Doubling the Angle from the Ladder 2

$\sin x=\frac {H}{L}$ and $\cos x =\frac {D}{L}$

In the new triangle, $x$ would get turned into $2x$ , keeping $L$ constant.

Since the expressions for $\sin 2x$ and $\cos 2x$ in terms of $\sin x$ and $\cos x$ are polynomial functions with rational coefficients, the values of trigonometric functions for double the angle will still be rational.

$\sin 2x=2\sin x \cos x=\frac {2HD}{L^2}= \frac {H'}{L}$

$\cos 2x = \cos ^{2} x - \sin ^{2} x = \frac {D^2-H^2}{L^2}= \frac {D'}{L}$

Hence, $H'=\frac {2HD}{L}$ and $D'= \frac {D^2-H^2}{L}$

For every integral Pythagorean triplet, there will be some proportion of it which will have the above two values integral, i.e. for every triplet $a,b,c$ there will be a $ka,kb,kc$ for which the above two values are integral.

For Pythagorean triplets, let's us take:

$D = k(m^2 - n^2); H = k(2mn), L = k(m^2 + n^2)$ , where $m,n,k \in \mathbb{N}$ and $m > n$

Let $tan(x) = \frac{H}{D} = \frac{2mn}{m^2 - n^2} \Rightarrow x = arctan(\frac{2mn}{m^2 - n^2})$ , and $2x = 2 \cdot arctan(\frac{2mn}{m^2 - n^2})$ . Taking our doubled angle, $2x$ , we now obtain:

$tan(2x) = \frac{H'}{D'} = tan(2 \cdot arctan(\frac{2mn}{m^2 - n^2}) = \frac{2tan(arctan(\frac{2mn}{m^2 - n^2})}{1 - tan^{2}(arctan(\frac{2mn}{m^2 - n^2})} = \frac{4mn}{m^2-n^2} \cdot \frac{(m^2 - n^2)^2}{(m^2 - n^2)^2 - (2mn)^2} = \frac{4mn(m^2 - n^2)}{(m^2 - n^2)^2 - (2mn)^2}.$ (i)

The expression in (i) requires $(m^2 - n^2)^2 - (2mn)^2 \neq 0 \Rightarrow m^2 - n^2 \neq 2mn$ , which is always true for the leg lengths of Pythagorean triplets. Hence, there are infinitely many triplets that satisfy this $L, D, H, D', H' \in \mathbb{N}$ for a doubled angle $x$ .