只需这一个恒等式,令 y=2k−1xy=2^{k-1}xy=2k−1x,就能为题目中每一项生成裂项所需的形式。
利用 cos2y=2cos2y−1\cos 2y=2\cos^2y-1cos2y=2cos2y−1 及 sin2y=2sinycosy\sin 2y=2\sin y\cos ysin2y=2sinycosy,可得 coty−cot2y=cosysiny−cos2ysin2y=cosysiny−2cos2y−12sinycosy=2cos2y−(2cos2y−1)2sinycosy=12sinycosy=1sin2y\cot y-\cot 2y=\frac{\cos y}{\sin y}-\frac{\cos 2y}{\sin 2y}=\frac{\cos y}{\sin y}-\frac{2\cos^2y-1}{2\sin y\cos y}=\frac{2\cos^2y-(2\cos^2y-1)}{2\sin y\cos y}=\frac{1}{2\sin y\cos y}=\frac{1}{\sin 2y}coty−cot2y=sinycosy−sin2ycos2y=sinycosy−2sinycosy2cos2y−1=2sinycosy2cos2y−(2cos2y−1)=2sinycosy1=sin2y1。