Elementary proof of tangent half angle formula 
Hi all, I am interested to find elementary proof of tangent half angle formula.
My solutions are the following:
Triangle $AOB$ is such that $|AB|=1$ and $\angle AOB=\theta$. We then extend $OB$ to $P$ and $Q$ such that $|OP|=|OQ|=1$. Thus we will have two isosceles triangles: $AOP$ and $AOQ$.
From the picture, $\tan{\left(\frac{\theta}{2}\right)}=\frac{AB}{BP}=\frac{\sin{\left(\theta\right)}}{1+\cos{\left(\theta\right)}}\ \ \ =\frac{BQ}{AB}=\frac{1-\cos{\left(\theta\right)}}{\sin{\left(\theta\right)}}$
Could You guys please check my solution. I am also wondering if there are other elementary solutions, please share, thanks!
 A: I like your proofs, so I'll just offer a refinement of their presentation:

$$\frac{\sin\theta}{1+\cos\theta}=\frac{|AB|}{|PB|}=\tan\frac{\theta}{2} = \frac{|QB|}{|AB|}=\frac{1-\cos\theta}{\sin\theta}$$
A: Here's another proof:
$\tan(\frac{A}{2}) = \frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} = \frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} \cdot \left(\frac{2\sin\frac{A}{2}}{2\sin\frac{A}{2}}\right) = \frac{2\sin^2(\frac{A}{2})}{2\sin(\frac{A}{2})\cos(\frac{A}{2})} = \frac{2 \cdot \frac{1-\cos A}{2}}{\sin \left(2 \cdot \frac{A}{2}\right)} = \frac{1-\cos A}{\sin A}$
And yes, to me, your proof is correct.
A: Another easy way with your upper triangle:
By the whole triangle $\;\Delta APB\;$ and then a little algebra + trigonometry,
$$\tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta}=\frac{\sin\theta}{1+\cos\theta}\cdot\frac{1-\cos\theta}{1-\cos\theta}=\frac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta}=$$$$\frac{\sin\theta(1-\cos\theta)}{\sin^2\theta}=\frac{1-\cos\theta}{\sin\theta}$$
Your solution is correct, too.
