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