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Show that: $$ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$$

My try:

As we can see that the LHS is in form of $\cos 8 \theta$ which must be converted into $\cos \theta $ in order to solve the equation.

There are several questions similar on this site but those aren't helping me much regarding this question.

So, I tried converting $\cos 8 \theta$ into $\cos \theta$ and then putting the value in the equation under square root and then further solving it into RHS.

So I got $\cos 8 \theta$ as:

$$ 2\cdot \{ 2\cdot [ 4 \cos ^4 \theta - 4.cos^2 \theta]^2 \}$$ So, I don't know how to solve it further.

Please help, putting the value of $\cos 8 \theta$ in place of that doesn't helps me much.

Thanks in Advance

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    $\begingroup$ There is the trig identity $\sqrt{2+2\cos\alpha}=\pm2\cos(\alpha/2)$, but the sign ambguity here is relevant. Your formula is false whenever $2\cos\theta<0$ because a square-root is automatically non-negative. If you assume $\theta\in[-\pi/16,\pi/16]$, then you only need to apply that formula thrice. $\endgroup$ – Jyrki Lahtonen May 25 '18 at 12:27
  • $\begingroup$ @JyrkiLahtonen Btw, square roots are always positive, because $\sqrt {x} = | x^\frac{1}{2} |$, take a look on any good book, specially for IIT JEE/NCERT Ones. $\endgroup$ – Abhas Kumar Sinha May 25 '18 at 12:30
  • $\begingroup$ @JyrkiLahtonen Thank you, your hint helps :) $\endgroup$ – Abhas Kumar Sinha May 25 '18 at 12:31
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    $\begingroup$ No problem :-)${}$ $\endgroup$ – Jyrki Lahtonen May 25 '18 at 12:35
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    $\begingroup$ Let $\theta = \pi.$ Then $\cos(8\theta) = \cos(8\pi) = 1,$ so $\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = \sqrt{2 + \sqrt{2 + \sqrt{4}}} = \sqrt{2 + \sqrt{4}} = \sqrt{4} = 2,$ but $2\cos\theta = -2,$ so the identity is false. You have to place some conditions on $\theta$ (such as in J.G.'s answer) to make the identity true. $\endgroup$ – David K May 25 '18 at 13:24
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$$\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta}}}$$ Firstly, $2+2cos8\theta=2+2(2cos^24\theta-1)=4cos^24\theta$ $$\therefore\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta}}}=\sqrt{2+\sqrt{2+2cos4\theta}}$$ Then, $2+2cos4\theta=2+2(2cos^22\theta-1)=4cos^22\theta$ $$\therefore\sqrt{2+\sqrt{2+2cos4\theta}}=\sqrt{2+2cos2\theta}$$ Finally, $2+2cos2\theta=2+2(2cos^2\theta-1)=4cos^2\theta$ $$\therefore\sqrt{2+\sqrt{2+2cos2\theta}}=\sqrt{4cos^2\theta}=2cos\theta$$

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Since $2(1+\cos x)=(2\cos x/2)^2$, $$\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}=\sqrt{2+\sqrt{2+2\cos 4\theta }}=\sqrt{2+2\cos 2\theta}=2\cos\theta,$$provided $\cos\theta,\,\cos 2\theta,\,\cos 4\theta>0$. In fact this implies $$\cos 2\theta =\sqrt{\frac{1+\cos 4\theta}{2}}\ge\frac{1}{\sqrt{2}},\,\cos\theta=\sqrt{\frac{1+\cos 2\theta}{2}}\ge\frac{\sqrt{2+\sqrt{2}}}{2}.$$

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  • $\begingroup$ you are genius, I was a bit confused with the formula, thanks for answering my question $\endgroup$ – Abhas Kumar Sinha May 25 '18 at 12:33
  • $\begingroup$ @AbhasKumarSinha that was simple but thank you for compliment $\endgroup$ – Abhishek May 25 '18 at 12:34
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Hint: use $$1+\cos 2x = 2\cos^2x$$

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  • $\begingroup$ thanks, that helps :) $\endgroup$ – Abhas Kumar Sinha May 25 '18 at 12:32
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You should notice that this is a (three times) repeated application of

$$\sqrt{2+2\cos 2\theta}=2\cos\theta$$ which is established by

$$\sqrt{2+2\cos2\theta}=\sqrt{2+4\cos^2\theta-2}.$$

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It's interesting to note that the equality holds for only a limited range of values as pointed out in J.G.'s answer. See graph below.

enter image description here

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  • $\begingroup$ Ugh oh, y = 2cos 8x, not y = 2cos x $\endgroup$ – Abhas Kumar Sinha May 28 '18 at 4:23
  • $\begingroup$ y = 2 cos x is RHS $\endgroup$ – hypergeometric May 28 '18 at 8:46

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