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The question asks to prove that - $$\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x) $$

I tried combining the first two or the last two fractions on the L.H.S to allow me to use the double angle formula and get $\sin 6x$ or $\sin 18x$ but that did not help at all.

I'm pretty sure that if I express everything in terms of $x$, the answer will ultimately appear but I'm also certain that there must be another simpler way.

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marked as duplicate by lab bhattacharjee trigonometry Mar 20 '16 at 14:07

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  • $\begingroup$ Does the identity tan x = sin 2x/(1+cos 2x) help? $\endgroup$ – Jeffrey L. May 26 '15 at 21:33
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In general, we have $$\sum_{k=1}^n \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2$$ We can prove this using induction. $n=1$ is trivial. All we need to make use of is that $$\tan(3t) = \dfrac{3\tan(t)-\tan^3(t)}{1-3\tan^2(t)}$$ For the inductive step, we have $$\sum_{k=1}^n \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2$$ Adding the last term, we obtain $$\sum_{k=1}^{n+1} \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2 + \dfrac{\sin\left(3^{n}x \right)}{\cos\left(3^{n+1}x\right)} = \dfrac{\tan\left(3^{n+1}x\right)-\tan(x)}2$$ where we again rely on $(\spadesuit)$ to express $\tan\left(3^{n+1}x\right)$ in terms of $\tan\left(3^{n}x\right)$.

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  • $\begingroup$ I'm not familiar with induction as of yet. I will try to read up on that but are you aware of any other approach? $\endgroup$ – MayankJain May 26 '15 at 19:58
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@MayankJain @user,

I don't know that the case for $n=1$ is trivial, especially for someone in a trig class currently. I will offer a proof without induction using substitution instead. Mayank, to see that this is the case for $n=1$, convert the RHS of the equation as follows:

$\dfrac{1}{2} \cdot \big{[}\dfrac{\sin 3x}{\cos 3x} - \dfrac{\sin x}{\cos x}]$ = (1) $\dfrac{1}{2} \cdot \big(\dfrac{\sin 3x \cos x - \cos 3x \sin x}{\cos 3x \cos x}\big{)} $ = (2) $ \dfrac{1}{2} \cdot \dfrac{\sin 2x}{\cos 3x \cos x}$ = (3) $\dfrac{1}{2} \cdot \dfrac{2\sin x \cos x}{\cos 3x \cos x}$ = $\dfrac{\sin x}{\cos 3x}$

This is pretty heavy on trig identities. We get equivalence (1) by multiplying out the fraction, equivalence (2) because $\sin(u - v) = \sin u \cos v - \cos u \sin v$, equivalence (3) because $\sin 2x = 2\sin x \cos x$, and, finally, (4) by cancellation.

Now, if you are unfamiliar with induction, substitution will help here as an alternative method. Since you can now derive the equivalence for the 'trivial' case, set $3x = u$ for the second case, and $9x = v$ for the third. Then you already know $\dfrac{\sin u}{\cos 3u} = \dfrac{1}{2} \cdot (\tan 3u - \tan u)$, and similarly, $\dfrac{\sin v}{\cos 3v} = \dfrac{1}{2} \cdot (\tan 3v - \tan v)$. So now we can re-write the original expression $\dfrac{\sin x}{\cos 3x} + \dfrac{\sin u}{\cos 3u} + \dfrac{\sin v}{\cos 3v}$ as $\dfrac{1}{2} \cdot (\tan(3x) - \tan x + \tan(9x) - \tan(3x) + \tan(27x) - \tan(9x))$ and everything cancels out except the desired expression: $\dfrac{1}{2} \cdot (\tan(27x) - \tan(x))$.

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For $\cos y\ne0,$ $$\tan3y-\tan y=\dfrac{\sin(3y-y)}{\cos3y\cos y}=\dfrac{2\sin y\cos y}{\cos3y\cos y}$$

$$\implies\tan3y-\tan y=2\dfrac{\sin y}{\cos3y}$$

Set $y=x,3x,9x$ and add to recognize the Telescoping series

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  • $\begingroup$ why is $\cos y \neq 0$? $\endgroup$ – pi-π Feb 7 '17 at 8:30
  • $\begingroup$ @tRiGoNoMeTrY, What is $y$ if $\cos y=0$ Put that value in $\tan3y-\tan y=\dfrac{2\sin y}{\cos3y}$ $\endgroup$ – lab bhattacharjee Feb 7 '17 at 8:34
  • $\begingroup$ $L.H.S=R.H.S=0$ $\endgroup$ – pi-π Feb 7 '17 at 8:36
  • $\begingroup$ @tRiGoNoMeTrY, $$\tan(3\cdot90^\circ)=?$$ $\endgroup$ – lab bhattacharjee Feb 7 '17 at 8:37
  • $\begingroup$ @ lab bhattacharjee $\tan 270=\frac {1}{0}$. $\endgroup$ – pi-π Feb 7 '17 at 8:40

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