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In my math text book the proof for the Sum to Product rule starts with a substitution:

let A = (u+v)/2, let B = (u-v)/2

...and then uses this substitution to prove that A+B = u and A-B = v

...then substitutes (u+v)/2 and (u-v)/2 in the Product-to-Sum equation to demonstrate that the Sum-to-Product rule is true.

As it appears to me from what is presented in the text book, the substitutions (u+v)/2 and (u-v)/2 were pulled out of thin air.

How was this determined to work? I doubt it was just a lot of guess and check until something worked.

Is there a point in math where one just needs to accept an equation for what it is and not worry about why or how it is?

I'm getting the feeling that understanding some of the how's and why's right now requires an understanding of math concepts that I haven't learned yet.

Any insight would be appreciated. Thanks!

Evan Mayfield
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    One is often looking for a path to a solution to a problem, which you could imagine as finding the way out of a maze ... one tries different routes, and finally they get one that works. When they explain to someone (i.e. you reading this derivation) how to solve the maze, they show the correct route without telling you all the false starts that they went through before they found it. And it may seem like magic to you -- how did they know to do that? -- but you haven't seen the work that went into figuring it out. – Ned May 22 '20 at 01:15
  • yeah. I'm trying to see that work. It exists somewhere, presumably. In order for it to be taken as gospel, it's work that can be recreated. – Evan Mayfield May 24 '20 at 18:22
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    Is your issue with this (1) you don't see why it is correct, i.e. what the point is, what it says; or (2) you understand that it's correct in what it's doing, but can't see where it came from and how you would come up with something like it, or (3) something else entirely. I was trying to answer (2), but maybe didn't address your actual question. – Ned May 24 '20 at 21:43
  • People often don't bother to explain how someone might have thought of a proof, unfortunately. Learning to see proofs from the right viewpoint that makes the proof seem obvious is one of the constant challenges of learning math. But, I recommend watching Summoning Salt's videos about the history of video game speed running if you want to be convinced that any trick that is waiting to be discovered, no matter how subtle, will eventually be discovered by humans who are sufficiently obsessed. – littleO May 25 '20 at 12:47
  • You might find this proof to be more clear: https://math.stackexchange.com/a/1342/40119 – littleO May 25 '20 at 12:52

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I will try to motivate the choice of substitution based on the known sum and difference results \begin{equation*} \sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B), \\ \sin(A-B) = \sin(A)\cos(B) - \cos(A)\sin(B). \end{equation*} Observe that these look quite similar to one another, with the only differences being the sign of the final terms. We might then be prompted to add or subtract these two identities to discover another potentially useful relationship since some cancellation will occur if we do so. Adding gives \begin{equation*} \sin(A+B) + \sin(A-B) = 2 \sin(A)\cos(B). \tag*{(1)} \end{equation*} This identity essentially tells us that if we have a sum of the sines of two angles, we can express it as a product, but in this form, the result is not always convenient.

To see more directly what this result says about a sum of the form $\sin(u) + \sin(v)$, we can let $u = A+B$ and $v = A-B$. Then $(1)$ becomes \begin{equation*} \sin(u)+\sin(v)=2\sin(A)\cos(B), \end{equation*} where $A$ and $B$ are related to $u$ and $v$. Yet if we are starting on the left-hand side with $u$ and $v$, it is desirable to express the right-hand side completely in terms of $u$ and $v$ if possible. We can express $A$ and $B$ in terms of $u$ and $v$ by rearranging our substitutions and eliminating one variable at a time to get* \begin{equation*} A = \frac{u+v}{2}, \qquad B = \frac{u-v}{2}. \end{equation*} Hence \begin{equation*} \sin(u) + \sin(v) = 2 \sin\mathopen{}\left(\frac{u+v}{2} \right)\mathclose{}\cos\mathopen{}\left(\frac{u-v}{2} \right)\mathclose{}. \end{equation*}

You can derive the other sum-to-product results in a similar way, by adding or subtracting the results involving sines or cosines of sums and differences, and then making the same substitutions.

As far as the last part of your question goes, in general I feel it is very good to worry about the why's and how's as you study mathematics, since it helps you hone your intuition and reinforces concepts as you learn them. This also often reveals ways to derive formulas on the spot as you need them, rather than simply memorising them, as is the case here.

Edit: *The process of rearranging the substitutions and solving for $u$ and $v$ in terms of $A$ and $B$ is as follows. Given the substitutions \begin{equation*} u = A + B \tag{2} \end{equation*} and \begin{equation*} v = A - B \tag{3}, \end{equation*} we can add equations $(2)$ and $(3)$ to get $u + v = 2A$ so that \begin{equation*} A = \frac{u+v}{2}. \end{equation*} Instead subtracting $(3)$ from $(2)$ gives $u - v = 2B$, i.e. \begin{equation*} B = \frac{u-v}{2}. \end{equation*}

Zac
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  • so this part:>>>>>> "We can express A and B in terms of u and v by rearranging our substitutions and eliminating one variable at a time to get " <<<< What's the math for making sin(u)+sin(v)=2sin(A)cos(B) into sin(u)+sin(v)=2sin(u+v)/2 * cos(u−v)/2. Why are the arguments inside the functions now divided by 2? Why are we taking the sum and difference of u and v? – Evan Mayfield May 24 '20 at 18:19
  • Please see my updated answer which should now clarify this. We are essentially solving the pair of simultaneous equations (that we have introduced) for $A$ and $B$ in terms of $u$ and $v$. In the process of solving these simultaneous equations, we end up summing $u$ and $v$ simply to eliminate $B$, and get $A$ in terms of $u$ and $v$. Similarly, we take their difference to eliminate $A$, thus giving us $B$ in terms of $u$ and $v$ only. – Zac May 25 '20 at 12:49
  • Thanks! This! Solved! – Evan Mayfield May 28 '20 at 17:59
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I doubt it was just a lot of guess and check until something worked.

The original proof of the formula was likely not discovered "guess and check" in the sense that someone had to "guess" two equations that would make it possible to derive the sum-to-product rule. After all, the first person to prove the sum-to-product formula did not have a trig textbook to read the formula itself from before proving it.

There undoubtedly was a lot of "guessing" in the early development of trigonometry in that people tried different substitutions and combinations of things to see if anything good came from it, not even knowing in advance what the formulas would look like when they were finally proved.

What we don't have is a simple turn-the-crank algorithm that will spit out new theorems of mathematics at predictable intervals. More typically, some inspired guesswork is involved.

On the other hand, once a successful "guess" has been made (discovering a new theorem along with a way to prove it), the discoverer has to organize the proof in such a way that it can be verified. In that sense, we don't "just accept an equation." An equation may be proved in terms of other equations, or may be the way of introducing one or more new named quantities to a proof. For example, if I have a statement involving variables $u$ and $v,$ and I write "let $A = \frac{u+v}2,$" I have just defined the meaning of the symbol $A$ in this context. If I also write "let $B = \frac{u-v}2,$" this defines the symbol $B$, and then the meaning of the expression $A + B$ is $\frac{u+v}2 + \frac{u-v}2.$ You tell me whether that expression is or is not equal to $u.$ Please do be concerned with "why" and "how" it is or is not equal.

David K
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