(a) 使用 De Moivre 定理证明

(b) 据此求方程

的根，并将其写成 $\cot(q\pi)$ 的形式，其中 $q$ 为有理数。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

用 de Moivre 定理，分别取六次方展开的实部和虚部。两式相除后，分子分母同除以 $\sin^6\theta$，即可写成余切的幂。

Write $c=\cos\theta$ and $s=\sin\theta$. By de Moivre’s theorem,

Taking the ratio and dividing its numerator and denominator by $s^6$ gives, where the cotangents are defined,

把多项式看成 (a) 中的分子减去分母。令 $x=\cot\theta$，在 $0<\theta<\pi$ 中解 $\cot6\theta=1$，就不会重复取到同一个 $x$。

With $x=\cot\theta$, multiplying the polynomial by $\sin^6\theta$ gives $\cos6\theta-\sin6\theta$. Its zeros satisfy $\cos6\theta=\sin6\theta$ and hence $\cot6\theta=1$, since sine and cosine cannot both be zero. Therefore

For $0<\theta<\pi$, the values are $k=0,1,\ldots,5$. The roots are

They are distinct because cotangent is strictly decreasing on $(0,\pi)$. These six roots account for the full degree.