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This question is related purely for my students of an high school and indirectly for me. The formulas below are the formulas of prostapheresis,

\begin{cases} \sin\alpha+\sin\beta=2\,\sin \dfrac {\alpha+\beta}{2}\, \cos \dfrac {\alpha-\beta}{2} \\ \sin\alpha-\sin\beta=2\sin \dfrac {\alpha-\beta}{2} \,\cos \dfrac {\alpha+\beta}{2}\\ \cos\alpha+\cos\beta=2\cos \dfrac {\alpha+\beta}{2}\,\cos \dfrac {\alpha-\beta}{2}\\ \cos\alpha-\cos\beta=-2 \,\sin \dfrac {\alpha+\beta}{2} \,\sin \dfrac {\alpha-\beta}{2} \end{cases}

and while I am able to find them, I am not able to find a technique to memorize them.

Is there a technique to be able to memorize them?

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    $\begingroup$ A good question since I often wonder how to do this myself, and I'm teaching a trig class this semester! I usually just derive identities on these on the fly using sorts of very compatible techniques (e.g. $e^{i\theta}$ or rotation matrices), but I imagine you mean some sort of memory technique or mnemonic. While I don't have anything to offer at this time, I hope you can find an answer (for both of us). $\endgroup$ – Eevee Trainer Jan 27 at 22:07
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    $\begingroup$ I was never able to remember these formulas. I always find them from the usual $\cos(a\pm b)$ and $\sin(a\pm b)$. $\endgroup$ – Jean-Claude Arbaut Jan 27 at 22:08
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    $\begingroup$ I personally prefer to memorize the equivalent ones, like $\cos (a+b)= \cos a \cos b - \sin a \sin b$ from which the above follow by replacing $a,b$ with their half-sum and half-difference. $\endgroup$ – G Cab Jan 27 at 22:09
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    $\begingroup$ @downvoters: To the hater or haters who will vote negatively: sooner or later we should leave this Earth. Leave one day at least something good for someone to remember you by. $\endgroup$ – Sebastiano Jan 27 at 22:19
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    $\begingroup$ Just memorize the third formula above, cos(a)+cos(b). They is what some used for multiplication prior to the invention of logarithms by Napier and Briggs. $\endgroup$ – richard1941 Feb 2 at 23:16
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This answer is not apt to the general high school public, but it can be useful for particularly curious students.

I like very much how these formulas are derived by Feynman in his “Beats” lecture (Lectures on Physics, volume 1, https://www.feynmanlectures.caltech.edu/I_48.html, section 48-1). He uses complex exponential, something that has already been mentioned in comments.

I have always loved his explanation. These apparently obscure formulas actually express the adding and the subtracting of two waves. Since there is a real and an imaginary part, this amounts to four real formulas. The physical phenomenon behind them is the “Beats” one, and it can be heard easily by picking two strings of a guitar. It can actually be used to tune it.

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  • $\begingroup$ I give you my upvote and thank you for your answer-advice. Very very nice the link: I have understood many parts but not all also if to my age 46 I teach with so much passion and I also research on relativistic electrodynamics also present in the link. Unfortunately, even though I have been teaching for many years, for me the real school no longer exists, the one I truly believed in. Thank you very much from Sicily and I will thank you further. $\endgroup$ – Sebastiano Jan 28 at 12:21
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    $\begingroup$ I am glad you enjoyed it. The world needs teachers with passion like you. (I was born and raised in Puglia, and I am in love with Sicilia. I’ve been there many times but still not enough. Buona serata). $\endgroup$ – Giuseppe Negro Jan 28 at 22:33
  • $\begingroup$ Dear Giuseppe, I'm sorry but I confide to you that I'm crying right now. The reasons are many: the school does not need teachers like me because with so much humility always go to the first place incompetent people. Can you please create a room if you have 5 minutes of time? Please so I write in Italian because I use the translator. $\endgroup$ – Sebastiano Jan 28 at 22:44
  • $\begingroup$ chat.stackexchange.com/rooms/119063/giuseppe-e-sebastiano can you go, please into the room. Thank you very much. $\endgroup$ – Sebastiano Jan 28 at 22:48
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    $\begingroup$ I've closed the room because I think you're busy or out of the site. My best regards to you to Puglia. "Sereno riposo a te e grazie" $\infty$. $\endgroup$ – Sebastiano Jan 28 at 22:59
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It is useful to know the principle that sum or difference to sine and cosine can be written in terms of products of sine and cosine. But I never memorize such identities per se. Whenever needed, you can derive them if you remember the formulas for $\sin(a\pm b)$ and $\cos(a\pm b)$. Or you can simply look at the known list: https://en.wikipedia.org/wiki/List_of_trigonometric_identities

If I need to take a close-book exam that requires memorizing these identities, some observations may be useful for a short-term memory.

