Talk:Sine of Integer Multiple of Argument/Formulation 9

For $n \in \Z_{>1}$:

If $n$ is odd, then:

$\sin n \theta = \map \cos {\paren {n - 1 } \theta} \paren { a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots \cfrac {} {a_{n-2} + \cfrac 1 {a_{n-1}}} }}} }$

where $a_0 = a_2 = a_4 = \ldots = 2 \sin \theta$ and

$a_1 = a_3 = a_5 = \ldots = -2 \sin \theta$ and

A terminal $a_{n - 1} = \sin \theta$ term if and only if $n$ is odd.

$\paren {n }$ TOTAL terms.

If $n$ is even, then:

$\sin n \theta = \map \cos {\paren {n - 1 } \theta} \paren { a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots \cfrac {} {a_{n-3} + \cfrac 1 {a_{n-2}}} }}} }$

where $a_0 = a_2 = a_4 = \ldots = 2 \sin \theta$ and

$a_1 = a_3 = a_5 = \ldots = -2 \sin \theta$

$\paren {n - 1 }$ TOTAL terms.

The Equation template does not seem to permit continued fractions which is why part of the latex code utilizes the equation template and other parts do not.

--Robkahn131 (talk) 20:54, 22 February 2021 (UTC)

a) the point is that your equation on the page goes down only to $a_{n - 2}$ and so it's not obvious where the $a_{n - 1}$ term fits in.

b) I will let you inspect the code so you can see for yourself what you did wrong. There's discussion of the matter on Help:Questions. Bottom line: there are in fact fairly good reasons for the strictness of our house rules -- one of those is the compatibility with the treatment of code in eqn templates.

As for being able to count to a TOTAL of $n - 1$ terms, how would I know? I can't count. --prime mover (talk) 21:23, 22 February 2021 (UTC)

$\sin 2 \theta = \map \cos {\theta} \paren { 2 \sin \theta }$

$\sin 3 \theta = \map \cos {2 \theta} \paren { 2 \sin \theta + \cfrac 1 {-2 \sin \theta + \cfrac 1 {\sin \theta }} }$

$\sin 4 \theta = \map \cos {3 \theta} \paren { 2 \sin \theta + \cfrac 1 {-2 \sin \theta + \cfrac 1 {2\sin \theta }} }$

$\sin 5 \theta = \map \cos {4 \theta} \paren { 2 \sin \theta + \cfrac 1 {-2 \sin \theta + \cfrac 1 {2\sin \theta + \cfrac 1 {-2 \sin \theta + \cfrac 1 {\sin \theta }} }} }$

When $n$ is even, there are $\paren{n - 1 }$ terms. Even terms are $2 \sin \theta$ and odd terms are $-2 \sin \theta$

When $n$ is odd, everything is the same, except there are $\paren{n }$ terms AND the very last term is different from all of the others. --Robkahn131 (talk) 22:57, 22 February 2021 (UTC)

I'm sorry, but I still don't understand.

When $n$ is odd, there is an extra term at the end which is not defined in the equation. It's like saying "$a = b + c + d$ where $e = 345$". --prime mover (talk) 23:00, 22 February 2021 (UTC)

The point is, $a_{n - 1}$ is not defined in the equation at all. would it be okay to replace it with $a_{n - 2} = -2 \sin \theta + \dfrac 1 {\sin \theta}$? Because as it is, the reader has to "guess" what you have in mind, and it is not obvious how that guess should pan out. --prime mover (talk) 23:03, 22 February 2021 (UTC)

Thanks for the feedback - agreed - not as clear as necessary. Does the latest clarify? --Robkahn131 (talk) 23:27, 22 February 2021 (UTC)

Not really, no. We have the capability of being precise and rigorous, and it doesn't achieve that. Now I understand what you mean here (see what I mean? It's unclear enough so that if you're not that clever, like me, it's difficult to understand what is meant) I'll get onto it and put it into a form that's more precise in its presentation. --prime mover (talk) 06:25, 23 February 2021 (UTC)

What I've put is now unambiguous, I think. All we need to do now is fix up the proof so it does what it's supposed to. I can't see how the conclusion about the terminal term actually follows from the previous line, which does not clearly make the connection between the index of the term currently being investigated and the last one. Perhaps I need to go back to school, but I'm having more and more trouble lately working out how to get "there" from "here" when I'm presented with an argument. I seem to need everything to be a proof by induction nowadays or I can't get the hang of it. --prime mover (talk) 07:22, 23 February 2021 (UTC)

Line 2 shows a fraction with $\dfrac {\map \sin {\paren {n - 2 } \theta} } {\map \cos {\paren {n - 1 } \theta} }$ and Line 8 shows a fraction with $\dfrac {\map \sin {\paren {n - 4 } \theta} } {\map \cos {\paren {n - 3 } \theta} }$

Numerator: $2 \paren{ k }$ goes to $2 \paren { k + 1 }$ and Denominator: $2 \paren{ k } - 1$ goes to $2 \paren { k + 1 } - 1 $ --Robkahn131 (talk) 14:54, 5 March 2021 (UTC)

$3$ is prime. $5$ is prime. $7$ is prime. Thus I have demonstrated that when $n$ is prime, then so is $n + 2$. Therefore all odd numbers $3$ or greater are prime. --prime mover (talk) 15:00, 5 March 2021 (UTC)