4. Given that $\triangle A B C$ is an isosceles right triangle, $\angle A=90^{\circ}$, and $\overrightarrow{A B}=\boldsymbol{a}+\boldsymbol{b}$, $\overrightarrow{A C}=a-b$. If $a=(\cos \theta, \sin \theta)(\theta \in \mathbf{R})$, then the area of $\triangle A B C$ is $\qquad$ .
(4) 1 Hint: From the given, $A B \perp A C,|A B|=|A C|$, so
$$
\left\{\begin{array}{l}
(\boldsymbol{a}+\boldsymbol{b}) \cdot (\boldsymbol{a}-\boldsymbol{b})=0, \\
|\boldsymbol{a}+\boldsymbol{b}|=|\boldsymbol{a}-\boldsymbol{b}|,
\end{array} \quad \text { that is } \left\{\begin{array}{l}
|\boldsymbol{a}|=|\boldsymbol{b}|, \\
\boldsymbol{a} \cdot \boldsymbol{b}=0 .
\end{array}\right.\right.
$$

Since $|\boldsymbol{a}|=1$, then $|\boldsymbol{b}|=1$.
According to the geometric meaning of vector addition and subtraction, we get $|a+b|=|a-b|=\sqrt{2}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2}|\boldsymbol{a}+\boldsymbol{b}| \cdot|\boldsymbol{a}-\boldsymbol{b}|=1$.