$\sin\frac{\pi}{10}$ 是 $f(x)=ax^3-bx+1$ 的零点
题目（8）： 若 $\sin\dfrac{\pi}{10}$ 是函数 $f(x)=ax^3-bx+1$（$a,b\in\mathbb{N}^*$）的一个零点，则 $f(1)=$（ ）
A. $2$ B. $3$ C. $4$ D. $5$
解：
$$\because \sin\frac{\pi}{10}=\cos\frac{2\pi}{5}=2\cos^2\frac{\pi}{5}-1=2\left(1-2\sin^2\frac{\pi}{10}\right)^2-1$$
$$=8\sin^4\frac{\pi}{10}-8\sin^2\frac{\pi}{10}+1$$
$$\therefore 8\sin^2\frac{\pi}{10}\left(\sin^2\frac{\pi}{10}-1\right)-\left(\sin\frac{\pi}{10}-1\right)=0$$
$$\therefore\left(\sin\frac{\pi}{10}-1\right)\left(8\sin^3\frac{\pi}{10}+8\sin^2\frac{\pi}{10}-1\right)=0$$
$$\therefore 8\sin^3\frac{\pi}{10}+8\sin^2\frac{\pi}{10}-1=0\quad\left(\sin\frac{\pi}{10}\neq 1\right)$$
$$\because 8\sin^2\frac{\pi}{10}-2=4\left(1-\cos\frac{\pi}{5}\right)-2=2-4\cos\frac{\pi}{5}$$
$$=2-2\cdot\frac{2\cos\frac{\pi}{5}\sin\frac{\pi}{5}}{\sin\frac{\pi}{5}}=2\,\frac{\sin\frac{\pi}{5}-\sin\frac{2\pi}{5}}{\sin\frac{\pi}{5}}=2\,\frac{-2\cos\frac{3\pi}{5}\sin\frac{\pi}{10}}{2\sin\frac{\pi}{10}\cos\frac{\pi}{10}}$$
$$=-2\,\frac{\sin\frac{2\pi}{10}}{\cos\frac{\pi}{10}}=-4\sin\frac{\pi}{10}$$
$$\therefore 8\sin^3\frac{\pi}{10}-4\sin\frac{\pi}{10}+1=0$$
$$\therefore a=8,\ b=4$$
$$\therefore f(1)=a-b+1=5$$
答案选 D。