Good question.
Here is one way of thinking about it.
You see that #-11/6# is very close to #-12/6# which would be equal to #-2#, so I would first try to rewrite as such:
#(-11pi)/6 = (-12pi)/6 + pi/6 = -2pi + pi/6#
You are asked to compute the sine and cosine of that angle, so this can be written:
#cos((-11pi)/6) = cos(-2pi +pi/6)#
and
#sin((-11pi)/6) = sin(-2pi +pi/6)#
Now, picture yourself the trigonometric circle.
Going around a full circle is #2 pi# (radian), so if you add #2pi#, or any multiple of #2pi# if you're going full-circle many times, to whatever angle, you come back to where you started from (of course, it's the same thing if you subtract #2pi#!).
So,
#cos(-2pi + pi/6) = cos(pi/6)#
and
#sin(-2pi + pi/6) = sin(pi/6)#
and these are equal to
#cos(pi/6)=sqrt(3)/2#
and
#sin(pi/6)=1/2#.
I hope this helps.
