Is there a way to use the Weierstrass Substitution for $n\theta$? So, I was looking at the Wikipedia page for the tangent half-angle formula (or Weierstrass substitution) and I noticed something: Is there a similar set of substitutions as the ones below:
$$\cos\theta = \frac{1 - t^2}{1 + t^2}$$
$$\sin\theta = \frac{2t}{1 + t^2}$$
$$\tan\theta = \frac{2t}{1 - t^2}$$
but have $n\theta$ instead of $\theta$?
Edit: The reason why I asked this question is because I was trying to solve the following question below (I made the question up!):
Find the value of $\theta$ and hence the $s_\infty$ in this geometric sequence:
$$\frac{1}{2}\sin^4 4\theta, \frac{1}{6}\cos^3 8\theta, \frac{1}{24}\tan^2 12\theta$$
and I ended up with this equation (by trying to find $r$ and using the fact that $\frac{u_2}{u_1}$ $=$ $\frac{u_3}{u_2}$ as well as using power-reduction formulas and sum-to-product formulas):
$2\cos 24\theta(3\cos 8\theta - 2) + \cos 16\theta (8 cos 4\theta - 1) + \cos 4\theta - 4 = 0$
I was thinking at this point to use the Weierstrass substitution for  $\cos$ $n\theta$ but I couldn't find any analogue for it, hence the posting of this question.
Edit 2: Although the solution posted by Somos is great for personal use, I have decided to uncheck the solution because I want a solution to the question asked above that I can explain easily to A-Level Mathematics students that don't want to hear about anything beyond the syllabus. So if there are any solutions that stay within the boundaries of A-Level Mathematics, please feel free to post it.
Edit 3: The reason why I have checked Somos's solution is that the solution has helped me out the most, not because it solves the problem.
 A: You asked about a generalization of the
Weierstrass substitution.
Here is one approach using multiple angles.
Given $\,t := \tan(\theta),\; s := \sec(\theta) = \sqrt{1+t^2}\,$ then
$$ \sin(\theta) = \frac{t}{s},\;\;
\cos(\theta) = \frac{1}{s},\;\; \tan(\theta)=t, \tag{1} $$
$$ \sin(2\theta) = \frac{2t}{s^2},\;\;
\cos(2\theta) = \frac{1-t^2}{s^2},\;\; \tan(2\theta)=\frac{2t}{1-t^2},\tag{2}$$
$$ \sin(3\theta) = \frac{3t-t^3}{s^3},\;\;
\cos(3\theta) = \frac{1-3t^2}{s^3},\;\;
\tan(3\theta) = \frac{3t-t^3}{1-3t^2}. \tag{3} $$
This generalizes to
$$ \sin(n\theta) = \frac{S_n(t)}{ s^n},\;\;
\cos(n\theta) = \frac{C_n(t)}{ s^n},\;\;
\tan(n\theta) = \frac{ S_n(t)}{C_n(t)} \tag{4} $$
where $\,S_n(t),C_n(t)\,$ are polynomials in $t$. Use the
Trigonometric Addition Formulas to get
$$ \sin(n\theta) = \sin((n-1)\theta)\cos(\theta)
 +\cos((n-1)\theta)\sin(\theta), \tag{5}$$
$$ \cos(n\theta) = \cos((n-1)\theta)\cos(\theta)
 -\sin((n-1)\theta)\sin(\theta). \tag{6}$$
Use the equations in $(4)$ to get
$$ S_n(t) = S_{n-1}(t) + C_{n-1}(t)\,t, \tag{7}$$
$$ C_n(t) = C_{n-1}(t) - S_{n-1}(t)\,t. \tag{8}$$
For example,
$$ S_4(t) = S_3(t) + C_3(t)\,t = (3t-t^3) + (1-3t^2)\,t = 4t-4t^3,
\tag{9} $$
$$ C_4(t) = C_3(t) - S_3(t)\,t = (1-3t^2) - (3t-t^3)\,t = 1-6t^2+t^4,
\tag{10} $$
$$ S_8(t) = 8t-56t^3+56t^5-8t^7, \quad
C_8(t) = 1-28t^2+70t^4-28t^6+t^8. \tag{11}$$
The coefficients of $\,S_n(t)\,$ alternate in sign and are the odd indexed
terms of the $n$-th row of
Pascal's triangle.
Similarly, for $\,C_n(t)\,$ with the even indexed terms. For example,
the $8$-th row is
$$1\quad 8\quad 28\quad 56\quad 70\quad 56\quad 28\quad 8\quad 1. \tag{9}$$

NOTE: In equations $(1),(3)$ and $(4)$ for odd $\,n\,$ there is a little
glitch with the $\,\sqrt{1+t^2}.\,$ The function $\,\tan(\theta)\,$ has
period $\,\pi\,$ while $\,\sin(\theta)\,$ and $\,\cos(\theta)\,$ both have
period $\,2\pi.\,$ Thus, the sign of the square root must be chosen to be
the appropriate one depending on the quadrant of the angle $\,\theta\,$
but this adjustment is only needed if $\,n\,$ is odd.
