Come trova il valore esatto di #sin ((7pi) / 12) #?

#sin(7pi/12)=sin(pi/3+pi/4)#
#=sin(pi/3)cos( pi/4)+cos(pi/3)sin(pi/4)#
#=sqrt3/2xx1/sqrt2+1/2xx1/sqrt2=(sqrt3+1)/(2sqrt2)#