[tex]\sin (\frac{\pi}{12})\cos (\frac{7\pi}{12})-\cos (\frac{\pi}{12})\sin (\frac{7\pi}{12})[/tex]
Let:
[tex]\begin{gathered} A=\frac{\pi}{12} \\ B=\frac{7\pi}{12} \end{gathered}[/tex]
Using the sine difference identity:
[tex]\begin{gathered} \sin (A)\cos (B)-\cos (A)\sin (B)=\sin (A-B) \\ so\colon \\ \sin (\frac{\pi}{12})\cos (\frac{7\pi}{12})-\cos (\frac{\pi}{12})\sin (\frac{7\pi}{12})=\sin (\frac{\pi}{12}-\frac{7\pi}{12}) \\ \sin (\frac{\pi}{12}-\frac{7\pi}{12})=\sin (-\frac{6\pi}{12}) \\ \sin (-\frac{\pi}{2}) \end{gathered}[/tex]
Answer:
[tex]\sin (-\frac{\pi}{2})[/tex]
