$$\left(\;\cos 2\alpha + \cos 2\beta + 2 \cos( \alpha+\beta ) \;\right)^2 \;\;+\;\; \left(\;\sin 2\alpha + \sin 2\beta + 2 \sin( \alpha+\beta )\; \right)^2 \\[6pt]

= \;\;\left(\;2\;\left(\;1 + \cos(\alpha-\beta)\;\right)\;\right)^2

$$

= \;\;\left(\;2\;\left(\;1 + \cos(\alpha-\beta)\;\right)\;\right)^2

$$

**Note:** The simplest form of the hypotenuse length is, of course, \( 4 \cos^2\left(\frac{\alpha-\beta}{2}\right) \).

Augmenting this trigonograph to illustrate introduces too much clutter for my taste.

I’d rather find a more-direct alternative construction.

Motivated by this question on the Mathematics Stack Exchange.