Oh my, #pi/4# again. What a surprise.

Question writers, questions like this are your big chance to get away from the two cliche triangles of trig.

I prefer the notation #text{Arc}text{sin}(x)# for the principal value and #arcsin(x)# for all values. Let's run the problem both ways.

Of course,

#sin(pi/4) = cos (pi/4) = 1/sqrt{2} #

#text{Arc}text{sin}(cos (pi/4) ) = text{Arc}text{sin}( 1/sqrt{2} ) = pi/4
#

#x = arcsin(cos (pi/4)) # is equivalent to

# sin(x) = cos(pi/4)#

I always remember #cos x =cos a# has solutions #x=pm a + 2pi k quad# integer #k#.

# cos(pi/2 - x) = cos(pi/4)#

#pi/2 - x = pm pi/4 + 2pi k#

# x = pi/2 pm pi/4 + 2pi k #

# x = { pi/4, {3pi}/4 } + 2pi k#

We can rewrite that as

#x = pi/4 + pi k #

#arcsin(cos (pi/4)) = pi/4 + pi k, quad# integer #k#