The answer is: -(sqrt6-sqrt2)/4−√6−√24
The formula of alf‐angle is:
cos(alpha/2)=+-sqrt((1+cosalpha)/2)cos(α2)=±√1+cosα2, 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
