回復 8# wdemhueebhee 的帖子
選擇第 6 題：
令 \(x=2\sin\theta\)
則 \(dx = 2\cos\theta d\theta\)
\(\displaystyle \int_0^1 \frac{1}{\sqrt{4-x^2}}dx = \int_0^{\pi/6} \frac{2\cos\theta}{\sqrt{4-4\sin^2\theta}}d\theta=\int_0^{\pi/6}d\theta=\frac{\pi}{6}\)
填充第 2 題：
\(-\sin^2\theta-\left(x+\cos\theta\right)^2=0\Rightarrow x=-\left(\cos\theta\pm i\sin\theta\right)\)
\(\Rightarrow a^k+b^k = \left[-\left(\cos\theta+ i\sin\theta\right)\right]^k+\left[-\left(\cos\theta- i\sin\theta\right)\right]^k\)
\(= \left[-\left(\cos\theta+ i\sin\theta\right)\right]^k+\left[-\left(\cos\left(-\theta\right)+ i\sin\left(-\theta\right)\right)\right]^k\)
\(= \left(-1\right)^k\left(\cos k\theta+ i\sin k\theta\right)+\left(-1\right)^k\left(\cos \left(-k\theta\right)+ i\sin\left(-k\theta\right)\right)\)
\(= \left(-1\right)^k\left(\cos k\theta+ i\sin k\theta\right)+\left(-1\right)^k\left(\cos k\theta- i\sin k\theta\right)\)
\(=\left(-1\right)^k \cdot 2\cos k\theta\)
