Note that, if&nbsp;\(z = \cos \theta + i \sin \theta\), that is, \(|z| = 1\), then&nbsp;\(z^* = \frac{1}{z} = \cos \theta - i \sin \theta\).
\[z + z^* = 2 \mathrm{Re}(z) \quad \text{and} \quad z - z^* = 2i \mathrm{Im}(z)\]
Therefore,
\[z + \frac{1}{z} = 2 \cos \theta \quad \text{and} \quad z - \frac{1}{z} = 2i \sin \theta\]
Furthermore, de Moivre's Theorem shows us that, for any \(n \in \mathbb{Z}\),
\[z^n = \cos n\theta + i \sin n\theta \quad \text{and}\]
\[\frac{1}{z^n} = (\cos \theta + i \sin \theta)^{-n} = \cos(-n\theta) + i \sin(-n\theta) = \cos n\theta - i \sin n\theta\]
Hence,
\[z^n + \frac{1}{z^n} = 2 \cos n\theta \quad \text{and} \quad z^n - \frac{1}{z^n} = 2i \sin n\theta\]

Note that, if \(z = \cos \theta + i \sin \theta\), that is, \(|z| = 1\), then \(z^* = \frac{1}{z} = \cos \theta - i \sin \theta\).

\[z + z^* = 2 \mathrm{Re}(z) \quad \text{and} \quad z - z^* = 2i \mathrm{Im}(z)\]

\[z + \frac{1}{z} = 2 \cos \theta \quad \text{and} \quad z - \frac{1}{z} = 2i \sin \theta\]

Furthermore, de Moivre's Theorem shows us that, for any \(n \in \mathbb{Z}\),

\[z^n = \cos n\theta + i \sin n\theta \quad \text{and}\]

\[\frac{1}{z^n} = (\cos \theta + i \sin \theta)^{-n} = \cos(-n\theta) + i \sin(-n\theta) = \cos n\theta - i \sin n\theta\]

\[z^n + \frac{1}{z^n} = 2 \cos n\theta \quad \text{and} \quad z^n - \frac{1}{z^n} = 2i \sin n\theta\]

Powers of sines and cosines

Powers of sines and cosines