1. The problem is to find the values of $\sin \frac{5\pi}{8}$ and $\cos \frac{7\pi}{12}$ using half-angle formulas. 2. The half-angle formulas are: $$\sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}}$$ $$\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}$$ The sign depends on the quadrant of $\frac{\theta}{2}$. 3. For $\sin \frac{5\pi}{8}$, note that $\frac{5\pi}{8} = \frac{1}{2} \times \frac{5\pi}{4}$, so $\theta = \frac{5\pi}{4}$. 4. Calculate $\cos \frac{5\pi}{4}$: $$\cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$$ 5. Apply the half-angle formula for sine: $$\sin \frac{5\pi}{8} = \pm \sqrt{\frac{1 - \cos \frac{5\pi}{4}}{2}} = \pm \sqrt{\frac{1 - (-\frac{\sqrt{2}}{2})}{2}} = \pm \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}$$ 6. Determine the sign: $\frac{5\pi}{8}$ is in the second quadrant where sine is positive, so: $$\sin \frac{5\pi}{8} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}$$ 7. For $\cos \frac{7\pi}{12}$, note that $\frac{7\pi}{12} = \frac{1}{2} \times \frac{7\pi}{6}$, so $\theta = \frac{7\pi}{6}$. 8. Calculate $\cos \frac{7\pi}{6}$: $$\cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$$ 9. Apply the half-angle formula for cosine: $$\cos \frac{7\pi}{12} = \pm \sqrt{\frac{1 + \cos \frac{7\pi}{6}}{2}} = \pm \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$$ 10. Determine the sign: $\frac{7\pi}{12}$ is in the second quadrant where cosine is negative, so: $$\cos \frac{7\pi}{12} = - \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$$ Final answers: $$\sin \frac{5\pi}{8} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}$$ $$\cos \frac{7\pi}{12} = - \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$$