Answer:
[tex]\sin(\frac{19}{12}\pi)\\\\=\sin(\frac{15}{12}\pi+\frac{4}{12}\pi)\\\\=\sin(\frac{15}{12}\pi)\sin(\frac{1}{3}\pi)+\cos(\frac{15}{12}\pi)\cos(\frac{1}{3}\pi)\\\\=-\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{1}{2}\\\\=\frac{-\sqrt{6}-\sqrt{2}}{4}\\\\=-\frac{\sqrt{2}(\sqrt{3}+1)}{4}[/tex]
A=2
B=3
I genuinely hope this helped you answer the question. Have a nice day.
