I noticed a relation between the cosines of $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$ and the square roots of $4, 3, 2, 1, 0$

Question

Yesterday I've noticed some relationships with cosine and square root. Anything interesting about it?

I was trying to find the smallest width on an hexagon with radius $1.0$ and I noticed that I could get it both by $\sqrt 3$ and the or two times the cosine of $30^\circ$.

I wondered if there would be more matches and so I notice these even though might look silly...

$$\begin{align} 2 \cos 90^\circ &= \sqrt 0 \\ 2 \cos 60^\circ &= \sqrt 1 \\ 2 \cos 45^\circ &= \sqrt 2 \\ 2 \cos 30^\circ &= \sqrt 3 \\ 2 \cos \phantom{9}0^\circ &= \sqrt 4 \end{align}$$

Is there anything interesting to know about these?

Thanks

Answer 1

This rules we used in high school with fingers do apply for both sine and cosine see the pictures

See also the following

Answer 2

Your list is sometimes given in trig books as a mnemonic device. Looking a little deeper the pattern that the angles in a regular $n$-gon give algebraic values for the $\cos$ function is showing itself, though the pattern of simple surds does not continue, e.g. $\cos (\frac{2\pi}{5})=\frac{1}{4}(\sqrt{5}-1)$

Answer 3

The last figure was also teached to me mnemotecnically to memorize the value of trigonometric functions on certain angles, notice that you can also calculate the tangent by ignoring dividing 2 this way:

Take the 0 and the 4 of the fist column (0 radians), divide and take the root, you have: $tan (0)= \sqrt(\dfrac{0}{4})=0$

For the second column we have $\dfrac{\pi}{6}$, take the 1 and the 3, divide and take the root again: $tan (\dfrac{\pi}{6})=\sqrt(\dfrac{1}{3})=\dfrac{\sqrt(1)}{\sqrt(3)}=\dfrac{1}{\sqrt 3}$

Similarily, $tan(\dfrac{\pi}{4})=\sqrt(\dfrac{2}{2})=\sqrt(1)=1$

$tan(\dfrac{\pi}{3})=\sqrt(\dfrac{3}{1})=\sqrt 3$

Finally, take for the 4th column ($\dfrac{\pi}{2}$) 4 and $0$. $tan(\dfrac{\pi}{2})=\sqrt(\dfrac{4}{0})!$ wich is in agreement of the fact that $tan$ is not defined on $\dfrac{\pi}{2}$ and thus cannot be.