Finding the area of a triangle on a unit circle Assume that $0< \theta < \pi$. For three points $A(1,0)$, $B(\cos(\theta),\sin(\theta))$ and $C(\cos(2\theta),\sin(2\theta))$ on a unit circle, the area of triangle $ABC$ is: ???
I drew out the unit circle and tried to get the dimensions. I ended up drawing a triangle underneath the main triangle to try and get the base, but that ended up giving me a square root and I couldn't find a way to get rid of it.
The answer is: $\sin(1-\cos)$

The three side lengths of $\triangle ABC$ are:
$$BC = AB = \sqrt{(\cos \theta - 1)^2 + (\sin \theta - 0)^2}$$
$$= \sqrt{\cos^2 \theta - 2 \cos \theta + 1 + \sin^2 \theta}$$
$$= \sqrt{2 - 2 \cos \theta}$$
$$AC = \sqrt{(\cos(2\theta) - 1)^2 + (\sin(2\theta)-0)^2}$$
$$= \sqrt{\cos^2(2\theta) - 2\cos(2\theta) + 1 + \sin^2(2\theta)}$$
$$= \sqrt{2 - 2\cos(2\theta)}$$
$$= \sqrt{2 - 2(1 - 2\sin^2 \theta)}$$
$$= \sqrt{4\sin^2\theta}$$
$$= 2 \sin \theta$$
(In that last step, we can assume that $\sin\theta$ is positive because were are given $0 < \theta < \pi$.)
If we cut this isosceles triangle in half by bisecting $\angle B$, we get two right triangles with a hypotenuse of $\sqrt{2 - 2\cos\theta}$ and a base leg of $\sin \theta$.  The other leg, the height of the triangle, is:
$$\sqrt{(\sqrt{2 - 2\cos\theta})^2 - \sin^2 \theta}$$
$$= \sqrt{2 - 2\cos\theta - \sin^2 \theta}$$
$$= \sqrt{2 - 2\cos\theta - (1 - \cos^2 \theta)}$$
$$= \sqrt{1 - 2\cos\theta + \cos^2 \theta}$$
$$= \sqrt{(1 - \cos\theta)^2}$$
$$= 1 - \cos\theta$$.
We thus have:
$$\text{area of } \triangle ABC = \frac{1}{2} \text{(base)}\text{(height)}$$
$$=\frac{1}{2} (2 \sin \theta)(1 - \cos \theta)$$
$$=(\sin \theta)(1 - \cos \theta)$$
