First we need to remember the product-to-sum trigonometry formula :
[tex]\sin \mleft(a\mright)\sin \mleft(b\mright)=\frac{1}{2}\mleft(\cos \mleft(a-b\mright)-\cos \mleft(a+b\mright)\mright)[/tex]So, in our question, we have:
[tex]\sin (\frac{\pi}{4})\sin (\frac{\pi}{6})=\frac{1}{2}(\cos (\frac{\pi}{12})-\cos (\frac{5\pi}{12}))[/tex]