# 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.

$$\begin{array}[cc] \;n & \sin_n(1) \\ 0 & 1.0 \\ 1 & 0.9375 \\ 2 & 0.89233017 \\ 3 & 0.86740446 \\ 4 & 0.8545371 \\ 5 & 0.84802556 \\ 6 & 0.84475327 \\ 7 & 0.8431133 \\ \end{array}$$

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.