# Is there a way to derive sin/cos series from multiple angle formulas?

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:

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.

• 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 $\cos(x)$? May 25 at 22:38
• @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 May 25 at 23:04
• The binomial expansion is a Taylor Series. So it shoud not be surprising to see factorials in the denominator. May 25 at 23:18

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 $$\frac{\epsilon}{2k}$$.