Given that, #cot8x=-2 rArr tan8x=1/(cot8x)=-1/2.#
Recall that, # sin2theta=(2tantheta)/(1+tan^2theta).........(star^1),#
# cos2theta=(1-tan^2theta)/(1+tan^2theta)..........(star^2),#
# tan2theta=(2tantheta)/(1-tan^2 theta)..............(star^3).#
With #theta=8x" in "(star^1),# we have,
#sin(2(8x))=sin16x=(2tan8x)/{1+tan^2 8x)=(2(-1/2))/(1+1/4)=-4/5.#
Sub.ing, #theta=4x" in "(star^3),# we get,
#tan(2(4x))=tan
8x=(1-tan^2 4x)/(1+tan^2 4x)." But, "tan8x=-1/2.#
#:. -1/2=(1-tan^2 4x)/(1+tan^2 4x).#
#:. 2tan^2 4x-2=tan^2 4x+1, or, tan^2 4x=3.#
# :. tan 4x=+-sqrt3.#
But, # pi/2 lt 8x lt pi rArr pi/4 lt 4x lt pi/2. :. tan4x=+sqrt3.#
#:. sec^2 4x=1+tan^2 4x=4 ;. sec4x=+-2 :. cos4x=+-1/2.#
#"But, "pi/4 lt 4x lt pi/2. :. cos4x=+1/2.#