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What is the purpose of the $\mp$ symbol in mathematical usage?

Just as the title explains. I've seen my professor actually differentiating between those two. Do they not mean the same?

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marked as duplicate by MJD, The Chaz 2.0, lhf, Henry, t.b. Jun 20 '12 at 4:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

@mark Seems like I did duplicate, but I wasn't able to find that question since the title of the one you posted was not something I had in mind. – drum Jun 19 '12 at 20:22
My comment was intended only to be helpful, not to rebuke you. I know it can be very difficult to find related threads before asking a question. I only knew about this earlier thread because I happened to participate in it. – MJD Jun 19 '12 at 20:22
@drum: I'm sure you made a good-faith effort to search before posting. It can be quite difficult to find something on here, even when we know what we're looking for! – The Chaz 2.0 Jun 19 '12 at 20:24
up vote 17 down vote accepted

If you write $$ \cos(a \pm b) = \cos a \cos b \mp \sin a \sin b, $$ then + on the left side corresponds to minus on the right side, and - on the left side corresponds to + on the right side.

Standing alone, they mean the same.

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I don't think anyone ever uses $\mp$ without $\pm$ in the same formula. If I saw that done, I think I would be seriously confused. – MJD Jun 19 '12 at 20:21
@ Mark Dominus: I agree. I have never seen $\mp$ alone, and I would find it strange if I ever saw it. – N.U. Jun 19 '12 at 20:29
+1. Whenever I discuss the cosine formula, I describe the "$\mp$" in this context as the "co-sign" of "$\pm$". Not only does this make the point of your answer to the OP's question, but it makes possible this "cheerleader" mnemonic for the angle sum/difference formulas for sine and cosine: "sine! cosine! SIGN! cosine! sine!!! ... cosine! cosine! CO-SIGN! sine! sine!!!" :) (Conveniently, @talmid's answer has both formulas.) – Blue Jun 19 '12 at 20:29
Now that I think about it, it's kind of funny that using $\pm$ twice in the same formula is ambiguous; it could mean the two signs should be the same but more commonly means they could be anything! – MartianInvader Jun 19 '12 at 23:55

If it stands alone, say $a \pm b$, then it means the same as $a \mp b$. However, if they both occur in the same statement, such as $a\pm b \mp c$, then you may pick the "top" row or the "bottom" row of operators. In this case $a+b-c$ and $a-b+c$ would what is intended. But $a+b+c$ and $a-b-c$ would not be allowed.

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In general, we use $\pm$, but when we want to correlate a change of sign we also use $\mp$. For example: $2 (x \pm y) = 2x \pm 2y$, meaning that $2(x+y) = 2x + 2y$, and that $2(x-y) = 2x-2y$. Now, if we wanted the second sign to be the opposite of the first, we use $\mp$. For example: $-2(x \pm y) = 2x \mp 2y$ would mean that $-2(x+y) = -2x - 2y$ and $-2(x-y) = -2x + 2y$.

That is, whenever we have an expression involving $\pm$ or $\mp$, it's actually an abbreviation for two expressions: one in which we read all the top symbols ($+$ in $\pm$ and $-$ in $\mp$), and another one in which we read all the bottom symbols. Common examples:

$\sin (x\pm y) = \sin x \cos y \pm \cos x \sin y$ means $\begin{cases} \sin (x+ y) = \sin x \cos y +\cos x \sin y \\ \sin (x- y) = \sin x \cos y - \cos x \sin y \end{cases}$ $\cos (x \pm y) = \cos x \cos y \mp \sin x \sin y$ means $\begin{cases} \cos (x + y) = \cos x \cos y - \sin x \sin y\\ \cos (x - y) = \cos x \cos y + \sin x \sin y\end{cases}$

Now, when we don't have any changes of sign, like in $\sin (x\pm y) = \sin x \cos y \pm \cos x \sin y$ (all the top symbols are $+$), we could also write $\sin (x\mp y) = \sin x \cos y \mp \cos x \sin y$, and it would be the same, but this isn't common usage. The symbol $\mp$ only appears when there's already a $\pm$, but we want to establish a correspondence between opposite signs in an equation.

Note that it's only a matter of style; we could dispose completely of $\mp$ and used $\pm -$ instead, e.g., $\cos (x \pm y) = \cos x \cos y \pm (- \sin x \sin y)$.

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I like your new edition. – drum Jun 19 '12 at 20:24

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