Talk:Sextuple Angle Formulas/Sine

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I think the main theorem and corollary should be switched to be consistent with other pages. --Robkahn131 (talk) 01:01, 27 February 2021 (UTC)

I've been sitting here for half an hour trying to work out an argument to back up "no", and having difficulty.
A corollary is a result which follows simply and/or trivially from another result.
The existing corollary to this page is actually the one which is misnamed, because it is not a corollary of Sextuple Angle Formulas/Sine but is derived from a completely different result.


When $N = 3$, the corollary would have a $\cos^2 \theta$ term
\(\ds \sin 3 \theta\) \(=\) \(\ds \sin \theta \paren {3 - 4 \sin^2 \theta }\) N = 3 Main Result: Triple Angle Formulas/Sine
\(\ds \sin 3 \theta\) \(=\) \(\ds \sin \theta \paren {4 \cos^2 \theta - 1 }\) N = 3 Corollary: Does not exist yet


When $N = 4$, the corollary has a $\cos^3 \theta$ term
\(\ds \sin 4 \theta\) \(=\) \(\ds \sin \theta \paren {4 \cos \theta - 8 \sin^2 \theta \cos \theta }\) N = 4 Main Result: Quadruple Angle Formulas/Sine
\(\ds \sin 4 \theta\) \(=\) \(\ds \sin \theta \paren {8 \cos^3 \theta - 4 \cos \theta }\) N = 4 Corollary: Quadruple Angle Formulas/Sine/Corollary 1


When $N = 5$, the corollary has a $\cos^4 \theta$ term
\(\ds \sin 5 \theta\) \(=\) \(\ds \sin \theta \paren {5 - 20 \sin^2 \theta + 16 \sin ^4 \theta }\) N = 5 Main Result: Quintuple Angle Formulas/Sine
\(\ds \sin 5 \theta\) \(=\) \(\ds \sin \theta \paren {16 \cos^4 \theta - 12 \cos^2 \theta + 1 }\) N = 5 Corollary: Quintuple Angle Formulas/Sine/Corollary


So When $N = 6$, the corollary "should" have a $\cos^5 \theta$ term, but it is switched
\(\ds \sin 6 \theta\) \(=\) \(\ds \sin \theta \paren {32 \cos^5 \theta - 32 \cos^3 \theta + 6 \cos \theta }\) N = 6 Main Result: Sextuple Angle Formulas/Sine
\(\ds \sin 6 \theta\) \(=\) \(\ds \sin \theta \paren {6 \cos \theta - 32 \sin^2 \theta \cos \theta + 32 \sin^4 \theta \cos \theta }\) N = 6 Corollary: Sextuple Angle Formulas/Sine/Corollary


Regarding your other point about the corollary being derived from a separate result...
Once we make the switch, the corollary would/could look like this...
\(\ds \sin 6 \theta\) \(=\) \(\ds \sin \theta \paren {6 \cos \theta - 32 \sin^2 \theta \cos \theta + 32 \sin^4 \theta \cos \theta }\) Main Result
\(\ds \) \(=\) \(\ds \sin \theta \paren {6 \cos \theta - 32 \paren{\sin^2 \theta } \cos \theta + 32 \paren {\sin^2 \theta }^2 \cos \theta }\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {6 \cos \theta - 32 \paren{1 - \cos^2 \theta } \cos \theta + 32 \paren {1 - \cos^2 \theta }^2 \cos \theta }\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \sin \theta \paren {6 \cos \theta - 32 \paren{1 - \cos^2 \theta } \cos \theta + 32 \paren {1 - 2 \cos^2 \theta + \cos^4 \theta} \cos \theta }\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {\paren {6 - 32 + 32 } \cos \theta + \paren {32 - 64 } \cos^3 \theta + 32 \cos^5 \theta }\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {32 \cos^5 \theta - 32 \cos^3 \theta + 6 \cos \theta }\) Desired End Corollary with the $\cos^5 \theta$ term

--Robkahn131 (talk) 15:35, 27 February 2021 (UTC)


The existing results were posted up only because they appeared as exercises in various texts. Sorry, but I don't see the need for an avalanche of more or less trivial applications of ever more complicated and contrived formulas of no particular profundity just because we've been able to fill in the numbers and turn the handle. --prime mover (talk) 16:18, 27 February 2021 (UTC)