How do you evaluate sin (13π/12) ?

13pi/12 lies between pi and 2pi, which means sin (13pi/12) < 0

Recall the double angle identity,

sin^2 (x) = (1 - cos (2x)) / 2

If we let x = 13pi/12, then

sin (13pi/12) = - sqrt[ (1 - cos (13pi/6)) / 2]

where we took the negative square root because we expect a negative value.

Now, because cosine has a period of 2pi, we have

cos (13pi/6) = cos (2pi + pi/6) = cos (pi/6) = sqrt[3]/2

Then

sin (13pi/12) = - sqrt[ (1 - sqrt[3]/2) / 2]

sin (13pi/12) = - sqrt[2 - sqrt[3]]/2