1. Let the magnitudes of the two vectors be |A|=|B|=R and the angle between them be $\theta$.

2. Magnitude of the sum $\vec{A}+\vec{B}$:

   Using $|\vec{A}+\vec{B}|^2=|\vec{A}|^2+|\vec{B}|^2+2|\vec{A}||\vec{B}|\cos\theta$

   we get $|\vec{A}+\vec{B}|^2=R^2+R^2+2R^2\cos\theta=2R^2(1+\cos\theta)$

   Now use $1+\cos\theta=2\cos^2\left(\frac{\theta}{2}\right)$

   so, $|\vec{A}+\vec{B}|^2=4R^2\cos^2\left(\frac{\theta}{2}\right)$

   Hence, $|\vec{A}+\vec{B}|=2R\cos\left(\frac{\theta}{2}\right)$

   So option A is correct.

3. Magnitude of the difference $\vec{A}-\vec{B}$:

   Using $|\vec{A}-\vec{B}|^2=|\vec{A}|^2+|\vec{B}|^2-2|\vec{A}||\vec{B}|\cos\theta$

   we get $|\vec{A}-\vec{B}|^2=R^2+R^2-2R^2\cos\theta=2R^2(1-\cos\theta)$

   Now use $1-\cos\theta=2\sin^2\left(\frac{\theta}{2}\right)$

   so, $|\vec{A}-\vec{B}|^2=4R^2\sin^2\left(\frac{\theta}{2}\right)$

   Hence, $|\vec{A}-\vec{B}|=2R\sin\left(\frac{\theta}{2}\right)$

4. Check options:

   • A: $|\vec{A}+\vec{B}|=2R\cos\left(\frac{\theta}{2}\right)$ ✅ correct
   • B: $|\vec{A}-\vec{B}|=2R\cos\left(\frac{\theta}{2}\right)$ ❌ incorrect
   • C: $|\vec{A}-\vec{B}|=\sqrt{2}R\sin\left(\frac{\theta}{2}\right)$ ❌ incorrect
   • D: $|\vec{A}+\vec{B}|=2R\sin\left(\frac{\theta}{2}\right)$ ❌ incorrect

5. Therefore, the single correct answer is: $\boxed{A}$

Comparison with stored correct answer:

• Stored correct answer: A
• Derived answer: A
• They agree.