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]#