Question

sin ET in 1 (-- cos 12 Answer here

Answer 1

$\begin{gathered} \sin (\alpha)\sin (\beta)=\frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta)) \\ In\text{ this case} \\ \alpha=\frac{\pi}{4} \\ \beta=\frac{\pi}{6} \\ \alpha+\beta=\frac{\pi}{4}+\frac{\pi}{6}=\frac{5\pi}{12} \\ \alpha-\beta=\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12} \\ \text{Hence} \\ \sin (\frac{\pi}{4})\sin (\frac{\pi}{6})=\frac{1}{2}(\cos (\frac{\pi}{12})-\cos (\frac{5\pi}{12})) \end{gathered}$