sin(π/12)=sin(π/3-π/4)=sin(π/3)cos(π/4)-cos(π/3)sin(π/4)=(√3/2)(√2/2)-(1/2)(√2/2)=(√6-√2)/4.
ALTERNATIVELY
sin(π/12)=sin((1/2)(π/6)); sin(π/6)=1/2.
sin(x+x)=sin(2x)=2sin(x)cos(x); let 2x=π/6, then x=π/12. Let y=sin(x):
1/2=2y√(1-y^2); 1/4=4y^2(1-y^2); y^2(1-y^2)=1/16.
y^4-y^2=-1/16; y^4-y^2+1/4=1/4-1/16=3/16.
(y^2-1/2)^2=3/16; y^2-1/2=±√3/4; y^2=1/2±√3/4. Therefore y^2=(2±√3)/4. But sin(2x)>sin(x) so we need to take the smaller root: y^2=(2-√3)/4 because (2+√3)/4 is approximately 0.9 which is bigger than 0.5.
Let y=√a-√b, then y^2=a+b-2√(ab)=1/2-√3/4. Equating rational and irrational parts: a+b=1/2 and 2√(ab)=√3/4.
√(ab)=√3/8 and ab=3/64, so b=3/(64a); a+3/(64a)=1/2; a^2+3/64=a/2; a^2-a/2=-3/64.
a^2-a/2+1/16=1/16-3/64=1/64; (a-1/4)^2=1/64; a-1/4=±1/8, a=3/8 and b=1/8 or vice versa.
So y=√a-√b=√(3/8)-√(1/8)=√(6/16)-√(2/16)=(√6-√2)/4=sin(π/12).