trigonometry - Generative method for reducing trigonometric argument coefficients to unity

Suppose I have a term
$$f(\beta x) = C$$
where $\beta \in \mathbb{Z}^{+},\ f \in \{\sin,\cos\}$.

And I want to find an algebraically equivalent sum

$$\begin{align*}\sum_{i=1}^{n} K_i \sin^{a_i}(x)\cos^{b_i}(x) = C\end{align*}$$
where $K_i \in \mathbb{Z},\ a_i,b_i \in \mathbb{Z}^{+}$, and $\forall\ i,j \in [1,2,..,n],\ (a_i = a_j) \wedge (b_i = b_j) \iff i = j$.

In other words, there is exactly one integer coefficient $K_i$ for each unique pair of positive integer exponents $a_i, b_i$. This transformation is useful for tangent half-angle substitution. Sum and double-angle identities ensure that this transformation is always possible, using recursive binary splitting.

For example, we'd begin,
$$\begin{align*}\sin(47x) & = \sin(32x + 15x) \\ & = \sin(32x)\cos(15x) + \cos(32x)\sin(15x)\end{align*}$$
then repeat the procedure for $\sin(15x)$ and $\cos(15x)$, and so on. All of the factors like $\sin(2^{k} x)$ reduce using double-angle formulas. So we can get to nice expressions like these:

$$\begin{align*}\sin(8x) =\ & 2^3 \sin(x)\cos(x) \\ -\ & 2^4 \sin^3(x)\cos(x) \\ -\ & 2^6 \sin^3(x)\cos^3(x) \\ +\ & 2^7 \sin^5(x)\cos^3(x)\end{align*}$$

$$\begin{align*}\sin(7x) =\ & \sin(x) \\ -\ & 2\sin^3(x) \\ +\ & 6\cos^2(x)\sin(x) \\ -\ & 16\cos^4(x)\sin^3(x) \\ +\ & 48\cos^2(x)\sin^5(x) \\ -\ & 32\cos^2(x)\sin^3(x)\end{align*}$$

I can do these by hand, but term count explodes so awfully... until everything combines at the end. It's about as efficient as expanding $(3x + 5)^{97}$ using the distributive property!! Thankfully, we have the binomial theorem.

Question: Is there any reason to think there is a "generative pattern" for constructing these expressions, analogous to the binomial theorem for the distributive property? If there is, what is a systematic way I could go about finding this pattern?

The radix-conversion-like nature of the manual algorithm doesn't seem to lend itself to a direct algebraic abstraction. But the procedure is so elegantly repetitive that I'm inclined to think there must be a way to write these directly without all the recursion.

This is a personal project. If and when I find a solution, I will post it here.

Answer

Following up on my comment under the question, using $e^{in\theta} = (e^{i\theta})^n$, $e^{i\theta} = \cos\theta + i\sin\theta$, and the Binomial theorem, we have

$$\cos(n\theta) + i\sin(n\theta) = (\cos\theta + i\sin\theta)^n = \sum_{k\,=\,0}^{n} i^k\,{n \choose k} \cos^{n-k}\!\theta \sin^k\!\theta$$

Splitting the sum into even and odd powers of $\sin\theta$,

$$\cos(n\theta) + i\sin(n\theta) = \sum_{r\,=\,0}^{\lfloor n\,/\,2 \rfloor} i^{2r}\,{n \choose 2r} \cos^{n-2r}\!\theta \sin^{2r}\!\theta + \sum_{r\,=\,0}^{\lceil n\,/\,2 \rceil - 1} i^{2r+1}\,{n \choose 2r+1} \cos^{n-2r-1}\!\theta \sin^{2r+1}\!\theta$$

from which we get

$$\cos(n\theta) = \sum_{r\,=\,0}^{\lfloor n\,/\,2 \rfloor} (-1)^{r}\,{n \choose 2r} \cos^{n-2r}\!\theta \sin^{2r}\!\theta$$

$$\sin(n\theta) = \sum_{r\,=\,0}^{\lceil n\,/\,2 \rceil - 1} (-1)^{r}\,{n \choose 2r+1} \cos^{n-2r-1}\!\theta \sin^{2r+1}\!\theta$$