Question

Prove that :-
$\tt \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos8 \: \theta} } } = 2 \cos \theta$
Given that ;
$\theta \: \epsilon \:( - \dfrac{ \pi}{8} ,\dfrac{ \pi}{8} )$
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Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \: \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos8 \: \theta} } }$

can be rewritten as

$\rm \: = \: \sqrt{2 + \sqrt{2 + \sqrt{2(1 + \cos8 \: \theta)} } }$

We know,

$\boxed{\sf{ \: \: cos2x = {2cos}^{2}x - 1 \: }}$

So, using this identity, we get

$\rm \: = \: \sqrt{2 + \sqrt{2 + \sqrt{2( {2cos}^{2} \: 4\theta}) } }$

$\rm \: = \: \sqrt{2 + \sqrt{2 + \sqrt{ {(2cos4\theta)}^{2} \:} } }$

As it is given that,

$\rm \: \theta \: \epsilon \: \bigg( - \dfrac{ \pi}{8} ,\dfrac{ \pi}{8}\bigg )\rm\implies \:4\theta \: \epsilon \:\bigg( - \dfrac{ \pi}{2} ,\dfrac{ \pi}{2}\bigg )$

$\rm\implies \:cos4\theta > 0$

$\rm\implies \: \sqrt{ {cos}^{2} 4\theta} = |cos4\theta| = cos4\theta$

So, using this, above expression can be rewritten as

$\rm \: = \: \sqrt{2 + \sqrt{2 + 2\cos4\: \theta}}$

$\rm \: = \: \sqrt{2 + \sqrt{2(1 + \cos4\: \theta)}}$

$\rm \: = \: \sqrt{2 + \sqrt{2( {2cos}^{2}2 \theta)}}$

$\rm \: = \: \sqrt{2 + \sqrt{( {2cos2\theta)}^{2}}}$

As it is given that,

$\rm \: \theta \: \epsilon \: \bigg( - \dfrac{ \pi}{8} ,\dfrac{ \pi}{8}\bigg )\rm\implies \:2\theta \: \epsilon \:\bigg( - \dfrac{ \pi}{4} ,\dfrac{ \pi}{4}\bigg )$

$\rm\implies \:cos2\theta > 0$

$\rm\implies \: \sqrt{ {cos}^{2} 2\theta} = |cos2\theta| = cos2\theta$

So, using this, above expression can be rewritten as

$\rm \: = \: \sqrt{2 + 2cos2\theta}$

$\rm \: = \: \sqrt{2 (1+ cos2\theta)}$

$\rm \: = \: \sqrt{2( {2cos}^{2} \theta)}$

$\rm \: = \: \sqrt{ {(2cos\theta)}^{2} }$

As it is given that,

$\rm \: \theta \: \epsilon \:\bigg( - \dfrac{ \pi}{8} ,\dfrac{ \pi}{8} \bigg)$

$\rm\implies \:cos\theta > 0$

$\rm\implies \: \sqrt{ {cos}^{2} \theta} = |cos\theta| = cos\theta$

So, using this, we get

$\rm \: = \: 2 \: cos\theta \:$

Hence,

$\rm\implies \:\boxed{\sf{ \:\rm \: \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos8 \: \theta} } } = 2cos\theta \: }}$

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Additional Information

$\boxed{\sf{ \:sin2\theta = 2sin\theta \: cos\theta \: = \: \frac{2tan\theta}{1 + {tan}^{2} \theta} \: }}$

$\boxed{\sf{ \:cos2\theta = {cos}^{2}\theta - {sin}^{2}\theta = 1 - 2 {sin}^{2}\theta = {2cos}^{2}\theta - 1 \: }}$

$\boxed{\sf{ \:tan2\theta = \frac{2tan\theta}{1 - {tan}^{2} \theta} \: }}$

$\boxed{\sf{ \:sin3\theta = 3sin\theta - {4sin}^{3}\theta \: }}$

$\boxed{\sf{ \:cos3\theta = {4cos}^{3}\theta - 3cos\theta \: }}$

$\boxed{\sf{ \:tan3\theta = \frac{3tan\theta - {tan}^{3} \theta}{1 - {3tan}^{2} \theta} \: }}$

Answer 2

Step-by-step explanation:

$\begin{aligned} \tt \: Sol _{n}. \qquad \: L.H.S. & \tt=\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}} \\ \\ &=\sqrt{2+\sqrt{2+\sqrt{2\left(2 \cos ^{2} 4 \theta\right)}}} \\ \\ &=\sqrt{2+\sqrt{2+2 \cos 4 \theta}}\\ \\ &=\sqrt{2+\sqrt{2(1+\cos 4 \theta)}} \\ \\ &=\sqrt{2+\sqrt{2\left(2 \cos ^{2} 2 \theta\right)}}\\ \\ &=\sqrt{2+2 \cos 2 \theta} \\ \\ &=\sqrt{2(1+\cos 2 \theta)}\\ \\ &=\sqrt{2 \cdot\left(2 \cos ^{2} \theta\right)}\\ \\ &=\sqrt{4 \cos ^{2} \theta}=2|\cos \theta| \\ \\ &=2 \cos \theta \text { as } \theta \in\left[\frac{-\pi}{8}, \frac{\pi}{8}\right] \\ \\ & \Rightarrow \cos \theta>0 \Rightarrow|\cos \theta|\\ \\ &=\cos \theta \\ \\ & = \text{ R. H. S}\end{aligned}$