Question

$\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}=a\cos \theta ,0<\theta <\pi /16$
.Find $a$

Answer 1

L.H.S. $=\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 8\theta )}}}$
$=\sqrt{2+\sqrt{2+\sqrt{2\left(2{\cos }^{2}4\theta \right)}}}$ $[\because 1+\cos 8\theta =2{\cos }^2\frac{8\theta }{2}]$
$=\sqrt{2+\sqrt{2+\sqrt{\left(4{\cos }^{2}4\theta \right)}}}$
$=\sqrt{2+\sqrt{2+2\cos 4\theta }}$
$=\sqrt{2+\sqrt{2(1+\cos 4\theta )}}$
$=\sqrt{2+\sqrt{2\left(2{\cos }^{2}2\theta \right)}}$ $[\because 1+\cos 4\theta =2{\cos }^{2}2\theta ]$
$=\sqrt{2+2\cos 2\theta }=\sqrt{2(1+\cos 2\theta )}=\sqrt{2\left(2{\cos }^2\theta \right)}=2\cos \theta =$ R.H.S.
Ans: 2