Evaluating $\frac{1}{\sin(2x)} + \frac{1}{\sin(4x)} + \frac{1}{\sin(8x)} + \frac{1}{\sin(16x)}$ 
Evaluate
$$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(4x)} + \dfrac{1}{\sin(8x)} + \dfrac{1}{\sin(16x)}$$

It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.
Rewriting and using trigonometric identities
$$\dfrac{1}{\sin(2x)} + \dfrac{1}{\sin(2x) \cos (2x)} + \dfrac{1}{ 2\big [2\sin (2x)\cos (2x)\cos (4x)\big ]} + \dfrac{1}{\sin(16x)}$$
What am I missing?
Regards

$$\frac{1}{\sin2x}=\frac{\sin x}{\sin 2x \sin x}=\frac{\sin (2x-x)}{\sin 2x \sin x}=\frac{\sin 2x \cos x - \cos 2x \sin x}{\sin 2x \sin x}=\cot x - \cot2x$$
$$\frac{1}{\sin4x}=\cot 2x - \cot4x$$
$$\frac{1}{\sin8x}=\cot 4x - \cot8x$$
$$\frac{1}{\sin16x}=\cot 8x - \cot16x$$
$$\frac{1}{\sin2x}+\frac{1}{\sin4x}+\frac{1}{\sin8x}+\frac{1}{\sin16x}=\cot x-\cot 16x$$
