Let, #color(red)(22.5°=theta)#
#:.2theta=45°#
Now we get, #2theta=45°#
#:.tan(2theta)=tan45°#
#:.tan(2theta)=1#
#:.color(red)(sin(2theta)=cos(2theta))#......(1).
Now, by formula of multiple and sub-multiple angles #rarr#
#sin(2theta)=2sinthetacostheta#........(2).
#cos(2theta)=cos^2(theta)-sin^2(theta)#.......(3).
Now substituting #(3)# in #(1)# #rarr#
#:.sin(2theta)=cos^2(theta)-sin^2(theta)#
#:.(sin^2(2theta))/(4sin^2theta)-sin^2(theta)=1/sqrt(2)#.......From (2)#.
#:.(1-8sin^4(theta))/(8sin^2(theta))=1/sqrt(2)#
#:.color(red)(8sin^4(theta)+4sqrt(2)sin^2(theta)-1=0)#
#:.sin^2theta=(-4sqrt2+-sqrt((4sqrt2)^2-4(8)(-1)))/(2(8))#
#:.sin^2theta=(2-sqrt2)/4,-(2+sqrt2)/4#
#:.sintheta=sqrt(2-sqrt2)/2, sqrt(-(2+sqrt2))/2#
Now ignoring the imaginary value
of #sintheta# and substituting #theta=22.5°# we get,
#color(red)(sin22.5°=sqrt(2-sqrt2)/2)#. (Answer).
Hope it Helps:)
