An extremely beautiful trigonometric pattern

There is a pattern which is quite common: \(\sin 0^\circ=\dfrac{\sqrt{0}}{2}, \sin 30^\circ=\dfrac{\sqrt{1}}{2}, \sin 45^\circ=\dfrac{\sqrt{2}}{2}, \sin 60^\circ=\dfrac{\sqrt{3}}{2}, \sin 90^\circ=\dfrac{\sqrt{4}}{2}\)

This pattern is useful because it helps us to memorize the trig functions of the main angles. However, when we start to look for the sine of some angles like $15^\circ$ or $67.5^\circ$, we may struggle for some time. However, there is a beautiful pattern: $\sin 0^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{4}} \quad \sin 15^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{3}} \quad \sin 22.5^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{2}} \\ \sin 30^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{1}} \quad \sin 45^{\circ} = \frac{1}{2}\sqrt{2\pm\sqrt{0}} \quad \sin 60^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{1}} \\ \sin 67.5^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{2}} \quad \sin 75^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{3}} \quad \sin 90^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{4}}$ That's amazing, isn't it? But I want to tell you, there's a reason behind. I'll show you that.

Firstly, from the first pattern, we know that: $\cos 0^\circ=\dfrac{\sqrt{4}}{2}\quad\cos 30^\circ=\dfrac{\sqrt{3}}{2}\quad\cos 45^\circ=\dfrac{\sqrt{2}}{2}\\\cos 60^\circ=\dfrac{\sqrt{1}}{2}\quad\cos 90^\circ=\dfrac{\sqrt{0}}{2}\quad\cos 120^\circ=-\dfrac{\sqrt{1}}{2}\\\cos 135^\circ=-\dfrac{\sqrt{2}}{2}\quad\cos 150^\circ=-\dfrac{\sqrt{3}}{2}\quad\cos 180^\circ=-\dfrac{\sqrt{4}}{2}$ Then, we'll use the half-angle formula $\sin^2 \dfrac{\theta}{2} = \dfrac{1-\cos\theta}{2}$

$\because 0^\circ \le\theta\le 180^\circ \rightarrow 0^\circ \le\dfrac{\theta}{2}\le 90^\circ \\ \therefore \sin \dfrac{\theta}{2} = \sqrt{\dfrac{1-\cos\theta}{2}}\ge 0 \\ \text{The }\cos\theta\text{ we want are all in the form }\pm\dfrac{\sqrt{n}}{2}\text{ where }n=0,1,2,3,4 \\ \quad\sin \dfrac{\theta}{2}=\sqrt{\dfrac{1\mp\frac{\sqrt{n}}{2}}{2}}=\sqrt{\dfrac{2\mp\sqrt{n}}{4}}=\dfrac{1}{2}\sqrt{2\mp\sqrt{n}}$

Therefore, we get the the result below: $\sin 0^\circ=\sin \dfrac{1}{2}\left(0^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{4}}\quad\sin 15^\circ=\sin \dfrac{1}{2}\left(30^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{3}}\quad\sin 22.5^\circ=\sin \dfrac{1}{2}\left(45^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{2}}\\\sin 30^\circ=\sin \dfrac{1}{2}\left(60^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{1}}\quad\sin 45^\circ=\sin \dfrac{1}{2}\left(90^\circ\right)=\dfrac{1}{2}\sqrt{2\mp\sqrt{0}}\quad\sin 60^\circ=\sin \dfrac{1}{2}\left(120^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{1}}\\\sin 67.5^\circ=\sin \dfrac{1}{2}\left(135^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{2}}\quad\sin 75^\circ=\sin \dfrac{1}{2}\left(150^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{3}}\quad\sin 90^\circ=\sin \dfrac{1}{2}\left(180^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{4}}$

I am so proud of myself that I can prove this beautiful pattern! (I first found this pattern when I see this) I hope it will help you guys when you are doing some work about trigonometry.