(a) 使用棣莫弗定理证明

(b) 由此求方程

的根，并将其表示为 $\operatorname{cosec}(q\pi)$ 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。

取五次方的虚部，再用 $\cos^2\theta=1-\sin^2\theta$ 得到只含 $\sin\theta$ 的多项式。取倒数后，将分子分母同除以 $\sin^5\theta$。

Write $c=\cos\theta$ and $s=\sin\theta$. De Moivre’s theorem gives $(c+\mathrm{i}s)^5=\cos5\theta+\mathrm{i}\sin5\theta$. Expanding,

Taking imaginary parts,

Taking the reciprocal where both cosecants are defined gives

用 (a) 把方程化为 $\csc5\theta=2$。选择 $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$，避免同一个正弦值被重复使用。

With $x=\csc\theta$, the equation becomes $x^5=2(5x^4-20x^2+16)$. The denominator $5x^4-20x^2+16$ cannot vanish at a root, since the equation would then give $x=0$, whereas its denominator at $0$ is $16$. Using (a), this gives $\csc5\theta=2$ and hence $\sin5\theta=\frac12$. Thus

On $[-\frac\pi2,\frac\pi2]$, the five solutions give

The five values are distinct, because sine is strictly increasing on the chosen interval and none of these sines is zero. They account for all roots of the degree-five polynomial.