計算
(1) 因為\(\displaystyle \overline{OA}=\overline{OB}=40,\overline{OC}=\overline{OD}=36\)
且\(\angle{AOC}=\angle{BOD}=\theta\)
故\(\displaystyle \triangle{AOC}\cong  \triangle{BOD} (SAS)\)

(2)\(\overline{BC}^2+\overline{AC}^2=\overline{BC}^2+\overline{BD}^2=\overline{CD}^2\)

在\(\triangle{COD}\)中，\(\overline{CD}^2=36^2+36^2-2\times 36^2 \times cos\angle{AOB}=324\)

10. 定義 \(\displaystyle f_n(x)=cosxcos2x\cdots cos(nx)\)

有 \(\displaystyle f_n(x)=f_{n-1}(x)cos(nx)\)

一次微分得到 \(\displaystyle f'_n(x)=f'_{n-1}(x)cos(nx)-nf_{n-1}(x)sin(nx)\)

二次微分且 \(x=0\)代入得到  \(f''_n(0)=f''_{n-1}(0)-n^2\)，且   \( f''_1(0)=-1\)

即可解出 \(f''_n(0)\)的一般式為\(\displaystyle \frac{-n(n+1)(2n+1)}{6}\)

所求即為 \(\displaystyle \frac{n(n+1)(2n+1)}{6}\geq 2025\)

n=17時為1785，n=18時為2109滿足題目
故n=18

4. 設\(\displaystyle \alpha= \frac{4}{3},\beta=\frac{2}{3}\)

有\(2f(2)=2=\displaystyle f(\frac{2}{3})f(\frac{4}{3})\)

又\(\displaystyle 2f(\frac{4}{3})=f^2(\frac{2}{3})\)

因此可得 \(\displaystyle 4= f^3(\frac{2}{3}) \Rightarrow \displaystyle f(\frac{2}{3})=\sqrt[3]{4}\)