Question

Prove that :
$\sqrt{2+\sqrt{2+\sqrt{2+2.cos8\theta }}}=2cos\theta$

Answer 1

$cos2\Theta =2co{s}^2\Theta -1$

$LHS\to$ $\sqrt{2+\sqrt{2+\sqrt{2+2cos8\Theta }}}$

$=\sqrt{2+\sqrt{2+\sqrt{2+2cos2\left(4\Theta \right)}}}$

$=\sqrt{2+\sqrt{2+\sqrt{2+2(2co{s}^{2}4\Theta -1)}}}$

$=\sqrt{2+\sqrt{2+\sqrt{2+4co{s}^{2}4\Theta -2}}}$

$=\sqrt{2+\sqrt{2+2cos4\Theta }}=\sqrt{2+\sqrt{2+2cos2\left(2\Theta \right)}}$

$=\sqrt{2+\sqrt{2+2(2co{s}^{2}2\Theta -1)}}=\sqrt{2+\sqrt{2+4co{s}^{2}2\Theta -2}}$

$=\sqrt{2+2cos2\Theta }$

$=\sqrt{2+2(2co{s}^2\Theta -1)}=\sqrt{2+4co{s}^2\Theta -2}$

$=\sqrt{4co{s}^2\Theta }$

$=2cos\Theta =RHS$

$Q.E.D$