Find the area of triangle APB, where P is a point $(a\cos\theta, b\sin\theta)$ on an ellipse and $A, B$ are its radii points $(a,0) (0,b)$ 
A point $P(a\cos\theta, b\sin\theta)$ sits on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The points $A$ and $B$ have coordinates $(a,0)$ and $(0,b)$ respectively. Show that the area of triangle $APB$ is:
  $$
\frac{1}{2}AB(\cos\theta + \sin\theta -1)
$$

The length of $AB$ is $\sqrt{a^2 + b^2}$; and $C$, its mid-point, has co-ordinates of $(\frac{a}{2}, \frac{b}{2})$. So the area of $APB$ must be $\frac{1}{2}\cdot AB \cdot CP$, where:
$$
CP = \sqrt{\left(a\cos\theta - \frac{a}{2}\right)^2+\left(b\sin\theta - \frac{b}{2}\right)^2}
$$
But this gets rather messy. Am I going wrong somewhere?

A triangle with two sides given by vectors ${\bf x}, {\bf y}$ has area
$$\frac{1}{2}|{\bf x} \times {\bf y}|.$$ In this case we may take our vectors to be
\begin{align}
    {\bf x} &= (a \cos \theta, b \sin \theta) - (a, 0) = (a(\cos \theta - 1), b \sin \theta)\\
    {\bf y} &= (0, b) - (a, 0) = (-a, b).
\end{align}
Substituting and simplifying gives that the area is
$$\frac{1}{2}ab|\cos \theta + \sin \theta - 1|,$$
and checking shows that we can drop the absolute value symbols when $0 \leq \theta \leq \frac{\pi}{2}$.
