Come trova il valore esatto di #cos (pi / 12) #?

#cos (pi/12)=cos(pi/4 -pi/6)#
#cos(A-B)=cosAcosB+sinAsinB#
#cos(pi/4 -pi/6)=cos (pi/4)cos(pi/6)+sin(pi/4)sin(pi/6)#
#cos(pi/4 -pi/6)=sqrt2/2*sqrt3/2+sqrt2/2*1/2=sqrt6/4+sqrt2/4#
#cos (pi/12)=sqrt6/4+sqrt2/4#