iterated double angle formula -- speed of convergence The double angle formulas (101) and (102), combined with the best linear approximations at $0$ (103) and (104) can be used to estimate $\sin$ or $\cos$ by expanding an expression $n$ times and shrinking the angle to $\frac{\theta}{2^n}$ and then applying the linear approximation.
$$ \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \tag{101} $$
$$ \cos(\theta) = \cos^2\left(\frac{\theta}{2}\right) - \sin^2\left(\frac{\theta}{2}\right) \tag{102} $$
$$ \sin(\theta) \approx \theta \tag{103} $$
$$ \cos(\theta) \approx 1 \tag{104} $$
It seems to converge slowly, maybe adding a bit of accuracy or less per iteration.
How quickly does it converge to the true value of $\sin(\theta)$ ?
Here's a sample illustrating the relatively slow convergence.
$$
\begin{array}[cc]
\;n & \sin_n(1) \\
0 & 1.0 \\
1 & 0.9375 \\
2 & 0.89233017 \\
3 & 0.86740446 \\
4 & 0.8545371 \\
5 & 0.84802556 \\
6 & 0.84475327 \\
7 & 0.8431133 \\
\end{array}
$$
 A: The error decays like a constant multiple of $2^n$, approximately halving at each step.
To calculate the speed, we first put this estimate into a more conventional form. Note that if we use the complex form $e^{i\theta}=\cos\theta+i\sin\theta$, this estimate becomes $e^{i\theta}\approx \left(1+\frac{i\theta}{2^n}\right)^{2^n}$, so
$$\sin_n(1) = \operatorname{Im}\left(\left(1+\frac{i}{2^n}\right)^{2^n}\right)=\operatorname{Im}\left(\exp\left(2^n\ln\left(1+\frac{i}{2^n}\right)\right)\right)$$
Estimate that logarithm with the power series; $\ln\left(1+\frac{i}{2^n}\right)=\frac{i}{2^n}+\frac{1}{2\cdot 2^{2n}}+O\left(\frac1{2^{3n}}\right)$.
Multiply by $2^n$ and exponentiate for $(\cos 1+i\sin 1)\cdot \left(1+\frac1{2\cdot 2^n}+O(2^{-2n})\right)\cdot (1+O(2^{-2n}))$. We extract the imaginary part of that for
$$\sin_n(1)=\sin 1+\frac1{2^n}\cdot\frac{\sin 1}{2}+O(2^{-2n})$$
There it is - the precise decay rate of the error, including the constant factor $\frac{\sin 1}{2}$.
That doesn't quite match your reported data, being off by a factor of $2$. Looking at that data, it looks like you've underreported the number of steps for each estimate by $1$. (This is now corrected in the question.) I get
$$\sin 1=2\sin\frac12\cos \frac12\approx 2\cdot\frac12\cdot 1=1$$
for $n=1$ step and
$$\sin 1= 2\sin\frac12\cos\frac12=4\sin\frac14\cos\frac14(\cos^2\frac14-\sin^2\frac14)\approx 4\cdot \frac14\cdot (1-\frac1{16})=0.9375$$
for $n=2$ steps.
