looking for reference or nice proof of trig lemma Math people:
I am looking for a reference or a nice proof of the following fact.  I have proven it myself, but my proof is messy: let $\theta \in (0,1]$ and $\alpha \in (0, \frac{1}{2}\theta^2]$.  Let $P$ and $S$ be two points on the unit circle $x^2 + y^2 = 1$ such that the arc length of ${\stackrel{\frown}{PS}}$ is $\theta$.  Let $Q$ and $R$ be on ${\stackrel{\frown}{PS}}$ with $Q$ between $P$ and $R$, such that ${\stackrel{\frown}{PQ}}$ and ${\stackrel{\frown}{RS}}$ have equal arc length, and the arc length of ${\stackrel{\frown}{QR}}$ is $\alpha$.  Let $T$ and $U$ be the points on the chord ${\overline {PS}}$ such that $\Delta PQT$ and $\Delta SRU$ are right triangles. See the picture below.

Obviously I would not include a hand-drawn picture in a paper I would submit to a journal.  
I have proven that the the triangles have disjoint closures.  
I proved it the following way: I assumed $P = (\cos(\frac{1}{2}\theta),\sin(\frac{1}{2}\theta))$, $S = (\cos(\frac{1}{2}\theta),-\sin(\frac{1}{2}\theta))$.  Clearly it is good enough to take $\alpha = \frac{1}{2}\theta^2$.  Then $Q = (\cos(\frac{1}{4}\theta^2),\sin(\frac{1}{4}\theta^2))$.  I defined $V = (\cos(\frac{1}{2}\theta, 0)$.  It suffices to show $\angle VQP$ is obtuse.  I showed ${\vec{QP}} \cdot {\vec{QV}} < 0$.  This required a long calculation that was not pretty.  I am looking for a reference of this fact or a more elegant proof, perhaps a geometric proof.  
 A: No one has answered in a long time, so I'll answer myself.  I'll start as I indicated in my question. It actually is not that bad.  Using standard trig identities,
$${\vec{QP}}\cdot{\vec{QV}} = (\cos(\frac{1}{4}\theta^2)-\cos(\frac{1}{2}\theta))^2-
      (\sin(\frac{1}{2}\theta)-\sin(\frac{1}{4}\theta^2))\sin(\frac{1}{4}\theta^2)=\ldots=
     \frac{3}{2}+\frac{1}{2}\cos(\theta)-\frac{1}{2}\cos(\frac{\theta}{2}+\frac{\theta^2}{4})-\frac{3}{2}\cos(\frac{\theta}{2}-\frac{\theta^2}{4}). $$
Using the Maclaurin series of $\cos$ and properties of alternating series,
$1-\frac{x^2}{2}<\cos x < 1-\frac{x^2}{2}+\frac{x^4}{24}$ for all $x \in (0,1)$.  Both 
$\frac{\theta}{2}+\frac{\theta^2}{4}$ and $\frac{\theta}{2}-\frac{\theta^2}{4}$ are between $0$ and $1$.  Therefore
$${\vec{QP}}\cdot{\vec{QV}} < 
   \frac{3}{2}+\frac{1}{2}(1-\frac{\theta^2}{2}+\frac{\theta^4}{24}) -
   \frac{1}{2}(1-(\frac{1}{2}(\frac{\theta}{2}+\frac{\theta^2}{4})^2)-
  \frac{3}{2}(1-(\frac{1}{2}(\frac{\theta}{2}-\frac{\theta^2}{4})^2) =
    -\frac{1}{8}\theta^3+\frac{1}{24}\theta^4<0.$$
