Question

Prove the following

$\sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8x} } } = 2cosx$
​

Answer 1

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

Consider LHS

$\sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8x} } }$

can be rewritten as

$\rm \: = \sqrt{2 + \sqrt{2 + \sqrt{2(1 + cos8x)} } }$

We know,

$\boxed{ \rm{ \:1 + cos2x = {2cos}^{2}x \: }}$

$\rm \: = \sqrt{2 + \sqrt{2 + \sqrt{2. {2cos}^{2}4 x} } }$

$\rm \: = \sqrt{2 + \sqrt{2 + \sqrt{ {(2cos4x)}^{2}} } }$

$\rm \: = \sqrt{2 + \sqrt{2 + 2cos4x } }$

$\rm \: = \sqrt{2 + \sqrt{2(1 + cos4x) } }$

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

$\rm \: = \sqrt{2 + 2cos2x}$

$\rm \: = \sqrt{2(1 + cos2x)}$

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

$\rm \: = 2cosx$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8x} } } = 2cosx \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered} { \boxed{ \begin{array}{c} \underline{\underline{ \color{orange} \text{Additional \: lnformation}}} \\& \rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}}\end{gathered}$

Answer 2

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

Consider LHS

$\sqrt{2 + \sqrt{ 2 + \sqrt{2 + 2 \cos(8x) } } }$

Can be written as

$\large{\sf{ \sqrt{2 \sqrt{2 \sqrt{2(1 + \cos(8x) } } } }}$

We know,

$1 + \cos(2x) = {2 \cos}^{2}x$

$\sqrt{2 + \sqrt{2 + \sqrt{2.2 { \cos}^{2} 4x} } }$

$\sqrt{2 + \sqrt{2 + \sqrt{(2 { \cos4x) }^{2} } } }$

$\sqrt{2 + \sqrt{2 + 2 \cos4x} }$

$\sqrt{2 \sqrt{2(1 + \cos4x} }$

$\sqrt{2 + { \sqrt{2(2 { \cos}^{2}2x } } }$

$\sqrt{2 + 2 \cos2x}$

$\sqrt{2(1 + \cos2x}$

$\sqrt{2 (2 \cos {}^{2}x) }$

$2 \cos x$