4. Given $\triangle A B C$ is an isosceles right triangle, $\angle A$ $=90^{\circ}$, and $\overrightarrow{A B}=a+b, \overrightarrow{A C}=a-b$.

If $a=(\cos \theta, \sin \theta)(\theta \in \mathbf{R})$, then $S_{\triangle A B C}$ $=$ . $\qquad$
4. 1.

From the problem, we know that $A B \perp A C,|A B|=|A C|$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
(a+b) \cdot(a-b)=0, \\
|a+b|=|a-b|
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
|a|=|b|, \\
a \cdot b=0 .
\end{array}\right.
\end{array}
$$

Since $|a|=1$, we have $|b|=1$.
According to the geometric meaning of vector addition and subtraction,
$$
|a+b|=|a-b|=\sqrt{2} \text {. }
$$

Therefore, $S_{\triangle A B C}=\frac{1}{2}|a+b| \cdot|a-b|=1$.