# Is there a simple pattern to memorize the sine of $0^\circ$, $15^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $75^\circ$, $90^\circ$?

We know there is a nice pattern to memorize the sine of $$0^\circ$$, $$30^\circ$$, $$45^\circ$$, $$60^\circ$$, $$90^\circ$$ as follows.

\begin{align} \sin 0^\circ &= \tfrac12\sqrt0\\ \sin 30^\circ &= \tfrac12\sqrt1\\ \sin 45^\circ &= \tfrac12\sqrt2\\ \sin 60^\circ &= \tfrac12\sqrt3\\ \sin 90^\circ &= \tfrac12\sqrt4 \end{align}

We also know that

\begin{align} \sin 15^\circ &= \tfrac14(\sqrt6-\sqrt2)\\ \sin 75^\circ &= \tfrac14(\sqrt6+\sqrt2) \end{align}

# Question

If I want to combine these two groups, is there a simple nice pattern available for us to easily rote memorize them?

• (+1) for this nice pattern, I was not aware of. – Peter Aug 27 '19 at 13:51
• The sines below apparently do not fit in the pattern above. – Peter Aug 27 '19 at 13:53
• One way: youtube.com/watch?v=hTGhCGrPLmw - Not sure if it helps. – NoChance Aug 27 '19 at 14:32
• In this answer, I give a unified form for sines of multiples of $3^\circ$. No particularly "nice" patterns, though. – Blue Aug 28 '19 at 22:41
• @Blue: great info, entirely covers my small answer. – Oleg567 Aug 29 '19 at 12:23

Hmm, this pattern works:

$$\sin 0^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{4}}, \\ \sin 15^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{3}}, \\ \color{gray}{\sin 22.5^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{2}},} \\ \sin 30^{\circ} = \frac{1}{2}\sqrt{2-\sqrt{1}}, \\ \sin 45^{\circ} = \frac{1}{2}\sqrt{2\pm\sqrt{0}}, \\ \sin 60^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{1}}, \\ \color{gray}{\sin 67.5^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{2}},} \\ \sin 75^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{3}}, \\ \sin 90^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{4}}.$$

• Brilliant! That's why Maths are beautiful – Isaac YIU Math Studio Aug 28 '19 at 21:08
• "... of which fact you have discovered a truly marvelous sequence. The narrowness of this margin doesn’t permit [you to include the steps]." – Money Sets You Free Aug 29 '19 at 8:15

This answer is inspired from Oleg567's answer. (I should say, I will explain why does this beautiful sequence appears.)

Firstly we know that: (It should be the really amazing and beautiful sequence) $$\cos 0^\circ=\dfrac{\sqrt{4}}{2}\quad\cos 30^\circ=\dfrac{\sqrt{3}}{2}\quad\cos 45^\circ=\dfrac{\sqrt{2}}{2}\\\cos 60^\circ=\dfrac{\sqrt{1}}{2}\quad\cos 90^\circ=\dfrac{\sqrt{0}}{2}\quad\cos 120^\circ=-\dfrac{\sqrt{1}}{2}\\\cos 135^\circ=-\dfrac{\sqrt{2}}{2}\quad\cos 150^\circ=-\dfrac{\sqrt{3}}{2}\quad\cos 180^\circ=-\dfrac{\sqrt{4}}{2}\\$$ 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: $$\sin 0^\circ=\sin \dfrac{1}{2}\left(0^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{4}}\quad\sin 15^\circ=\sin \dfrac{1}{2}\left(30^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{3}}\\\sin 22.5^\circ=\sin \dfrac{1}{2}\left(45^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{2}}\quad\sin 30^\circ=\sin \dfrac{1}{2}\left(60^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{1}}\\\sin 45^\circ=\sin \dfrac{1}{2}\left(90^\circ\right)=\dfrac{1}{2}\sqrt{2\mp\sqrt{0}}\quad\sin 60^\circ=\sin \dfrac{1}{2}\left(120^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{1}}\\\sin 67.5^\circ=\sin \dfrac{1}{2}\left(135^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{2}}\quad\sin 75^\circ=\sin \dfrac{1}{2}\left(150^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{3}}\\\sin 90^\circ=\sin \dfrac{1}{2}\left(180^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{4}}$$