  • If you know $\sin(-x)=-\sin(x)$, then the second identity comes immediately from the first one.
  • For the rest: $$ \begin{align} \color{green}{\sin}\alpha+\color{green}{\sin}\beta=2\,\color{green}{\sin}\dfrac {\alpha+\beta}{2}\, \color{green}{\cos}\dfrac {\alpha-\beta}{2} \\ \cos\alpha\color{red}{+}\cos\beta=2\color{red}{\cos} \dfrac {\alpha+\beta}{2}\,\color{red}{\cos} \dfrac {\alpha-\beta}{2}\\ \cos\alpha\color{blue}{-}\cos\beta=\color{blue}{-}2 \,\color{blue}{\sin} \dfrac {\alpha+\beta}{2} \,\color{blue}{\sin} \dfrac {\alpha-\beta}{2} \end{align} $$
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  • $\begingroup$ In the meantime thank you very much. I remember very well the sum or difference to sine and cosine but the principal problem for me it is where I found the minus sign :-(. However with the colours it is very nice. $\endgroup$ – Sebastiano Jan 27 at 22:25
  • $\begingroup$ @Sebastiano Well, the special minus sign only appears in the last identity. If for an exam, I would simply memorize it separately. $\endgroup$ – user9464 Jan 27 at 22:28
  • $\begingroup$ And I have to remember that it applies when I have $\cos\alpha\color{magenta}{-}{\cos\beta}.$ $\endgroup$ – Sebastiano Jan 27 at 22:30
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    $\begingroup$ @Sebastiano yes, for an exam. If not for an exam, nobody prevents you look it up from a cheatsheet. :-) $\endgroup$ – user9464 Jan 27 at 22:33
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    $\begingroup$ "Whenever needed, you can derive them..." - The easiest way to derive almost all the trigonometric identities on-the-fly is using Euler's Formula, $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. Simplify using the normal rules of exponents, then equate the real and imaginary parts. $\endgroup$ – BlueRaja - Danny Pflughoeft Jan 28 at 8:14
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I am speaking of the angle summation formulas: $$ \eqalign{ & \cos \left( {a \pm b} \right) = \cos a\cos b \mp \sin a\sin b \cr & \sin \left( {a \pm b} \right) = \sin a\cos b \pm \cos a\sin b \cr} $$ Then e.g. summing the equations for $\cos$ $$\cos(a+b)+ \cos(a-b)=2\cos a \cos b$$ After which you can apply $$ \left\{ \matrix{ \alpha = {{a + b} \over 2} \hfill \cr \beta = {{a - b} \over 2} \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ a = \alpha + \beta \hfill \cr b = \alpha - \beta \hfill \cr} \right. $$

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  • $\begingroup$ Yes yes, this is the proof that also I adopt (the classic proof). I accept also your answer. $\endgroup$ – Sebastiano Jan 27 at 22:49
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    $\begingroup$ sorry, I saw now @mrsamy answer which is identical: I keep my answer however as a "reinforcement" (affter upvoting @mrsamy) $\endgroup$ – G Cab Jan 27 at 22:50
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You can memorize the pattern $$f(a)+\varepsilon f(b)=2\delta g\left(\frac{a+b}{2}\right)h\left(\frac{a-b}{2}\right)$$ where $f$, $g$ and $h$ are either $\sin$ or $\cos$, and $\varepsilon$ and $\delta$ are either $1$ or $-1$. Given $f$ and $\varepsilon$, you then need a strategy to find $g$, $h$ and $\delta$.

Step 1: Find $g$ and $h$

Specializing this for $b=a$ we get $$f(a)+\varepsilon f(a)=2\delta g\left(a\right)h\left(0\right)$$ and for $b=-a$ we get $$f(a)+\varepsilon f(-a)=2\delta g\left(0\right)h\left(a\right)$$

Note that if the function that is given $0$ as input is $\sin$ then the right-hand side is $0$ for all $a$, while it if is $\cos$, the right-hand side is non-zero for some $a$. Using the (a)symmetry of $f$, we can easily determine if the left-hand side is zero for all $a$, and hence what $g$ and $h$ are:

  • If $f(a)+\varepsilon f(a)=0$ for all $a$ then $h=\sin$ and otherwise $h=\cos$;
  • If $f(a)+\varepsilon f(-a)=0$ for all $a$ then $g=\sin$ and otherwise $g=\cos$.

