1. **State the problem:** Simplify the expression $$\sin\left(\frac{5\pi}{4}\right) \cos\left(\frac{3\pi}{4}\right) + \sin\left(\frac{5\pi}{3}\right) \cos\left(\frac{13\pi}{6}\right)$$. 2. **Recall the sine and cosine values for special angles:** - $$\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$ because $$\frac{5\pi}{4}$$ is in the third quadrant where sine is negative. - $$\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$ because $$\frac{3\pi}{4}$$ is in the second quadrant where cosine is negative. - $$\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$ because $$\frac{5\pi}{3}$$ is in the fourth quadrant where sine is negative. - $$\cos\left(\frac{13\pi}{6}\right) = \cos\left(2\pi + \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$ since cosine is periodic with period $$2\pi$$. 3. **Substitute these values into the expression:** $$\left(-\frac{\sqrt{2}}{2}\right) \left(-\frac{\sqrt{2}}{2}\right) + \left(-\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right)$$ 4. **Multiply the terms:** $$\frac{\sqrt{2} \times \sqrt{2}}{2 \times 2} + \left(-\frac{\sqrt{3} \times \sqrt{3}}{2 \times 2}\right) = \frac{2}{4} - \frac{3}{4}$$ 5. **Simplify the fractions:** $$\frac{2}{4} - \frac{3}{4} = \frac{2 - 3}{4} = -\frac{1}{4}$$ 6. **Final answer:** $$\boxed{-\frac{1}{4}}$$