An extremely beautiful trigonometric pattern

There is a pattern which is quite common: sin0=02,sin30=12,sin45=22,sin60=32,sin90=42\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 1515^\circ or 67.567.5^\circ, we may struggle for some time. However, there is a beautiful pattern: sin0=1224sin15=1223sin22.5=1222sin30=1221sin45=122±0sin60=122+1sin67.5=122+2sin75=122+3sin90=122+4\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: cos0=42cos30=32cos45=22cos60=12cos90=02cos120=12cos135=22cos150=32cos180=42\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 sin2θ2=1cosθ2\sin^2 \dfrac{\theta}{2} = \dfrac{1-\cos\theta}{2}

0θ1800θ290sinθ2=1cosθ20The cosθ we want are all in the form ±n2 where n=0,1,2,3,4sinθ2=1n22=2n4=122n\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: sin0=sin12(0)=1224sin15=sin12(30)=1223sin22.5=sin12(45)=1222sin30=sin12(60)=1221sin45=sin12(90)=1220sin60=sin12(120)=122+1sin67.5=sin12(135)=122+2sin75=sin12(150)=122+3sin90=sin12(180)=122+4\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.

Note by Isaac Yiu Math Studio
2 weeks, 3 days ago

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