By \(\mathbb{R}\) we denote the set of real numbers and by \(\mathbb{N}\) the set of positive integers. By trigonometric functions here we mean the functions cosine and sine; that is with \(\theta \in \mathbb{R}\), \[ \theta \mapsto \cos \theta \quad \text{and} \quad \theta \mapsto \sin \theta. \] By Multiple-Angle Functions we mean the trigonometric functions in which the variable is multiplied by a positive integer; that is with \(n\in\mathbb{N}\) and \(\theta \in \mathbb{R}\). \[ \theta \mapsto \cos(n\theta) \quad \text{and} \quad \theta \mapsto \sin(n\theta). \]
Below are my versions of the Power-reduction formulae: \begin{align*} (\cos \theta)^{2m} & = \frac{1}{4^{m}} \binom{2m}{m} + \frac{2}{4^{m}} \sum_{k=1}^{m} \binom{2m}{m-k} \cos(2k\theta), \ \ m\in\mathbb{N}, \\[6pt] (\cos \theta)^{2m+1} & =\frac{1}{4^{m}} \sum_{k=0}^{m} \binom{2m+1}{m-k} \cos\bigl((2k+1)\theta\bigr), \ \ m\in\mathbb{N}\cup\{0\}, \\[6pt] \bigl(\sin \theta\bigr)^{2m} & = \frac{1}{4^{m}} \binom{2m}{m} + \frac{2}{4^{m}} \sum_{k=1}^{m} (-1)^{k} \binom{2m}{m-k} \cos (2k \theta ), \ \ m\in\mathbb{N}, \\[6pt] \bigl(\sin \theta\bigr)^{2m+1} & = \frac{1}{4^{m}} \sum_{k=0}^{m} (-1)^{k} \binom{2m + 1}{m - k} \sin\bigl((2k + 1)\theta\bigr), \ \ m\in\mathbb{N}\cup\{0\}. \end{align*}
The special cases for the first four instances of each of the above formulas are given below.
The first four even powers of \(\cos\theta\): \begin{align*} \bigl(\cos \theta\bigr)^{2} & = \frac{1}{2} + \frac{1}{2} \cos(2\theta),\\[6pt] \bigl(\cos \theta\bigr)^{4} & = \frac{3}{8} + \frac{1}{2} \cos(2\theta) + \frac{1}{8} \cos(4\theta),\\[6pt] \bigl(\cos \theta\bigr)^{6} & = \frac{5}{16} + \frac{15}{32} \cos(2\theta) + \frac{3}{16} \cos(4\theta) + \frac{1}{32} \cos(6\theta), \\[6pt] \bigl(\cos \theta\bigr)^{8} & = \frac{35}{128} + \frac{7}{16} \cos(2\theta) + \frac{7}{32} \cos(4\theta) + \frac{1}{16} \cos(6\theta) + \frac{1}{128} \cos(8\theta). \end{align*}
The first four odd powers of \(\cos\theta\): \begin{align*} \bigl(\cos \theta\bigr)^{3} & = \frac{3}{4} \cos(\theta) + \frac{1}{4} \cos(3\theta),\\[6pt] \bigl(\cos \theta\bigr)^{5} & = \frac{5}{8} \cos(\theta) + \frac{5}{16} \cos(3\theta) + \frac{1}{16} \cos(5\theta),\\[6pt] \bigl(\cos \theta\bigr)^{7} & = \frac{35}{64} \cos(\theta) + \frac{21}{64} \cos(3\theta) + \frac{7}{64} \cos(5\theta) + \frac{1}{64} \cos(7\theta), \\[6pt] \bigl(\cos \theta\bigr)^{9} & = \frac{63}{128} \cos(\theta) + \frac{21}{64} \cos(3\theta) + \frac{9}{64} \cos(5\theta) + \frac{9}{256} \cos(7\theta) + \frac{1}{256} \cos(9\theta). \end{align*}
The first four even powers of \(\sin\theta\): \begin{align*} \bigl(\sin \theta\bigr)^{2} & = \frac{1}{2} - \frac{1}{2} \cos(2\theta),\\[6pt] \bigl(\sin \theta\bigr)^{4} & = \frac{3}{8} - \frac{1}{2} \cos(2\theta) + \frac{1}{8} \cos(4\theta),\\[6pt] \bigl(\sin \theta\bigr)^{6} & = \frac{5}{16} - \frac{15}{32} \cos(2\theta) + \frac{3}{16} \cos(4\theta) - \frac{1}{32} \cos(6\theta), \\[6pt] \bigl(\sin \theta\bigr)^{8} & = \frac{35}{128} - \frac{7}{16} \cos(2\theta) + \frac{7}{32} \cos(4\theta) - \frac{1}{16} \cos(6\theta) + \frac{1}{128} \cos(8\theta). \end{align*}
