and

$$z^{-n} = \cos n\theta - j \sin n\theta$$

以便

$$z^n + z^{-n} = 2 \cos n\theta \quad (3.20a)$$

$$z^n - z^{-n} = 2j \sin n\theta \quad (3.20b)$$

利用这些结果, $\cos^n \theta$ and $\sin^n \theta$ 可以用倍角的正弦和余弦来表示, 如例所示 3.23.

Example 3.23

按倍角的正弦和余弦展开

(a) $\cos^5 \theta$      (b) $\sin^6 \theta$

Solution (a) Using (3.20a) with $n = 1$ ,

$$(2 \cos \theta)^5 = \left(z + \frac{1}{z}\right)^5 = z^5 + 5z^3 + 10z + \frac{10}{z} + \frac{5}{z^3} + \frac{1}{z^5}$$

以便

$$32 \cos^5 \theta = \left(z^5 + \frac{1}{z^5}\right) + 5 \left(z^3 + \frac{1}{z^3}\right) + 10 \left(z + \frac{1}{z}\right)$$

which, 关于使用 (3.20a) with $n = 5$ , 3 and 1, gives

$$\cos^5 \theta = \frac{1}{32}(2 \cos 5\theta + 10 \cos 3\theta + 20 \cos \theta) = \frac{1}{16}(\cos 5\theta + 5 \cos 3\theta + 10 \cos \theta)$$

(b) Using (3.20b) with $n = 1$ ,

$$(2j \sin \theta)^6 = \left(z - \frac{1}{z}\right)^6 = z^6 - 6z^4 + 15z^2 - 20 + \frac{15}{z^2} - \frac{6}{z^4} + \frac{1}{z^6}$$

which, 注意到 $j^6 = -1$ , gives

$$-64 \sin^6 \theta = \left(z^6 + \frac{1}{z^6}\right) - 6 \left(z^4 + \frac{1}{z^4}\right) + 15 \left(z^2 + \frac{1}{z^2}\right) - 20$$

Using (3.20a) with $n = 6$ , 4 and 2 则给出

$$\begin{aligned} \sin^6 \theta &= -\frac{1}{64}(2 \cos 6\theta - 12 \cos 4\theta + 30 \cos 2\theta - 20) \ &= \frac{1}{32}(10 - 15 \cos 2\theta + 6 \cos 4\theta - \cos 6\theta) \end{aligned}$$

Conversely, 棣莫弗定理可用于展开 $\cos n\theta$ and $\sin n\theta$ , where $n$ 是一个正整数, 作为以……为变量的多项式 $\cos \theta$ and $\sin \theta$ . 由该定理

$$\cos n\theta + j \sin n\theta = (\cos \theta + j \sin \theta)^n$$

我们得到, writing $s = \sin \theta$ and $c = \cos \theta$ 为方便起见,

$$\cos n\theta + j \sin n\theta = (c + js)^n = c^n + jnc^{n-1}s + j^2 \frac{n(n-1)}{2!}c^{n-2}s^2 + \dots + j^n s^n$$