find the exact value of sin(5pi/12)

here's what I have, can you tell me if I'm doing it correctly and if I've simplified. I'm more than a little confused.

[sin(5pi/6)]/2 = +/- √[(1-cosx)/2]

= +/- √{[1-cos(5pi/6)]/2}

= +/- √ ({1-[-√(3)/2]}/2)

= +/- √ {[1+√(3)/2]/2}

And that's as far as I got. I'm not sure if I'm doing it correctly or if i am, i have no idea how to simplify further. :/

Thanks ahead of time, all help is very much appreciated.