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My question is about the rules of proving trig identities rather than showing that both sides are equal to each other.

For example for a problem like this:

$(sin(x) + cos(x))^2 = 1 + sin (2x)$

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.

So 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?

Again, I'm not asking how to do the problem as I already know this, but I want to know the rules involved when proving a trig identity problem.

Thanks.

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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 '17 at 0:08
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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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