計算證明題4.
設數列\(\langle\;a_n\rangle\;\)滿足\(\displaystyle a_n=\int_0^1 (1-x^2)^{\frac{n}{2}}dx\)，\(n=0,1,2,3,\ldots\)。
(1)證明：\(\displaystyle a_n=\frac{n}{n+1}a_{n-2},n\ge 2\)。
(2)試求\(\displaystyle \lim_{n\to \infty}\frac{a_{n+1}}{a_n}\)的值。
[解答]
計算4(1)
\(a_{n}=\int^{1}_{0}(1-x^{2})^{n/2}dx\)
\( \ \ \ =\int^{\pi/2}_{0}\cos^{n}{\theta}d{\sin{\theta}}(=\int^{\pi/2}_{0}\cos^{n+1}{\theta}d{\theta}\))
\( \ \ \ =\cos^{n}{\theta}\sin{\theta}|^{\pi/2}_{0}+\int^{\pi/2}_{0}n\cos^{n-1}{\theta}\sin^{2}{\theta}d{\theta}\)
\( \ \ \ =n\int^{\pi/2}_{0}\cos^{n-1}{\theta}(1-\cos^{2}{\theta})d{\theta}\)
\( \ \ \ =n\int^{\pi/2}_{0}\cos^{n-1}{\theta}d{\theta}-n\int^{\pi/2}_{0}\cos^{n+1}{\theta}d{\theta}\)
\( \ \ \ =na_{n-2}-na_{n}\)
整理得
\(\displaystyle a_{n}=\frac{n}{n+1}a_{n-2}\)



計算4(2)
考慮 \(0\leq \theta \leq \pi/2\Rightarrow0\leq \cos{\theta}\leq 1\Rightarrow\cos^{n}{\theta}\) 遞減\(\Rightarrow a_{n}\) 遞減
即 \(\displaystyle \frac{a_{n+2}}{a_{n}}\leq \frac{a_{n+1}}{a_{n}}\leq \frac{a_{n+1}}{a_{n+1}}=1\)
又因 \(\displaystyle \lim\limits_{n\rightarrow \infty}\frac{a_{n+2}}{a_{n}}=1\)
由夾擠定理得 \(\displaystyle \lim\limits_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=1\)

計算證明題2.
在坐標平面上，\(A,B,C\)三點形成直角三角形，其中\(\angle C=90^{\circ}\)，\(\overline{AB}=60\)，又過\(A\)與\(B\)兩點的中線方程式分別為\(y=x+2021\)與\(y=2x+110\)。試求三角形\(ABC\)的面積。
[解答]
計算2另解
設 \(\overline{BC}=2a\)，中點為 \(M_{A} \)、\(\overline{AC}=2b\)，中點為 \(M_{B} \)、重心為 \(G\)，
\(\displaystyle \tan{M_{A}GB}=\frac{2-1}{1+2*1}=\frac{1}{3}\)，\(\displaystyle \sin{AGB}=\frac{1}{\sqrt{10}}\)、\(a^{2}+b^{2}=900\)
由中線定理得
\(\overline{AM_{A}}^{2}=3b^{2}+900\)、\(\overline{BM_{B}}^{2}=3a^{2}+900\)
由 \(\displaystyle \Delta AGB=\frac{1}{3}\Delta ABC\) 得
\(\displaystyle \frac{1}{2}\times \frac{2}{3}\overline{AM_{A}}\times \frac{2}{3}\overline{BM_{A}}\times \sin{AGB}=\frac{1}{3}\times \frac{1}{2}\times (2a)\times(2b)\)
整理得 \(ab=200\)，
\(\Delta ABC=\frac{1}{2}\times (2a)\times(2b)=2ab=400\)

如圖

tan兩線夾角\(\displaystyle =\frac{m_1-m_2}{1+m_1m_2}\)