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My question is about the rules of proving trigonometric identities like $(\sin(x) + \cos(x))^2 = 1 + \sin (2x),$ rather than about particular proofs.

From what I've been told, you start with one side and by only using that one side you must obtain the other side. And this is how these problems are done. Do you have to do this for both sides for it to be a proof?

Are you allowed to move terms from the RHS to the LHS and vice versa?

What are the rules for proving trigonometric identities?

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  • $\begingroup$ There is no justification for any such "rules". From a computational perspective, it's probably better to transform $A = B$ to $A-B = 0$ and try to simplify $A-B$ to $0$, rather than working with only one side. $\endgroup$ Oct 26, 2017 at 0:08
  • $\begingroup$ I have downvoted as to me, it isn't clear what the question is asking. Also voting to close. $\endgroup$ Apr 26, 2023 at 11:42

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If you work only with the LHS and manage to simplify it to the point where it's identically equal to the RHS, then you're done.

Similarly, If you work only with the RHS and manage to simplify it to the point where it's identically equal to the LHS, then you're done.

But there are other options.

For example, you could simplify the LHS to reach an expression $A$, and then, if you manage to simplify the RHS to also reach $A$, then you're done, since you've proved both LHS and RHS are equal to $A$.

What you're not allowed to do is start with the claimed identity, and reduce the equation to some equation which is clearly true, unless . . .

Unless what?

Unless each step of the reduction is reversible. As long as the steps are reversible, you can start with the goal and try to reduce it to a known identity. If you choose this approach, to make the reversibility clear to the reader, each step should be connected to the previous one by the symbol $\iff$ (if and only if).

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You are allowed to move terms around while following commutative and associative rules (and other axioms in arithmetic). So yes, you are allowed to add or subtract or multiply by any terms you wish.

You should continue until both sides are definitely equal to each other. (for example, until you can deduce 0 = 0)

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The idea is to pick a side to work with and essentially 'scrap' the other side momentarily. I would, for instance, pick $(\cos(x)+\sin (x))^2$ in your example, and algebraically/trigonometrically manipulate the expression until you get $\sin(2x)+1$

However, it is equally valid to, say, prove that $(\cos(x)+\sin (x))^2 - \sin (2x) = 1$ as this statement and the original statement are equivalent. In this case, I would again fix the LHS and manipulate the expression until it equals $1$, then add $\sin(2x) $ to both sides to get the identity you originally wanted.

It is also valid to assume the identity and exchange terms between sides as you go along until you reach something resembling $A = A $, as this will show the equation is consistent. However, if you are going to use this approach, I would start by saying "assume the identity is true" (in similar fashion to proof by contradiction), as starting with an identity you have not shown to be true may seem circular to some (note that this is not necessary, but merely a word of caution)

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From what I've been told, you start with one side and by only using that one side you must obtain the other side. And this is how these problems are done.

This is one way that these identities are proved.

Do you have to do this for both sides for it to be a proof?

Of course not—just as the equality of $7+3\times2$ and $13$ is validly demonstrated by working in just one direction. Proving an identity or equality, unlike solving an equation, does not generally involve a sequence of equivalences $(\iff).$

Another method is to show that the LHS and RHS each equals the same expression (i.e., meeting in the middle).

Alternatively, we prove a trigonometric identity by starting with a known trigonometric identity, then manipulating it until we arrive at the required identity (here, the chain of reasoning is forward-directional).

Or we can show that an invertible modification of the LHS equals the same modification of the RHS.

Are you allowed to move terms from the RHS to the LHS and vice versa?

Do you mean to start with the trigonometric identity that you are required to prove? The only way this would be a valid proof is to show that it leads to a known identity via a chain of equivalences; in other words, every step must be explicitly shown to be reversible.

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  • $\begingroup$ "Another method is to show that the LHS and RHS each equals the same expression (i.e., meeting in the middle)." Isn't this the exact same as the "starting with one side and showing it equals the other side" method? Because when you do the method whereby "you start with one side and by only using that one side you must obtain the other side", then this necessarily involves some "meeting in the middle" expression, no? $\endgroup$ Apr 26, 2023 at 9:12
  • $\begingroup$ @AdamRubinson 1. Notice that the OP is even asking whether showing that L equals R is sufficient proof that, uhm, L equals R, or whether we also need to show that R equals L. $\quad$ 2. Of course working from L to R vs. working from R to L vs. working from L to M then R to M are all valid proofs of the same identity; the point is that the student is free to choose whichever of these feels easiest, but ought to avoid invalid proofs like merely demonstrating that the given identity implies a true statement (this would be akin to proving -3=3 by writing -3=3⟹(-3)^2=3^2⟹9=9). $\endgroup$
    – ryang
    Apr 26, 2023 at 9:33
  • $\begingroup$ Ok so you're saying that $a=b$ is in some way different to $b=a.$ I mean, I guess so. It's not really a fundamentally different approach to the proof... Anyway, I'm not interested in such things (which I consider trivial distinctions) so not really interested in continuing discussion. $\endgroup$ Apr 26, 2023 at 10:30
  • $\begingroup$ @AdamRubinson Indeed, the above comments are tangential to the OP's query. $\endgroup$
    – ryang
    Apr 26, 2023 at 11:23
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If you show that $f(\theta) = g(\theta)$ for all $\ \theta\in D,\ $ [here, $D$ is usually a subset of $\mathbb{R},\ $ but could be a subset of $\mathbb{C}$, or some other set], then since " Expression $f(\theta)$ is equal to Expression $g(\theta)$ for all $\ \theta\in D,\ $" is an equivalence relation, it immediately follows that $g(\theta) = f(\theta)$ for all $\ \theta\in D\ $ also.

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