Geometrically proving the half-angle formula for sine from a particular diagram

Question

I have attached a picture of the diagram I am using to prove the trig identity $\sin(\frac{\alpha}{2})= \sqrt{(\frac{1-\cos\alpha}{2})}$. I have that $\sin \alpha = \frac{DG}{OD}$ and $\sin\frac{\alpha}{2}=\frac{DE}{OD}$ as well as $\sin\frac{\alpha}{2}=\frac{EF}{OE}=EF$. I'm unsure of what segment could even represent $\sqrt{\frac{1-\cos\alpha}{2})}$?

Answer 1

Construct $DF$ let $DF$ intersect $OE$ at $P$

$\triangle OPF \cong \triangle OPD$ $\angle OPD\cong \angle OPF$ is a right angle.

$\sin \frac a2 = \frac {\text {opposite}}{\text{hypotenuse}} = \frac {PF}{OF} = \frac {PD}{OD}$

$OF = OD = \cos \frac a2\\ DP = PF = \cos \frac a2\sin \frac a2$

$DF= DP + PF = 2\cos \frac a2\sin \frac a2$

by the Pythagorean theorem $DF^2 = FG^2 + GD^2$

$OF = \cos \frac a2\\OG = OD\sin a= \cos \frac a2 \cos a\\FG = OF - OG = \cos a2 (1-cos a)\\GD = OD\sin a = \cos \frac a2\sin a$

$(2\cos \frac a2\sin \frac a2)^2 = (\cos \frac a2)^2(1-\cos a)^2 + ((\cos \frac a2)^2(\sin a)^2$

We can divide through by $(\cos \frac a2)^2$

$(2\sin \frac a2)^2 = (1-\cos a)^2 + (\sin a)^2$

$4\sin^2 \frac a2 = 1-2\cos a + cos^2 a + \sin^2 a = 2-2\cos a\\ \sin \frac a2 = \sqrt {\frac{1-\cos a}{2}}$

Answer 2

As "trigonographs" go, this one seems unsatisfying. Maybe I'm missing a cool geometric trick, but I don't see a way to tease-out the half-angle formulas from the elements given, without invoking a double-angle formula. Ideally, trigonographs are more self-evident.

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Here's a version of the diagram that I've "unfolded" a bit to reduce some overlap, replacing $\triangle DEH$ with $\triangle DEH^\prime$ and $\triangle ODG$ with $\triangle ODG^\prime$. Also, to avoid fractions, I'm using $\theta$ and $2\theta$ instead of $\alpha/2$ and $\alpha$. Side-lengths flow readly from $1$ to the rest of the figure.

Here, it's clear that $$\sin\theta\sin 2\theta = \cos\theta - \cos\theta \cos 2\theta \tag{1}$$ Then, since we "know" that $\sin 2\theta = 2\sin\theta\cos\theta$, we can write $$\sin\theta \cdot 2\sin\theta\cos\theta = \cos\theta\;\left(\;1-\cos 2\theta\;\right) \quad\to\quad \sin^2\theta = \frac12 \left(\;1-\cos 2\theta\;\right) \tag{2}$$

We also have $$\begin{align} \cos\theta\sin 2\theta = \sin\theta + \sin\theta \cos 2\theta &\quad\to\quad \cos\theta\cdot 2\sin\theta\cos\theta = \sin\theta\left(\;1+\cos 2\theta\;\right) \\[4pt] &\quad\to\quad \cos^2\theta = \frac12\left(\;1+\cos 2\theta\;\right) \end{align} \tag{3}$$

We get to the desired formula(s), but not so tidily as one might hope. I wonder: Does the source have a derivation that doesn't require re-writing $\sin 2\theta$?

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It's worth comparing the above to another trigonograph:

Here, a couple of leg-over-hypotenuse proportions from pairs of similar right triangles give well-known geometric mean relations:

$$\begin{align} \frac{2\sin\theta}{2} = \frac{1-\cos 2\theta}{2\sin\theta} &\quad\to\quad \sin^2\theta = \frac12\left(\;1-\cos 2\theta\;\right) \\[12pt] \frac{2\cos\theta}{2} = \frac{1+\cos 2\theta}{2\cos\theta} &\quad\to\quad \cos^2\theta = \frac12\left(\;1+\cos 2\theta\;\right) \end{align} \tag{4}$$

No extra trig identities are required, which would seem to give this trigonograph something of an intuitive advantage over the figure in the question.