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

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?



$$\sin(A-B)=\sin A \cos B-\cos A \sin B$$ $$\frac{1}{\sin 2x}=\dfrac{\sin(2x-x)}{\sin2x\sin x}=\cot x-\cot2x$$ $$\frac{1}{\sin 4x}=\dfrac{\sin(4x-2x)}{\sin4x\sin 2x}=\cot 2x-\cot4x$$ $$\frac{1}{\sin 8x}=\dfrac{\sin(8x-4x)}{\sin8x\sin 4x}=\cot 4x-\cot8x$$

$$\frac{1}{\sin 16x}=\dfrac{\sin(16x-8x)}{\sin16x\sin 8x}=\cot 8x-\cot16x$$

  • $\begingroup$ Where did that $\cot$ come from? $\endgroup$ – Busi Sep 6 '18 at 12:23
  • $\begingroup$ apply $\sin(a-b)$ formula $\endgroup$ – Deepesh Meena Sep 6 '18 at 13:42
  • $\begingroup$ @DeepeshMeena, Have you noticed the link? $\endgroup$ – lab bhattacharjee Sep 7 '18 at 7:44
  • $\begingroup$ which link can you provide me? $\endgroup$ – Deepesh Meena Sep 7 '18 at 9:42

$$\frac{1}{\sin2x}=\frac{\sin x}{\sin 2x \sin x}=\frac{\sin (2x-x)}{\sin 2x \sin x}=\frac{\sin 2x \cos x - \cos 2x \sin x}{\sin 2x \sin x}=\cot x - \cot2x$$

$$\frac{1}{\sin4x}=\cot 2x - \cot4x$$

$$\frac{1}{\sin8x}=\cot 4x - \cot8x$$

$$\frac{1}{\sin16x}=\cot 8x - \cot16x$$

$$\frac{1}{\sin2x}+\frac{1}{\sin4x}+\frac{1}{\sin8x}+\frac{1}{\sin16x}=\cot x-\cot 16x$$


Hint : Notice that $\frac{1}{\sin{(2x)})}$ is a common factor of all terms of the sum.

  • $\begingroup$ Then notice that $\frac{1}{\cos{2x}}$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems. $\endgroup$ – PackSciences Sep 6 '18 at 11:26

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