The answer is: #-(sqrt6-sqrt2)/4#
The formula of alf‐angle is:
#cos(alpha/2)=+-sqrt((1+cosalpha)/2)#, the #+-# in this case becomes #-# because the angle of #210°# is in the third quadrant and there the cosine is negative.
So:
#cos105°=-sqrt((1+cos210°))/2=-sqrt((1-sqrt3/2)/2)=-sqrt((2-sqrt3)/4)=-sqrt(2-sqrt3)/2#
or, using the formula of double radical, that says:
#sqrt(a+-sqrtb)=sqrt((a+sqrt(c))/2)+-sqrt((a-sqrt(c))/2)#, that is useful when #c=a^2-b# is a square.
So: #c=4-3=1#, and than:
#-sqrt(2-sqrt3)/2=-1/2[sqrt((2+1)/2)-sqrt(2-1)/2]=-1/2[sqrt(3/2)-sqrt(1/2)]=#
#=-1/2(sqrt3/sqrt2-sqrt1/sqrt2)=-1/2(sqrt3/sqrt2*sqrt2/sqrt2-1/sqrt2*sqrt2/sqrt2)=-1/2(sqrt6/2-sqrt2/2)=-(sqrt6-sqrt2)/4#
