Prove that `sqrt ( 2 + sqrt (2 + sqrt (2 + 2 cos 8 theta))) = 2 cos theta,` where ` theta in [ (-pi)/(8), (pi)/(8)]`

Explanation

LHS `=sqrt(2+sqrt(2+sqrt(2+2costheta)))=2costheta`
`=sqrt(2+sqrt(2+sqrt(2(2cos^(2)4theta))) ) [therefore 1+cos8theta=2cos^(@)(8theta)/(2)]`
`sqrt(2+sqrt(2+2cos 4theta))`
`=sqrt(2+sqrt(1+cos4theta))`
`=sqrt(2+sqrt(2(2cos^(2)2theta)))` [`therefore 1+cos4theta=2 cos^(2)2theta]`
`=sqrt(2+sqrt(4cos^(2)2theta))`
`sqrt(2(1+cos2theta))`
`sqrt(2(2cos^(2)theta)) [therefore 1+cos2theta=2cos^(2)theta]`
`=sqrt(4 cos^(2)theta)`
`=2cos theta`=RHS.
Hence proved.