Consider a pyramid $P-ABCD$ whose base $ABCD$ is square and whose vertex $P$ is equidistant from $A,B,C$ and $D$. If $AB=1$ and $\angle{APB}=2\theta$, then the volume of the pyramid is
$\text{(A) } \frac{\sin(\theta)}{6}\quad \text{(B) } \frac{\cot(\theta)}{6}\quad \text{(C) } \frac{1}{6\sin(\theta)}\quad \text{(D) } \frac{1-\sin(2\theta)}{6}\quad \text{(E) } \frac{\sqrt{\cos(2\theta)}}{6\sin(\theta)}$
As the base has area $1$, the volume will be one third of the height. Drop a line from $P$ to $AB$, bisecting it at $Q$.

Then $\angle QPB=\theta$, so $\cot\theta=\frac{PQ}{BQ}=2PQ$. Therefore $PQ=\tfrac12\cot\theta$.
Now turning to the dotted triangle, by Pythagoras, the square of the pyramid's height is \[PQ^2-(\tfrac12)^2=\frac{\cos^2\theta}{4\sin^2\theta}-\frac14=\frac{\cos^2\theta-\sin^2\theta}{4\sin^2\theta}=\frac{\cos 2\theta}{4\sin^2\theta}\] and after taking the square root and dividing by three, the result is $\fbox{E}$