The formulas of prostapheresis: memorization technique

Question

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?

Answer 1

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: $$ \begin{align} \color{green}{\sin}\alpha+\color{green}{\sin}\beta=2\,\color{green}{\sin}\dfrac {\alpha+\beta}{2}\, \color{green}{\cos}\dfrac {\alpha-\beta}{2} \\ \cos\alpha\color{red}{+}\cos\beta=2\color{red}{\cos} \dfrac {\alpha+\beta}{2}\,\color{red}{\cos} \dfrac {\alpha-\beta}{2}\\ \cos\alpha\color{blue}{-}\cos\beta=\color{blue}{-}2 \,\color{blue}{\sin} \dfrac {\alpha+\beta}{2} \,\color{blue}{\sin} \dfrac {\alpha-\beta}{2} \end{align} $$

Answer 2

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 $$ \left\{ \matrix{ \alpha = {{a + b} \over 2} \hfill \cr \beta = {{a - b} \over 2} \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ a = \alpha + \beta \hfill \cr b = \alpha - \beta \hfill \cr} \right. $$

Answer 3

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.

Answer 4

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: $$\cos(a)- \cos(b)=-2 \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$

Answer 5

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.

Answer 6

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" 

Answer 7

Let us say that I remember form $2\times$ product of some trigonometric function of $\frac{\alpha+\beta}2$ and $\frac{\alpha-\beta}2$.

I also remember that for small $x$ I have $\sin x\approx x$ and $\cos x$ is close to $1$.

If $\alpha$ and $\beta$ are small than \begin{align*} \sin\alpha+\sin\beta &\approx \alpha+\beta\\ 2\sin\frac{\alpha+\beta}2 &\approx \alpha+\beta \end{align*} So I have to multiply $2\sin\frac{\alpha+\beta}2$ by something close to one, from the choices I have at my disposal, it is only $\cos\frac{\alpha-\beta}2$.

This helps me remember $$\sin\alpha+\sin\beta = 2\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2$$ and using $\sin(-x)=-\sin x$ and $\cos(-x)=\cos x$, I get also a formula for $\sin\alpha-\sin\beta$.

For small $\alpha$, $\beta$, the value of $\cos\alpha+\cos\beta$ will be close to two. So I multiply by two the two things that are close to one $$\cos\alpha+\cos\beta = 2\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2.$$

And $\cos\alpha-\cos\beta$ will be close to zero, so I use the two small values. There is also the question about the sign. For small positive values $\cos$ is a decreasing function. So $\cos\alpha-\cos\beta$ has to be positive if $\beta>\alpha$. Also, for small angles such that $\beta>\alpha$ I have $\sin\frac{\alpha-\beta}2<0$. So to get the signs right, I use: $$\cos\alpha-\cos\beta = -2\sin\frac{\alpha+\beta}2\sin\frac{\alpha-\beta}2.$$

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Having said that, if I need these formulas and I did not work with them for a bit, I prefer to simply derive them (which can be done relatively quickly). But at least the first mnemonic mentioned above seems quite reasonable to me - I would use it if I had to work with these formulas frequently.