Respuesta :

[tex]\bf \textit{Product to Sum Identities} \\\\ sin(\alpha)cos(\beta)=\cfrac{1}{2}[sin(\alpha+\beta)\quad +\quad sin(\alpha-\beta)]\\\\ -------------------------------\\\\ sin\left( \frac{3\pi }{4} \right)cos\left( \frac{\pi }{12} \right)\implies \cfrac{1}{2}\left[ sin\left( \frac{3\pi }{4}+\frac{\pi }{12} \right) ~~+~~sin\left( \frac{3\pi }{4}-\frac{\pi }{12} \right)\right][/tex]

[tex]\bf \cfrac{1}{2}\left[ sin\left( \frac{9\pi+\pi }{12} \right) ~~+~~sin\left( \frac{9\pi-\pi }{12} \right)\right] \\\\\\ \cfrac{1}{2}\left[ sin\left( \frac{10\pi }{12} \right) ~~+~~sin\left( \frac{8\pi }{12} \right)\right]\implies \cfrac{1}{2}\left[ sin\left( \frac{5\pi }{6} \right) ~~+~~sin\left( \frac{2\pi }{3} \right)\right] \\\\\\ \cfrac{1}{2}\left[\cfrac{1}{2}~~+~~\cfrac{\sqrt{3}}{2} \right]\implies \cfrac{1}{2}\left[ \cfrac{1+\sqrt{3}}{2} \right]\implies \cfrac{1+\sqrt{3}}{4}[/tex]

Explanation: I got it right on Edmentum.

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