Problem Analysis

The question asks for the value of the expression $\sin \theta - \cos \theta$, given that $\sin \theta + \cos \theta = \frac{\sqrt{7}}{2}$.

Solving Trigonometric Expressions

We can solve this problem using algebraic manipulation combined with trigonometric identities.

Let $A = \sin \theta + \cos \theta$ and $B = \sin \theta - \cos \theta$. We are given $A = \frac{\sqrt{7}}{2}$ and need to find $B$.

Step-by-Step Calculation

1. Consider the square of the given expression: $A^2 = (\sin \theta + \cos \theta)^2$ $A^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta$
2. Using the identity $\sin^2 \theta + \cos^2 \theta = 1$: $A^2 = 1 + 2 \sin \theta \cos \theta$
3. Substitute the given value $A = \frac{\sqrt{7}}{2}$: $(\frac{\sqrt{7}}{2})^2 = 1 + 2 \sin \theta \cos \theta$ $\frac{7}{4} = 1 + 2 \sin \theta \cos \theta$
4. Calculate the value of $2 \sin \theta \cos \theta$: $2 \sin \theta \cos \theta = \frac{7}{4} - 1$ $2 \sin \theta \cos \theta = \frac{3}{4}$
5. Now consider the square of the expression we need to find: $B^2 = (\sin \theta - \cos \theta)^2$ $B^2 = \sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta$
6. Substitute $\sin^2 \theta + \cos^2 \theta = 1$ and the value of $2 \sin \theta \cos \theta$ from step 4: $B^2 = 1 - (\frac{3}{4})$ $B^2 = \frac{1}{4}$
7. Solve for B: $B = \pm \sqrt{\frac{1}{4}}$ $B = \pm \frac{1}{2}$

The possible values for $\sin \theta - \cos \theta$ are $\frac{1}{2}$ and $-\frac{1}{2}$. Since $\frac{1}{2}$ is an option, it is the required value.