Suppose theta= 11pi/12. How do you use the sum identity to find the exact value of sin theta?

The better way is, first we have to find the equivalent in degrees[tex]2\pi=360\º[/tex][tex]\frac{11\pi}{12}=345\º[/tex]now we can change this value to [tex]-15\º[/tex]how do we get an angle like this?![tex]30\º-45\º=-15\º[/tex]then[tex]sin(30\º-45\º)=sin(30\º)*cos(45\º)-sin(45\º)*cos(30\º)[/tex][tex]\begin{Bmatrix}sin(30\º)&=&\frac{1}{2}\\\\sin(45\º)&=&cos(45\º)&=&\frac{\sqrt{2}}{2}}\end{matrix}\\\\cos(30\º)&=&\frac{\sqrt{3}}{2}\end{matrix}[/tex]now we replace this values[tex]sin(-15\º)=\frac{1}{2}*\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}*\frac{\sqrt{3}}{2}[/tex][tex]sin(-15\º)=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4}[/tex][tex]\boxed{\boxed{sin(-15\º)=sin(345\º)=\frac{\sqrt{2}-\sqrt{6}}{4}}}[/tex]