## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{-1}{2}$
Recall the identity: $(f \cdot g) (x)=f(x) \cdot g(x)$ Let us consider that $f(x)=\sin x$ and $g(x)=\cos x$ We know from the unit circle that $\sin (\dfrac{3 \pi}{4}) =\dfrac{\sqrt {2}}{2}$ and $\cos (\dfrac{3 \pi}{4})=\dfrac{-\sqrt 2}{2}$ Thus, we have: $(f \cdot g) (60^{\circ}) =\sin (\dfrac{3 \pi}{4}) \cdot \cos (\dfrac{3 \pi}{4}) \\=\dfrac{\sqrt {2}}{2} \cdot [\dfrac{-\sqrt 2}{2} ]\\=\dfrac{-1}{2}$