The proofs that I've seen of the Sine Sum Formula $ \sin(a + b) = \sin(a)\cos(b) + \sin(b)\cos(a) $ are from Khan Academy and Socratic. Both of them begin with geometric constructions like this:
My question is how can you prove this geometric construction is allowed through Rigid Transformations. I don't see it as given that the hypotenuse can be 1, for any triangle created by angle A or created by angle B. I know that it's True, I just want to understand why we are arbitrarily able to stack right triangles like this and pick a hypotenus length, using Rigid Transformations (aka compass and straightedge).
For a given $c>0$ and acute angle $\theta$ (where $\theta$ is arbitrary in the interval $(0,90^{\circ}))$, you can construct a right triangle with hypotenuse $c$ and angle $\theta$. If you want to use rigid transformations, the construction goes like this. Draw $AC=c$, then rotate $C$ by $\theta$ degrees, and let its image be $C_1$. Reflect $C$ about the line $AC_1$, and let the reflection be $C_2$. Let lines $AC_1$ and $CC_2$ meet at $B$. Then your right triangle is $\triangle ABC$ with $\angle B=90^{\circ}$ and $\angle A=\theta$.
The proof you cited then could say something like this. Draw a right triangle with hypotenuse $1$ and acute angle $\beta$. Its legs are $\cos\beta$ and $\sin\beta$. Draw another right triangle with hypotenuse $\cos\beta$ (i.e. you use the leg of first triangle as a hypotenuse) and acute angle $\alpha$. That's how you arrive at the picture. The choice of $1$ for the length of the hypotenuse is irrelevant. You could choose any $c>0$. It's just that $c=1$ simplifies computations.