Ta có: ∀α thì:cotα−cot2α=cosαsin2α−cos2αsinαsinαsin2α=sinαsinαsin2α=1sin2αSuy ra:$\sum_{k=1}^n\frac{1}{\sin2^nx}=\sum_{k=1}^n(\cot2^{n-1}x-\cot2^nx)=\cot x-\cot2^nx$
Ta có:
∀α thì:
cotα−cot2α=cosαsin2α−cos2αsinαsinαsin2α=sinαsinαsin2α=1sin2αSuy ra:$\sum_{k=1}^n\frac{1}{\sin2^
kx}=\sum_{k=1}^n(\cot2^{
k-1}x-\cot2^
kx)=\cot x-\cot2^nx$