How do you evaluate arcsin(sqrt 2/2)?

##sin (pi/4) = sqrt(2)/2## is the length of one side of the isoceles right angled triangle with sides ##sqrt(2)/2## ##sqrt(2)/2## and ##1## which has internal angles ##pi/4## ##pi/4## and ##pi/2##.
(##pi/4## radians = ##45^o## and ##pi/2## radians = ##90^o## if you prefer)
To show this is right angled check with Pythagoras:
##(sqrt(2)/2)^2 + (sqrt(2)/2)^2##
##= sqrt(2)^2/2^2 + sqrt(2)^2/2^2##
##= 2/4 + 2/4 = 1/2+1/2 = 1 = 1^2##
So since ##sin (pi/4) = sqrt(2)/2## and ##pi/4## is in the
required range for ##arcsin## viz ##-pi/2 <= theta <= pi/2## we find ##arcsin (sqrt(2)/2) = pi/4##