1. **Problem Statement:** Prove that $$\frac{1}{2} \cos^{-1} x = \cos^{-1} \sqrt{\frac{1 + x}{2}}.$$\n\n2. **Recall the double-angle formula for cosine:** $$\cos 2\theta = 2\cos^2 \theta - 1.$$\nThis can be rearranged to express $$\cos^2 \theta$$ in terms of $$\cos 2\theta$$ as:\n$$\cos^2 \theta = \frac{1 + \cos 2\theta}{2}.$$\n\n3. **Set $$\theta = \frac{1}{2} \cos^{-1} x$$:**\nThen $$2\theta = \cos^{-1} x,$$ so $$\cos 2\theta = x.$$\n\n4. **Substitute into the formula:**\n$$\cos^2 \theta = \frac{1 + x}{2}.$$\nTaking the positive square root (since $$\theta$$ is in the range of inverse cosine), we get:\n$$\cos \theta = \sqrt{\frac{1 + x}{2}}.$$\n\n5. **Rewrite $$\theta$$:**\nSince $$\theta = \frac{1}{2} \cos^{-1} x,$$ then\n$$\cos \left( \frac{1}{2} \cos^{-1} x \right) = \sqrt{\frac{1 + x}{2}}.$$\n\n6. **Apply inverse cosine to both sides:**\n$$\frac{1}{2} \cos^{-1} x = \cos^{-1} \sqrt{\frac{1 + x}{2}}.$$\n\n**Final answer:**\n$$\boxed{\frac{1}{2} \cos^{-1} x = \cos^{-1} \sqrt{\frac{1 + x}{2}}}.$$