The first four odd powers of \(\sin\theta\): \begin{align*} \bigl(\sin \theta\bigr)^{3} & = \frac{3}{4}\sin(\theta) - \frac{1}{4} \sin(3\theta),\\[6pt] \bigl(\sin \theta\bigr)^{5} & = \frac{5}{8}\sin(\theta) - \frac{5}{16}\sin(3\theta) + \frac{1}{16} \sin(5\theta) ,\\[6pt] \bigl(\sin \theta\bigr)^{7} & = \frac{35}{64}\sin(\theta) - \frac{21}{64}\sin(3\theta) + \frac{7}{64}\sin(5\theta) - \frac{1}{64} \sin(7\theta), \\[6pt] (\sin \theta)^9 &= \frac{63}{128} \sin \theta - \frac{21}{64} \sin(3\theta) + \frac{9}{64} \sin(5\theta) - \frac{9}{256} \sin(7\theta) + \frac{1}{256} \sin(9\theta). \end{align*}
More formulas in the same spirit: \begin{align*} (\sin x)(\cos x) & = \frac{1}{2} \sin (2 x) \\ (\sin x)(\cos x)^2 & = \frac{1}{4} \sin (x) + \frac{1}{4} \sin (3 x) \\ (\sin x)(\cos x)^3 & = \frac{1}{4} \sin (2 x) + \frac{1}{8} \sin (4 x) \\ (\sin x)^3 & = \frac{3}{4} \sin (x) - \frac{1}{4} \sin (3 x) \\ (\sin x)^3(\cos x) & = \frac{1}{4} \sin (2 x) - \frac{1}{8} \sin (4 x) \\ (\sin x)^3 (\cos x)^2 & = \frac{1}{8} \sin (x) + \frac{1}{16} \sin (3 x) - \frac{1}{16} \sin (5 x) \\ (\sin x)^3(\cos x)^3 & = \frac{3}{32} \sin (2 x) - \frac{1}{32} \sin (6 x) \\ \end{align*}
\begin{align*} \cos\bigl((2m+1)\theta\bigr) & = (-1)^m \sum_{k=0}^{m} (-4)^k \frac{2m+1}{m+k+1} \binom{m+k+1}{2k+1} (\cos \theta)^{2k+1} \\[6pt] \cos\bigl(2m \theta\bigr) & = (-1)^m \sum_{k=0}^{m} (-4)^k \frac{m}{m+k} \binom{m+k}{m-k} (\cos\theta)^{2k} \\[6pt] \sin\bigl((2m+1)\theta\bigr) & = (-1)^m \sin(\theta) \sum_{k=0}^{m} (-4)^k \binom{m+k}{m-k} (\cos\theta)^{2k} \\[6pt] \sin\bigl(2m \theta\bigr) & = 2(-1)^{m-1} (\sin \theta) \sum_{k=1}^{m} (-4)^{k-1} \binom{m+k-1}{m-k} (\cos\theta)^{2k-1} \end{align*}
\begin{align*} \cos(3\theta) & = -3 (\cos\theta) + 4 (\cos\theta)^3 \\[6pt] \cos(5\theta) & = 5(\cos\theta) - 20(\cos\theta)^3 + 16(\cos\theta)^5 \\[6pt] \cos(7\theta) & = -7(\cos\theta) + 56(\cos\theta)^3 - 112(\cos\theta)^5 + 64(\cos\theta)^7 \\[6pt] \cos(9\theta) & = 9(\cos\theta) - 120(\cos\theta)^3 + 432(\cos\theta)^5 - 576(\cos\theta)^7 + 256(\cos\theta)^9 \end{align*}
\begin{align*} \cos(2\theta) & = - 1 + 2 (\cos\theta)^2 \\[6pt] \cos(4\theta) & = 1 - 8(\cos\theta)^2 + 8(\cos\theta)^4 \\[6pt] \cos(6 \theta) & = -1 + 18(\cos\theta)^2 - 48(\cos\theta)^4 + 32(\cos\theta)^6 \\[6pt] \cos(8\theta) & = 1 - 32(\cos\theta)^2 + 160(\cos\theta)^4 - 256(\cos\theta)^6 + 128(\cos\theta)^8 \end{align*}
\begin{align*} \sin(3\theta) & = (\sin\theta)\bigl(-1 + 4 (\cos\theta)^2 \bigr) \\[6pt] \sin(5\theta) & = (\sin\theta) \bigl(1 - 12(\cos\theta)^2 + 16(\cos\theta)^4 \bigr) \\[6pt] \sin(7\theta) & = (\sin\theta) \bigl(-1 + 24(\cos\theta)^2 - 80 (\cos\theta)^4 + 64 (\cos\theta)^6\bigr) \\[6pt] \sin(9\theta) & = (\sin\theta) \bigl(1 - 40(\cos\theta)^2 + 240 (\cos\theta)^4 - 448 (\cos\theta)^6 + 256 (\cos\theta)^8\bigr) \end{align*}
\begin{align*} \sin(2\theta) & = 2 (\sin\theta) \bigl(\cos\theta \bigr) \\[6pt] \sin(4\theta) & = 2 (\sin\theta) \bigl( -2 (\cos\theta) + 4 (\cos\theta)^3 \bigr) \\[6pt] \sin(6 \theta)& = 2 (\sin\theta) \bigl(3 (\cos\theta) - 16 (\cos\theta)^3 + 16 (\cos\theta)^5 \bigr) \\[6pt] \sin(8\theta)& = 2 (\sin\theta) \bigl(-4 (\cos\theta) + 40 (\cos\theta)^3 - 96 (\cos\theta)^5 + 64 (\cos\theta)^7 \bigr) \end{align*}