How do you use the half angle formula for ##sin ((5pi)/12)##?

What is ##(5pi)/12## half of?

It is half of its double: ##2 xx (5pi)/12 = 2/1*(5pi)/12 =(5pi)/6##

##sin(1/2x) = +-sqrt((1-cosx)/2)##

We know that ##(5pi)/12## is in quadrant I, so it has positive sine,

##sin((5pi)/12) = sin(1/2((5pi)/6)) = sqrt((1-cos((5pi)/6))/2)##

##= sqrt((1-(-sqrt3/2))/2)= sqrt((1+sqrt3/2)/2) = sqrt((2+sqrt3)/4) = sqrt(2+sqrt3)/2##