Ta có: $\forall \alpha$ thì:$\cot \alpha-\cot2\alpha=\frac{\cos\alpha\sin2\alpha-\cos2\alpha\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{1}{\sin2\alpha}$Suy ra:$\sum_{k=1}^n\frac{1}{\sin2^kx}=\sum_{k=1}^n(\cot2^{k-1}x-\cot2^kx)=\cot x-\cot2^nx$
Ta có: $\forall \alpha$ thì:$\cot \alpha-\cot2\alpha=\frac{\cos\alpha\sin2\alpha-\cos2\alpha\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{1}{\sin2\alpha}$Suy 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: $\forall \alpha$ thì:$\cot \alpha-\cot2\alpha=\frac{\cos\alpha\sin2\alpha-\cos2\alpha\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{1}{\sin2\alpha}$Suy ra:$\sum_{k=1}^n\frac{1}{\sin2^
kx}=\sum_{k=1}^n(\cot2^{
k-1}x-\cot2^
kx)=\cot x-\cot2^nx$