Prove that? `sqrt(2+sqrt(2+sqrt(2+2cos8theta` = `2costheta`

As `cos2A=2cos^2A-1`, we have

`cos8theta=2cos^2(4theta)-1`, `cos4theta=2cos^2(2theta)-1` and `cos2theta=2cos^2theta-1`

i.e. `1+cos8theta=2cos^2(4theta)`, `1+cos4theta=2cos^2(2theta)` and `1+cos2theta=2cos^2theta`

Hence `sqrt(2+sqrt(2+sqrt(2+2cos8theta)))`

= `sqrt(2+sqrt(2+sqrt(2(1+cos8theta))))`

= `sqrt(2+sqrt(2+sqrt(2xx2cos^2(4theta))))`

= `sqrt(2+sqrt(2+2cos(4theta)))`

= `sqrt(2+sqrt(2(1+cos(4theta)))`

= `sqrt(2+sqrt(2xx2cos^2(2theta)))`

= `sqrt(2+2cos2theta)`

= `sqrt(2(1+cos2theta))`

= `sqrt(2xx2cos^2theta)`

= `2costheta`

The given identity does not hold in general.

For example, with `theta = pi/6` we find:

`sqrt(2+sqrt(2+sqrt(2+2cos8 theta)))`

`=sqrt(2+sqrt(2+sqrt(2+2cos((4pi)/3))))`

`=sqrt(2+sqrt(2+sqrt(2+2(-1/2))))`

`=sqrt(2+sqrt(2+sqrt(1)))`

`=sqrt(2+sqrt(3)) ~~ 1.932`

But:

`2 cos(theta) = 2 cos(pi/6) = 2 (sqrt(3)/2) = sqrt(3) ~~ 1.732`

What is true?

A double angle formula for `cos` can be written:

`cos 2theta = 2 cos^2 theta - 1`

So:

`cos 8 theta = 2 cos^2 4theta - 1`

and:

`2+2cos 8 theta = 2+2(2 cos^2 4theta - 1) = 4 cos^2 4 theta`

So:

`sqrt(2+2cos 8 theta) = sqrt(4 cos^2 4 theta) = abs(2 cos 4theta)`

So if `cos 4 theta >= 0` then:

`sqrt(2+2cos 8 theta) = 2 cos 4 theta`

Then:

`2+2cos 4 theta = 2+2(2cos^2 2 theta - 1) = 4 cos^2 2 theta`

So:

`sqrt(2+2cos 4 theta) = sqrt(4 cos^2 2 theta) = abs(2cos 2 theta)`

So if `cos 2 theta >= 0` then:

`sqrt(2+2cos 4 theta) = 2cos 2 theta`

Then:

`2+2cos 2 theta = 2+2(2cos^2 theta - 1) = 4cos^2 theta`

So:

`sqrt(2+2cos 2 theta) = sqrt(4 cos^2 theta) = abs(2 cos theta)`

So if `cos theta >= 0` then:

`sqrt(2+2cos 2 theta) = 2 cos theta`

So we have shown that provided all of:

• `cos 4 theta >= 0`

• `cos 2 theta >= 0`

• `cos theta >= 0`

then:

`sqrt(2+sqrt(2+sqrt(2+2cos 8 theta))) = 2 cos theta`

These conditions will be satisfied for `theta in [-pi/8, pi/8]`