How to compute Dottie number accurately?

Using Newton's method,

$$\alpha = \alpha + \frac{\cos \alpha - \alpha}{\sin\alpha + 1}$$ Use this for a fixed-point iteration with chosen starting value ($\alpha_0 := \frac1{\sqrt2} = 0.7\color{red}{172}\ldots$ seems like a good choice)

Thus

$$\alpha_1 = \frac1{\sqrt 2} + \frac{\cos\frac1{\sqrt2}-\frac1{\sqrt2}}{\sin\frac1{\sqrt2} + 1} = 0.739\color{red}3\ldots \\ \alpha_2 = \frac1{\sqrt 2} + \frac{\cos\frac1{\sqrt2}-\frac1{\sqrt2}}{\sin\frac1{\sqrt2} + 1} + \frac{\cos\left(\frac1{\sqrt 2} + \frac{\cos\frac1{\sqrt2}-\frac1{\sqrt2}}{\sin\frac1{\sqrt2} + 1}\right) - \frac1{\sqrt 2} - \frac{\cos\frac1{\sqrt2}-\frac1{\sqrt2}}{\sin\frac1{\sqrt2} + 1}}{\sin\left(\frac1{\sqrt 2} + \frac{\cos\frac1{\sqrt2}-\frac1{\sqrt2}}{\sin\frac1{\sqrt2} + 1}\right) + 1} = 0.7390851\color{red}4\ldots$$

As you can see, it converges quickly. Only one more iteration gives

$$\alpha_3 = 0.73908513321516\color{red}1\ldots$$

Which is equal to $\alpha$ within the IEEE double precision standard. For $\alpha_0 = 0.7$ you need one more iteration for the same result, $0.739$ only requires two iterations and from $0.73908513$, one iteration is enough for double-precision.

Taylor series of order 2 gives a simple quadratic in $\alpha$: $$\alpha=1-\alpha^2/2\implies \alpha=0.\color{red}{73}2..$$ Of order 4 gives a bi-quadratic (there's a formula to solve roots of a polynomial of degree less than 5) in $\alpha^2$: $$\alpha=1-\alpha^2/2+\alpha^4/4\implies 0.\color{red}{739}2..$$ Fairly accurate for practical purposes wherein the correct value is $0.739085...$