Giải thích các bước giải:

Đặt $S_n=\dfrac{1}{\sin2x}+\dfrac{1}{\sin4x}+\dfrac{1}{\sin8x}+...+\dfrac{1}{\sin2^nx}$

Với $n=1\to S_1=\dfrac{1}{\sin2x}$

Mà $\cot x-\cot 2x=\dfrac{\cos x}{\sin x}-\dfrac{\cos2x}{\sin2x}=\dfrac{\cos x\sin2x-\sin x\cos2x}{\sin x\sin2x}$

$\to \cot x-\cot 2x=\dfrac{\cos x\sin2x-\sin x\cos2x}{\sin x\sin2x}$

$\to \cot x-\cot 2x=\dfrac{\sin(2x-x)}{\sin x\sin2x}$

$\to \cot x-\cot 2x=\dfrac{\sin x}{\sin x\sin2x}$

$\to \cot x-\cot 2x=\dfrac{1}{\sin2x}$

$\to S_1=\dfrac{1}{\sin2x}$ đúng

Giả sử $n=k$ đúng

$\to \dfrac{1}{\sin2x}+\dfrac{1}{\sin4x}+\dfrac{1}{\sin8x}+...+\dfrac{1}{\sin2^kx}=\cot x-\cot 2^kx$

Ta chứng minh $n=k+1$ cũng đúng

$\to \dfrac{1}{\sin2x}+\dfrac{1}{\sin4x}+\dfrac{1}{\sin8x}+...+\dfrac{1}{\sin2^{k+1}x}=\cot x-\cot 2^kx$

Mà:

$S_k= \dfrac{1}{\sin2x}+\dfrac{1}{\sin4x}+\dfrac{1}{\sin8x}+...+\dfrac{1}{\sin2^{k}x}$

$\to S_k+\dfrac{1}{\sin2^{k+1}x}= \dfrac{1}{\sin2x}+\dfrac{1}{\sin4x}+\dfrac{1}{\sin8x}+...+\dfrac{1}{\sin2^{k}x}+\dfrac{1}{\sin2^{k+1}x}$

$\to S_k+\dfrac{1}{\sin2^{k+1}x}= S_{k+1}$

$\to  S_{k+1}=\cot x-\cot 2^kx+\dfrac{1}{\sin2^{k+1}x}$

$\to  S_{k+1}=\cot x+\dfrac{1}{\sin2^{k+1}x}-\cot 2^kx$

$\to  S_{k+1}=\cot x+\dfrac{1}{\sin2^{k+1}x}-\dfrac{\cos2^kx}{\sin2^kx}$

$\to  S_{k+1}=\cot x+\dfrac{1}{\sin2^{k+1}x}-\dfrac{2\cos2^kx\cdot\cos2^kx}{2\sin2^kx\cdot\cos2^kx}$

$\to  S_{k+1}=\cot x+\dfrac{1}{\sin2^{k+1}x}-\dfrac{2(\cos2^kx)^2}{\sin(2\cdot 2^kx)}$

$\to  S_{k+1}=\cot x+\dfrac{1}{\sin2^{k+1}x}-\dfrac{2(\cos2^kx)^2}{\sin 2^{k+1}x}$

$\to  S_{k+1}=\cot x+\dfrac{1-2(\cos2^kx)^2}{\sin 2^{k+1}x}$

$\to  S_{k+1}=\cot x-\dfrac{2(\cos2^kx)^2-1}{\sin 2^{k+1}x}$

$\to  S_{k+1}=\cot x-\dfrac{\cos2^{k+1}x}{\sin 2^{k+1}x}$

$\to  S_{k+1}=\cot x-\cot(2^{k+1}x)$ đúng

Biểu thức đúng với mọi $n$ thỏa mãn đề

$\to đpcm$