$ \dfrac{2\sin x}{\cos 3x} + \dfrac{2\sin 3x}{\cos 9x} + \dfrac{2\sin 9x}{\cos 27x} = \tan 27 x- \tan x$.

I did not get even to start with which formula. I tried using multiple angle identities but did not find any suitable place to use those. So please help.


marked as duplicate by lab bhattacharjee trigonometry Jan 6 '16 at 12:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The question already exists $\endgroup$ – Archis Welankar Jan 6 '16 at 12:11
  • $\begingroup$ @ Archis where does it exist? I haven't posted it before $\endgroup$ – Ger Wyn Jan 6 '16 at 12:15
  • $\begingroup$ There are many people who ask questions please see tags. $\endgroup$ – Archis Welankar Jan 6 '16 at 12:21

See if this leads you anywhere. I've included an example for your first term:

$\dfrac{2\sin x}{\cos3x} = \dfrac{\dfrac{2\sin x\cos x}{\cos x}}{\cos3x} = \dfrac{2\sin x\cos x}{\cos x \times \cos3x} = \dfrac{2\sin x\cos x}{\dfrac{1}{2}(\cos4x + \cos2x)}$

One advantage of this is you may be able to begin condensing using the double angle formula. Also note that they're all in multiples of 3, i.e $x$ and $3x$, $3x$ and $9x$, etc. so a few substitutions may help.

EDIT: OP, I'd recommend you to take a look at the answers on this post when it was previously asked i.e here. Particularly Lab Bhatacharjee's post is simple and insightful, although I myself am not yet familiar with telescoping series (I am a sophomore in high school). Also, if you'd like an inductive approach, take a look at Leg's answer.


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