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I'm going through L. Harwood Clarke's A Notebook in Pure Mathematics from 1953 (William Heinemann Ltd.) and I find this in part $\text V$: "Trigonometry" (Formula 21):

$$\tan (A + B + C + \cdots) = \dfrac {s_1 - s_3 + s_5 - \cdots} {1 - s_2 + s_4 - \cdots}$$

stated with no comments.

The interesting thing is that he has not defined $s_1$, $s_2$, $s_3$, etc.

I've looked through the rest of the book and can't find anything that suggests what they might be.

The only thing that rings a bell is that the shape of the expressions on top and bottom of the RHS are reminiscent of the power series expansions of the sine and cosine functions respectively.

Context: This is not me trying to get someone to do my homework for me, honest, I'm not trying to cheat on my degree course or anything, please believe me. I was given this book by my wife who found it while going through her parents' things while trying to get their house on the market. She thought her father would have liked me to have it, so she gave it to me.

Martin Sleziak
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Prime Mover
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1 Answers1

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@Blue's comment is helpful, but let me expand on it. Since$$\cos\sum_j\theta_j+i\sin\sum_j\theta_j=\prod_j(\cos\theta_j+i\sin\theta_j),$$we have$$1+i\tan\sum_j\theta_j=\frac{\prod_j\cos\theta_j}{\cos\sum_j\theta_j}\prod_j(1+i\tan\theta_j).$$Though Clarke doesn't say so, they've clearly defined $s_k$ as the sum of all products of $k$ factors of the form $\tan\theta_j$ with distinct $j$s, so$$1+i\tan\sum_j\theta_j=\frac{\prod_j\cos\theta_j}{\cos\sum_j\theta_j}\sum_ki^ks_k.$$The imaginary part divided by the real part is then as in the stated equation.

J.G.
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  • That does make good sense. For my next trick, I need to establish an expression that says: "the sum of the product of $k$ factors of the form $\tan \theta_J$ with distinct $j$'s". I am struggling for a neat and clean way of expressing it. – Prime Mover Oct 30 '22 at 10:26
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    @PrimeMover I hope there's something neater than$$s_k:=\sum_{S\subseteq[1,,\cdots n]\land|S|=k}\prod_{j\in S}\tan\theta_j.$$ – J.G. Oct 30 '22 at 10:34
  • Exactly. Messy, isn't it? – Prime Mover Oct 30 '22 at 10:35
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    @PrimeMover You could try $s_k:=[x^k]\prod_j(1+x\tan\theta_j)$. – J.G. Oct 30 '22 at 10:36