Is it possible to express cos(40) as a radical?

[neilparker62]

TL;DR Summary

Trying to find an expression for cos(40 degrees) (using any combination of radicals)

What sort of number is cos(40) ? You can solve the equation:

$$ 4 \cos^{3}40 -3\cos40+0,5=0 $$

but you end up with complex numbers requiring a cube root. The polar angle gets divided by 3 and you end up needing cos(20) or cos(10) in your answer. No way (it seems) to express as a radical or combination of radicals in any form .

The solution from WA is interesting. There are 3 solutions - as a 'bonus' for solving for cos(40), you also get answers for cos(20) and cos(80).

 

[willem2]

cos(40) is the real part of z where z^9 is 1.
You can factor the polynomial Z^9 - 1. to get (z^6+z^3+1) (z^3-1) (see cyclotomic polynomial).
(z^6+z^3+1) is easy to solve, and will get you (cos(40)+sin(40)i) and (cos(80)+sin(80)i)

 

[neilparker62]

cos(40) is the real part of z where z^9 is 1.
You can factor the polynomial Z^9 - 1. to get (z^6+z^3+1) (z^3-1) (see cyclotomic polynomial).
(z^6+z^3+1) is easy to solve, and will get you (cos(40)+sin(40)i) and (cos(80)+sin(80)i)

I was hoping to find an expression for cos(40) in terms of radicals but it seems that won't be possible. I did find a wiki page on which angles could be expressed that way. Multiples of 10 don't seem to be on the list.

 

[pbuk]

I was hoping to find an expression for cos(40) in terms of radicals but it seems that won't be possible. I did find a wiki page on which angles could be expressed that way. Multiples of 10 don't seem to be on the list.

There are some multiples of 10° on the list, why do you think this is?

Spoiler

Degrees were invented by humans

Spoiler

$180° \equiv \pi$

 

[Vanadium 50]

I was hoping to find an expression for cos(40) in terms of radicals but it seems that won't be possible.

It is not. This would be true iff a regular nonagon is constructable. It isn't, so it isn't.

See
https://www.physicsforums.com/threads/exact-expression-for-sine-of-1-degree.993621/
https://www.physicsforums.com/threa...es-to-aops-for-prealgebra-precalculus.986266/