請參考

填充第8題
題目應是相異實根的個數吧?
\(f'\left( x \right)=3{{x}^{2}}+2ax+b=0\)之兩根為\({{x}_{1}},{{x}_{2}}\)
\(3{{\left[ f\left( x \right) \right]}^{2}}+2af\left( x \right)+b=0\)之根為\(f\left( x \right)={{x}_{1}}\)或\(f\left( x \right)={{x}_{2}}\)的根
畫圖可知\(f\left( x \right)={{x}_{1}}\)有兩相異實根，\(f\left( x \right)={{x}_{2}}\)有一根
故\(3{{\left[ f\left( x \right) \right]}^{2}}+2af\left( x \right)+b=0\)的相異實根個數為3

填充第12題
\(\begin{align}
  & \tan n\theta =\frac{C_{1}^{n}\tan \theta -C_{3}^{n}{{\tan }^{3}}\theta +C_{5}^{n}{{\tan }^{5}}\theta -\cdots }{C_{0}^{n}-C_{2}^{n}{{\tan }^{2}}\theta +C_{4}^{n}{{\tan }^{4}}\theta -\cdots } \\
& n=2k-1 \\
& \sum\limits_{j=0}^{2k-1}{\left| {{a}_{j}} \right|}=C_{0}^{2k-1}+C_{1}^{2k-1}+C_{2}^{2k-1}+C_{3}^{2k-1}+\cdots +C_{2k-1}^{2k-1}={{2}^{2k-1}} \\
\end{align}\)

113.04.25補充
There is a unique sequence of integers \(a_1, a_2, \cdots a_{2023}\) such that
\(\displaystyle tan2023x=\frac{a_1tanx+a_3 tan^3x+a_5tan^5x+\ldots+a_{2023}tan^{2023}x}{1+a_2tan^2x+a_4tan^4x+\ldots+a_{2022}tan^{2022}x}\)whenever \(\tan 2023x\) is defined. What is \(a_{2023}\)?
(A)\(-2023\)　(B)\(-2022\)　(C)\(-1\)　(D)1　(E)2023
(2023AMC12A，連結有解答https://artofproblemsolving.com/ ... Problems/Problem_25)

已知\(\displaystyle tan10\theta=\frac{a_1\cdot tan\theta+a_3\cdot tan^3\theta+a_5\cdot tan^5\theta+a_7\cdot tan^7\theta+a_9\cdot tan^9 \theta}{a_0+a_2\cdot tan^2\theta+a_4\cdot tan^4\theta+a_6\cdot tan^6\theta+a_8\cdot tan^8\theta+a_{10}\cdot tan^{10}\theta}\)，則\(\displaystyle \sum_{k=0}^{10}|\;a_k|\;=\)　　　。
(113中山女高，https://math.pro/db/thread-3834-1-1.html)

參考陳老師的文章
連結已失效h ttps://highscope.ch.ntu.edu.tw/wordpress/?p=17104