# [3 pi}/4 # is #135^circ = 3 times 45^circ #
Of course 30/60/90 and 45/45/90 are the only two triangles students are expected to know "exactly." We have to know the multiples of these triangles in the other quadrant. Here we have 45/45/90 in the second quadrant.
We start from #cos(45^circ)=sin(45^circ)=1/sqrt{2}#. The angle #135^circ# is supplementary to #45^circ#, so has the opposite cosine and the same sine.
#cos({3pi}/4)= - cos(pi - {3pi}/4)=- cos(pi/4)= - 1/sqrt{2}#
#sin({3pi}/4)=sin( pi - {3pi}/4) = sin(pi/4) = 1/sqrt{2}#
#tan({3pi}/4) = {sin({3pi}/4)}/{cos({3pi}/4)} = {1/sqrt{2}}/{-1/sqrt{2}}=-1#
#sec({3pi}/4)=1/cos({3pi}/4) = - sqrt{2}#
#csc({3pi}/4)=1/sin({3pi}/4) = sqrt{2}#
#cot({3pi}/4)=1/tan({3pi}/4) = -1#
