While I was randomly having fun with a calculator, I stumbled upon an interesting approximation: $\sin(1) \approx \frac{1}{\sqrt[4]{2}}$.
My initial step was to recognize that the sine function relates the angles in a right triangle, but this alone did not explain why $\sin(1)$ should be so close to $\frac{1}{\sqrt[4]{2}}$ as it doesn't make sense for the hypotenuse and opposite to be equal. So I looked at the taylor series expansion.
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$$
I plugged in $x = 1$ to this series and calculated several terms, which led to an approximation close to $\sin(1)$:
$$\sin(1) = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + ...$$
Next, I wanted to investigate the value of $\frac{1}{\sqrt[4]{2}}$. Using some algebraic manipulation:
$$\frac{1}{\sqrt[4]{2}}\approx 0.84089...$$
Surprisingly, this value is very close to the approximation I obtained for $\sin(1)$ using the Taylor series expansion.
Now, I'm keen to understand if there is any deeper mathematical significance or intuition behind this result, or if it is purely coincidental. Can anyone provide additional insights?
Let $\sin{\theta} = \dfrac{1}{\sqrt[4]{2}}$. Then $\theta = \arcsin{\dfrac{1}{\sqrt[4]{2}}} = \dfrac{1}{2}\arccos{(1 - \sqrt{2})}$ by the relation $\arccos{(1 - x)} = 2\arcsin\sqrt{\dfrac{x}{2}}$
Observe, $1 - \sqrt{2} < 0$. Hence, for some $\gamma$, which $\cos\gamma = 1 - \sqrt{2}$, $\dfrac{\pi}{2}<\gamma < \dfrac{3\pi}{2}$. Note that $\cos{\left(\gamma + \dfrac{\pi}{2}\right)} = -\sin\gamma$. Thus we could assume a first order taylor approximation of $-\sin{\gamma}$ centred at the origin, for $\cos\gamma, \space \gamma \sim \dfrac{\pi}{2}$, which would be
$$-\sin\gamma \approx -\gamma + \mathcal{O}(\gamma^3) \\\implies \cos(\gamma)_{\gamma \sim \pi/2} \approx \dfrac{\pi}{2} - \gamma + \mathcal{O}(\gamma^3)$$
Hence, inverting the approximation for $\gamma = \dfrac{1}{2}\arccos x$, we have $x \approx \dfrac{\pi}{2} - 2\gamma \iff \gamma \approx \dfrac{\pi - 2x}{4}$
Thus, $\gamma \approx \dfrac{\pi - 2(1 - \sqrt{2})}{4} = \dfrac{\pi}{4} + \dfrac{\sqrt{2}}{2} - 0.5$. Note that $\dfrac{1}{2}(\sqrt{2}) \approx \dfrac{1}{2} \cdot \left(1 +\dfrac{1}{2+\dfrac{1}{2 + \ddots_3}}\right) \approx \dfrac{99}{140} \approx 0.7071$ and using $\dfrac{\pi}{4}\approx \dfrac{355}{113 \cdot 4} \approx 0.7854$,
we see that
$\gamma \approx \dfrac{\pi}{4} + \dfrac{\sqrt{2}}{2} - 0.5 \approx 0.7854 + 0.7071 - 0.5 \approx 0.9925 \approx 1$