First of all we have to know that #\sin(pi/4)=1/\sqrt(2)# and #\cos(pi/4)=1/\sqrt(2)#. We can see this fact considering the diagonal of a square with side of #1#. The length of the diagonal, because of the Pitagora's Theorem, is #sqrt(1^2+1^2)=sqrt(2)#. The triangle has one angle of #90^\circ# (that is #\pi/2# and two of #45^\circ# that are #\pi/4# as in the image.
Using the definition of #sin# and #cos# it is easy to see that the value of both #sin# and #cos# of #pi/4# is #1/sqrt(2)#.
What about the #cos(7/4pi)#? This is simple if we see what #7/4pi# is on a circunference, as in the picture.
It is clear that #7/4pi# is nothing but #-pi/4# and the cosine has the property that it stay unchanged if you change the sign of the angle (and from the picture it is simple to see just looking at the cosine as the red line).
Then we have: #sin(pi/4)cos(7/4pi)=1/sqrt(2)cos(-pi/4)=1/sqrt(2)cos(pi/4)=1/sqrt(2)1/sqrt(2)=1/2#.
