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.