If `vec(a) and vec(b)` are unit vectors and `theta` is the angle between them, then what is `sin^(2)((theta)/(2))` equal to?

Options

A. `(|vec(a)+ vec(b)|^(2))/(4)`
B. `(|vec(a)- vec(b)|^(2))/(4)`
C. `(|vec(a) + vec(b)|^(2))/(2)`
D. `(|vec(a) - vec(b)|^(2))/(2)`

Answer

To solve the problem, we need to find the value of \( \sin^2\left(\frac{\theta}{2}\right) \) given that \( \vec{a} \) and \( \vec{b} \) are unit vectors and \( \theta \) is the angle between them.

### Step-by-Step Solution:

1. **Understanding Unit Vectors**:
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, their magnitudes are:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]

2. **Using the Dot Product**:
The dot product of two vectors can be expressed in terms of the angle between them:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) = 1 \cdot 1 \cdot \cos(\theta) = \cos(\theta) \]

3. **Using the Half-Angle Identity**:
We can use the half-angle identity for sine:
\[ \sin^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{2} \]

4. **Substituting the Value of \( \cos(\theta) \)**:
From the dot product, we know \( \cos(\theta) = \vec{a} \cdot \vec{b} \). Therefore, we can substitute this into the half-angle identity:
\[ \sin^2\left(\frac{\theta}{2}\right) = \frac{1 - \vec{a} \cdot \vec{b}}{2} \]

5. **Final Expression**:
Thus, the value of \( \sin^2\left(\frac{\theta}{2}\right) \) can be expressed as:
\[ \sin^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{2} = \frac{1 - \vec{a} \cdot \vec{b}}{2} \]

### Conclusion:
The final answer is:
\[ \sin^2\left(\frac{\theta}{2}\right) = \frac{1 - \vec{a} \cdot \vec{b}}{2} \]