三角函数恒等式

• \(\cos \frac{x}{2^n} \cdot \cos \frac{x}{2^{n-1}} \cdots \cos \frac{x}{2} = \frac{\sin x}{2^n \sin \frac{x}{2^n}}\)
• \(\sin x + \sin 2x + \cdots + \sin nx = \frac{\cos \frac{x}{2} - \cos(n + \frac{1}{2})x}{2 \sin \frac{x}{2}}\)
• \(\cos x + \cos 2x + \cdots + \cos nx = \frac{\sin (n+\frac{1}{2})x - \sin \frac{x}{2}}{2\sin \frac{x}{2}} = \frac{\sin(n+\frac{1}{2})x}{2 \sin \frac{x}{2}} - \frac{1}{2}\)
• \(\arctan \frac{1}{2k^2} = \arctan \frac{k}{k+1} - \arctan \frac{k-1}{k}\)

证明：(1) 将右侧分母乘到左侧去，用二倍角即可。

(2) 左边乘 \(\sin \frac{x}{2}\) ，令 \(I = \sin x + \sin 2x + \cdots + \sin nx\) ，使用 积化和差公式 得到

\[ \begin{align} \sin \frac{x}{2} I &= \sum\limits_{k = 1}^n \sin kx \sin \frac{x}{2} \\ &= - \frac{1}{2}\sum\limits_{k = 1}^n [\cos (k + \frac{1}{2})x - \cos(k - \frac{1}{2})x]\\ &= \frac{1}{2}\left[ \cos \frac{x}{2} - \cos(n + \frac{1}{2})x\right] \end{align}\]

(3)和(2)类似