Answer:
Step-by-step explanation:
2 sin a sin b=cos(a-b)-cos (a+b)
[tex]2 sin \frac{\pi }{4} sin \frac{\pi }{6}=cos (\frac{\pi }{4}-\frac{\pi }{6})-cos (\frac{\pi }{4} +\frac{\pi }{6} )\\=cos (\frac{3 \pi-2 \pi }{12} )-cos (\frac{3 \pi+2\pi}{12} )\\=cos (\frac{\pi }{12} )-cos( \frac{5 \pi}{12} )\\sin \frac{ \pi}{4} sin\frac {\pi}{6}=\frac{1}{2}[cos \frac{\pi}{12} -cos \frac{5 \pi}{12} ][/tex]
