# I've noticed some relationships with cosine and square root.

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...

$2 \times \cos 90^\circ = \sqrt 0$

$2 \times \cos 60^\circ = \sqrt 1$

$2 \times \cos 45^\circ = \sqrt 2$

$2 \times \cos 30^\circ = \sqrt 3$

$2 \times \cos 0^\circ = \sqrt 4$

Is there anything interesting to know about these?

Thanks

• Well, it doesn't look silly at all. It's a well known mnemonic of some important values of $\cos$ and $\sin$ as well. Jun 24, 2017 at 14:29
• You might notice that the pattern is not as regular as it first seems. The angle difference between $\sqrt 0$ and $\sqrt 1$ and between $\sqrt 3$ and $\sqrt 4$ is $30^\circ$, but the angle differences between $\sqrt 1, \sqrt 2,\sqrt 3$ are $15^\circ$. Jun 24, 2017 at 16:50
• @PaulSinclair You can see here the graph of the function $f(x) = (sin(x)*2)**2) i.imgur.com/5qrACzH.png Jun 24, 2017 at 22:16 • Jun 24, 2017 at 22:33 • @probiner - I'm sorry, but what was the point of posting those pictures? I can assure that I am quite well aware of what the graphs of trig function (and their squares) look like. Jun 25, 2017 at 17:16 ## 3 Answers Your list is sometimes given in trig books as a mnemonic device. Looking a little deeper the pattern that the angles in a regular$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)\$

• Did you end your answer? Looks cut. Someone mentioned this to me: purplemath.com/modules/specang.htm Jun 24, 2017 at 15:01
• @probiner thanks for pointing out the dangling conjunction. I was thinking about mentioning DeMoivre's Theorem but got interrupted. Jun 24, 2017 at 15:40

This rules we used in high school with fingers do apply for both sine and cosine see the pictures  • Cheers for the memorabilia :) Jun 30, 2017 at 14:31

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.