Step 2: Find $\delta$

Taking $a=\frac{\pi}{2}$ and $b=0$ yields

$$f\left(\frac{\pi}{2}\right)+\varepsilon f(0)=2\delta g\left(\frac{\pi}{4}\right)h\left(\frac{\pi}{4}\right)$$

Since both $\sin$ and $\cos$ evaluate to $\frac{\sqrt{2}}{2}$ at $\frac{\pi}{4}$, we therefore have

$$f\left(\frac{\pi}{2}\right)+\varepsilon f(0)=\delta $$

which gives us $\delta$.

Plugging other specific values for which both sides are easy to evaluate would also work as long as both $g\left(\frac{a+b}{2}\right)$ and $h\left(\frac{a-b}{2}\right)$ are non-zero (so you do not need to remember to take $a=\frac{\pi}{2}$ and $b=0$).

Example

$$\cos(a)- \cos(b)=2\delta g\left(\frac{a+b}{2}\right)h\left(\frac{a-b}{2}\right)$$

Taking $a=b$ on the left-hand side makes it $0$ so $h=\sin$. Taking $a=-b$ also makes it zero so $g=\sin$: $$\cos(a)- \cos(b)=2\delta \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$

Specializing to $b=0$ we get $$\cos(a)-\cos(0)=2\delta \sin\left(\frac{a}{2}\right)\sin\left(\frac{a}{2}\right)$$ so we just need to pick a value of $a$ such that $\sin\left(\frac{a}{2}\right)\not = 0$. Taking $a=\pi$ gives: $$\cos(\pi)-\cos(0)=2\delta \sin\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right)$$ i.e. $$-2=2\delta $$ which allows to conclude $\delta=-1$ and hence: $$\cos(a)- \cos(b)=-2 \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$

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  • $\begingroup$ Hi and also this answer it is very beautiful. Thank you very much for your effort and help. +1 $\endgroup$ – Sebastiano Jan 28 at 13:22
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If the students can recall

  1. the first formula \begin{equation} \sin\alpha+\sin\beta=2\,\sin \dfrac {\alpha+\beta}{2}\, \cos \dfrac {\alpha-\beta}{2}, \end{equation}
  2. the odd symmetry of $\sin(x)$ and basic differentiation rules,

then they can promptly derive the other three formulas. The second one follows by $(2)$, as already mentioned. Taking the derivative with respect to $\alpha$ in the first formula we obtain \begin{align} \cos(\alpha)=\cos\bigg(\frac{\alpha+\beta}{2}\bigg)\cos\bigg(\frac{\alpha-\beta}{2}\bigg)+\sin\bigg(\frac{\alpha+\beta}{2}\bigg)\sin\bigg(\frac{\alpha-\beta}{2}\bigg), \end{align} and switching $\alpha$ and $\beta$ the same formula for $\cos(\beta)$ follows. Now sum and subtract (using $(2)$ again) to obtain the two identities for the cosine.

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    $\begingroup$ I have appreciated also your answer. Thank you very much for your cooperation. $\endgroup$ – Sebastiano Feb 3 at 21:53
  • $\begingroup$ Why does the first formula look like something from my ham radio class about amplitude modulation? $\endgroup$ – richard1941 Feb 4 at 22:40
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I know a technique that my high school teacher taught me, and that could be particularly useful for Italian students, quite famous where I come from. It is the following: if in the second formula you apply commutativity and swipe $\cos$ with $\sin$ in the second member of the equation, obtaining $2\cos\big(\frac{\alpha+\beta}{2}\big) \sin\big(\frac{\alpha-\beta}{2}\big)$, then in the second member for all equations in the first argument $\alpha+\beta$ appears, while in the second argument $\alpha-\beta$ appears (reading from left to right). Now, there is a sentence that says "Cento rose sono meno belle" that presents vocals in the order by which you find $\cos$ and $\sin$ reading from left to right and top-down second members of formulas, reminding that in Italian $\sin$ can be used also as sen (meno (= less) indicates the sign of the second member of the last line). So using "Cento rose sono meno belle" the difficulty is reduced a lot, because it only needs to remember factors $2$ at the beginning and factor $\frac{1}{2}$ in the argument. For the first members of the equations I think that notice that they are pretty ordered (four $\sin$ then four $\cos$ with ordered $\alpha$ $\beta$ and alternated signs $+-+-$) should make it easy to remember them.

                         "Cento rose sono meno belle" 
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    $\begingroup$ Hi, and thank you very much much...for your answer that I vote up +1. $\endgroup$ – Sebastiano Feb 15 at 8:53

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