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Sketch

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!

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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.

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  • $\begingroup$ nice solution sir! $\endgroup$ – Rezha Adrian Tanuharja Mar 26 at 12:59
  • $\begingroup$ I tried to make one and I got $\dfrac{\sin x}{\cos x+1}$. In your case you'd get that by multiplying the numerator and denominator by $2\cos x$ instead. Why not use this one as the formula? $\endgroup$ – Plato Mar 28 at 14:45
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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.

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I like your proofs, so I'll just offer a refinement of their presentation:

enter image description here

$$\frac{\sin\theta}{1+\cos\theta}=\frac{|AB|}{|PB|}=\tan\frac{\theta}{2} = \frac{|QB|}{|AB|}=\frac{1-\cos\theta}{\sin\theta}$$

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