Here are the #sin#, #cos#, and #tan# half-angle identities:
#sin(theta/2)=+-sqrt((1-costheta)/2)#
#cos(theta/2)=+-sqrt((1+costheta)/2)#
#tan(theta/2)=+-sqrt((1-costheta)/(1+costheta))#
#color(white)(tan(theta/2))=sintheta/(1+costheta)#
#color(white)(tan(theta/2))=costheta/(1-sintheta)#
To figure out the #tan# of #22.5^@#, use the #tan# half-angle formula. I personally like the second one the best, so I'll use that one:
#color(white)=tan(22.5^@)#
#=tan(45^@/2)#
#=sin(45^@)/(1+cos(45^@))#
#=(sqrt2/2)/(1+sqrt2/2)#
#=sqrt2/(2+sqrt2)#
#=((sqrt2))/((2+sqrt2))color(red)(*((2-sqrt2))/((2-sqrt2)))#
#=((sqrt2)(2-sqrt2))/((2+sqrt2)(2-sqrt2))#
#=(2sqrt2-sqrt2^2)/((2+sqrt2)(2-sqrt2))#
#=(2sqrt2-2)/((2+sqrt2)(2-sqrt2))#
#=(2sqrt2-2)/(4+2sqrt2-2sqrt2-sqrt2^2)#
#=(2sqrt2-2)/(4color(red)cancelcolor(black)(color(black)+2sqrt2-2sqrt2)-sqrt2^2)#
#=(2sqrt2-2)/(4-sqrt2^2)#
#=(2sqrt2-2)/(4-2)#
#=(2sqrt2-2)/2#
#=(color(red)cancelcolor(black)2sqrt2-color(red)cancelcolor(black)2^1)/color(red)cancelcolor(black)2#
#=sqrt2-1#
This is the result. You can check your answer using a calculator:
Hope this helped!
