Answer :
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]
[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]