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I was reading about multiple angle formulas to expand $\sin{(nx)}$ or $\cos{(nx)}$ to be in terms of $\sin{x}$ and $\cos{x}$ on Wolfram MathWorld, and came across these formulas:

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These formulas look so much like the Taylor series expansions of the respective functions that I feel there must be a way to further transform these equations into ones representing the value of $\sin{x}$ and $\cos{x}$. I would expect some form of calculus to come into play, but I'm not sure where I would even start. In other words, I want to start with the above formulas and algebraically manipulate them to get:

$$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$$ $$\cos{x}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots$$

The formulas come from T. J. I'A. Bromwich and T. M. MacRobert's book An Introduction to the Theory of Infinite Series, 3rd ed.

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  • $\begingroup$ The expansions of $\sin(na)$ and $\cos(na)$ here are in terms of powers of $\sin(a)$, which is a binomilal expansion. Why do you believe that this is related to the Taylor expansion of $\sin(x)$ and $\cos(x)$? $\endgroup$
    – Mark Viola
    May 25 at 22:38
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    $\begingroup$ @MarkViola Mainly because all of the terms match except for the numerator coefficients, and $\sin{a} \approx a$ for small $a$. I'm thinking increasing $n$ while decreasing $a$ sort of demonstrates how the multiple angle formulae can certainly be used to approximate any $\sin{x}$ or $\cos{x}$ by breaking up $x$ into enough ($n$) tiny angles and repeating the multiple angle formula enough ($n$) times $\endgroup$ May 25 at 23:04
  • $\begingroup$ The binomial expansion is a Taylor Series. So it shoud not be surprising to see factorials in the denominator. $\endgroup$
    – Mark Viola
    May 25 at 23:18

1 Answer 1

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Yes, this is possible. Fix $t$ and write $a=t/n$ so that $x=\sin(t/n)$. In the last formula you cite from Bromwich (1991), for $n$ even $$\cos t=1-\frac{(nx)^2}{2!}+ \frac{(nx)^4}{4!} (1-2^2/n^2)-\frac{(nx)^6}{6!}(1-2^2/n^2)(1-4^2/n^2)+\ldots \tag{*}$$

As $n \to \infty$, we have $nx=n\sin(t/n) \to t$ as $n \to \infty$, so the series (*) converges to the Taylor series for cosine, term by term. To control the error, observe that truncating the series after $k$ terms incurs an error of order $\frac{t^{2k}}{(2k)!}$, uniformly in $n$, so we may choose $k$ to ensure this truncation incurs an error less than $\epsilon/2$. The first $k$ terms do indeed converge to the corresponding terms in the Taylor series as $n \to \infty$, so we may choose $n$ to ensure that the error in each of these terms is less than $\frac{\epsilon}{2k}$.

The other three formulas can be handled in the same way.

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