Question 873813
The angle sum identity for sine is
{{{sin(A+B)=sin(A)*cos(B)+cos(A)*sin(B)}}}
Since {{{2pi/3+pi/4=8pi/12+3pi/12=11pi/12}}} ,
you can use the angle sum identity (above),
and the values for sine and cosine of {{{2pi/3}}} and {{{pi/4}}}
to find {{{sin(11pi/12)=sin(2pi/3+pi/4)}}} .
{{{sin(pi/4)}}} and {{{cos(pi/4)}}} are easy. {{{pi/4}}} is {{{45^o}}} ,
and {{{sin(pi/4)=cos(pi/4)=sqrt(2)/2}}} .
{{{drawing(300,300,-0.8,0.2,-0.2,0.8,
triangle(0,0,-.707,0,0,0.707),
rectangle(0,0,-0.05,0.05),
arc(-0.707,0,0.4,0.4,-45,0),
arc(0,0.707,0.4,0.4,-270,-225),
locate(-0.36,-0.02,sqrt(2)/2),locate(0.03,0.38,sqrt(2)/2),
locate(-0.34,0.36,1),locate(-0.58,0.12,pi/4)
)}}}
{{{2pi/3}}} is a little harder because it is in quadrant II.
{{{drawing(300,300,-1.2,1.2,-1.2,1.2,
grid(0),circle(0,0,1),
arrow(0,0,0.65,1.126),arrow(0,0,-0.65,1.126),
red(arc(0,0,0.8,0.8,-60,0)),locate(0.18,0.3,red(pi/3)),
red(arc(0,0,0.8,0.8,-180,-120)),locate(-0.24,0.3,red(pi/3)),
blue(arc(0,0,1.5,1.5,-120,0)),locate(0.45,0.45,blue(2pi/3)),
red(circle(0.5,0.866,0.02)),locate(0.58,1.03,red(P(1/2,sqrt(3)/2)))
)}}}

{{{sin(2pi/3)=sin(pi-2pi/3)=sin(pi/3)=sqrt(3)/2}}}
{{{cos(2pi/3)=-cos(pi-2pi/3)=-cos(pi/3)=-1/2}}}
 
So, applying {{{sin(A+B)=sin(A)*cos(B)+cos(A)*sin(B)}}} with {{{A=2pi/3}}} and {{{B=pi/4}}}
{{{sin(11pi/12)=sin(2pi/3+pi/4)=sin(2pi/3)*cos(pi/4)+cos(2pi/3)*sin(pi/4)}}}
={{{(sqrt(3)/2)(sqrt(2)/2)+(-1/2)(sqrt(2)/2)=sqrt(3)*sqrt(2)/4-sqrt(2)/4}}}
That could be "simplified" different ways:
{{{sqrt(3)*sqrt(2)/4-sqrt(2)/4=highlight((sqrt(6)-sqrt(2))/4)}}}
or
{{{sqrt(3)*sqrt(2)/4-sqrt(2)/4=highlight((sqrt(3)-1)sqrt(2)/4)}}}
Those expressions are the exact value of {{{sin(11pi/12)}}}