Find the exact value of sin(13pi/12)?

13π/12 = 5π/4 - π/6

sin (13π/12) = sin (5π/4 - π/6)

sin (5π/4 - π/6) = sin (5π/4) cos (π/6) - cos (5π/4) sin (π/6)

sin (5π/4 - π/6) = (- √2/2)(√3/2) - (-√2/2(1/2)

sin (13π/12) = (√2/4)( - √3 + 1)

EDIT

As far as I can see I'm the only one who has given you a correct answer.

If you use your calculator, in radian mode, to find sin(13π/12) then you will obtain the same answer as for (√2/4)( - √3 + 1).

FURTHER EDIT

Despite what "Dragon.Jade" says, "JOS J"'s answer is incorrect.

"Dragon.Jade"'s reasoning contains an error.

sin(x/2) = ±√((1 - cos(x))/2)

Since 13π/12 is in QIII, sin(13π/12) is also negative. x/2 = 13π/12, x = 13π/6

sin(13π12) = -√((1 - cos(13π/6))/2)

sin(13π12) = -√((1 -√3/2)/2)

sin(13π12) = -√(((2 -√3)/2)/2)

sin(13π12) = -√((2 - √3)/4)

sin(13π12) = -1/2*√(2 - √3)

-((-1 + Sqrt[3])/(2 Sqrt[2]))

Hello,

sin(13π/12) = sin(π + π/12)
   = -sin(π/12) →→→ Because sin(a + π)=-sin(a)

Then let's use:
cos(2a) = 1 – 2.sin²(a)
2.sin²(a) = 1 – cos(2a)
sin²(a) = [1 – cos(2a)]/2
|sin(a)| = √{ [1 – cos(2a)]/2 }

With a=π/12, we obtain:
|sin(π/12)| = √{ [1 – cos(π/6)]/2 } →→→ π/6 is a standard angle whose sin and cos are known |sin(π/12)| = √{ [1 – (√3)/2]/2 } →→→ cos(π/6)=(√3)/2
sin(π/12) = √[(2 – √3)/4] →→→ Since π/12 is in first quadrant, sin(π/12)>0
sin(π/12) = [√(2 – √3)] / 2

Thus the expected result:
sin(13π/12) = -sin(π/12) = -[√(2 – √3)] / 2

= = = = = = = = = = = = = = = = = = = = = = = = = =
EDIT: As to Ray's point of view...

Please check
http://www.wolframalpha.com/input/?i=%28%E2%88%9A2%2F4%29%28+-+%E2%88%9A3+%2B+1%29

And you'll see that Ray's value (√2/4)(-√3 + 1)
can also be expressed as:

(-√3 + 1) / (2√2) →→→ JOS J 's answer
(√2 – √6) / 4
-√(2 – √3) / 2 →→→ coreyA and Dragon.Jade's answer

Regards,
Dragon.Jade :-)

13pi/12 = pi + pi/12 which takes you into the 3rd quadrant, where sine is -ve , so we want - sin(pi/12)
or if you find it easier to think of that as - sin30degrees, that's fine - just keep writing pi/12, although it would be far better for you to thoroughly familiarise yourself with all the common fractions of pi.. From that familiar right-angled triangle with side of 1, 2, rt3, you can see that is - 1/2.

I could be wrong, but isn't it 1/2?