you for giving such example. While it does not contradict my logic, it shows a reason for why my methodology can be impractical even in completely theoretical practice. – Bandoo Mar 17 '19 at 17:58
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As I think, you have obviously removed $\cos^2 x\sec^2 x$ from LHS to RHS (i.e. cross-manipulated the both sides) which may make you teacher object to you! A better and more standard way (to fit your teacher's will) is that simplify $\cos^2x\sec^2x$ and substitute it with $1$ (except when $\cos x=0$ since ambiguity happens) and get onto an always-true statement $\sin^2x+\cos^2x=1$. Therefore the answer is $$\Bbb R-\left\{k\pi+{\pi\over 2}\ |\ k\in\Bbb Z\right\}$$

Mostafa Ayaz
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