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 $0 \leq \theta \leq \frac{\pi}{2}$.