MHB Prove cos (π/100) is irrational

[anemone]

Prove that $\cos \dfrac{\pi}{100}$ is irrational.

 

[Dethrone]

Here is a really cheap proof, probably not what you're looking for, but something quick before I resume my homework :p

Spoiler

To prove that it is irrational, I will show that $\cos \dfrac{\pi}{100}$ can be expressed as a sum of two addends, one of which is irrational. It can be proven that if an addend is irrational, then the sum is also irrational. I will use the identity $\cos\left({A-B}\right)=2\cos\left({A}\right)\cos\left({B}\right)+(-\cos\left({A+B}\right))$.
Solving the two equations $A-B=\frac{\pi}{100}$ and $A+B=\frac{\pi}{4}$, then:$$\cos\left({\frac{\pi}{100}}\right)=\cos\left({\frac{13\pi}{100}-\frac{3\pi}{25}}\right)=2\cos\left({\frac{13\pi}{100}}\right)\cos\left({\frac{3\pi}{25}}\right)+(-\cos\left({\frac{\pi}{4}}\right))$$

Since $\cos\left({\frac{\pi}{4}}\right)$ is irrational (also can be proven), $\cos \dfrac{\pi}{100}$ is too.

 

[anemone]

Here is a really cheap proof, probably not what you're looking for, but something quick before I resume my homework :p

Spoiler

To prove that it is irrational, I will show that $\cos \dfrac{\pi}{100}$ can be expressed as a sum of two addends, one of which is irrational. It can be proven that if an addend is irrational, then the sum is also irrational. I will use the identity $\cos\left({A-B}\right)=2\cos\left({A}\right)\cos\left({B}\right)+(-\cos\left({A+B}\right))$.
Solving the two equations $A-B=\frac{\pi}{100}$ and $A+B=\frac{\pi}{4}$, then:$$\cos\left({\frac{\pi}{100}}\right)=\cos\left({\frac{13\pi}{100}-\frac{3\pi}{25}}\right)=2\cos\left({\frac{13\pi}{100}}\right)\cos\left({\frac{3\pi}{25}}\right)+(-\cos\left({\frac{\pi}{4}}\right))$$

Since $\cos\left({\frac{\pi}{4}}\right)$ is irrational (also can be proven), $\cos \dfrac{\pi}{100}$ is too.

This is exactly the elegant proof that I am looking for! Well done, Rido12, and thanks for participating!

 

[kaliprasad]

Here is a really cheap proof, probably not what you're looking for, but something quick before I resume my homework :p

Spoiler

To prove that it is irrational, I will show that $\cos \dfrac{\pi}{100}$ can be expressed as a sum of two addends, one of which is irrational. It can be proven that if an addend is irrational, then the sum is also irrational. I will use the identity $\cos\left({A-B}\right)=2\cos\left({A}\right)\cos\left({B}\right)+(-\cos\left({A+B}\right))$.
Solving the two equations $A-B=\frac{\pi}{100}$ and $A+B=\frac{\pi}{4}$, then:$$\cos\left({\frac{\pi}{100}}\right)=\cos\left({\frac{13\pi}{100}-\frac{3\pi}{25}}\right)=2\cos\left({\frac{13\pi}{100}}\right)\cos\left({\frac{3\pi}{25}}\right)+(-\cos\left({\frac{\pi}{4}}\right))$$

Since $\cos\left({\frac{\pi}{4}}\right)$ is irrational (also can be proven), $\cos \dfrac{\pi}{100}$ is too.

I cannot agree on this.

Spoiler

We know that $\cos\left({\frac{\pi}{4}}\right)$ is irrational but we do not know

$2\cos\left({\frac{13\pi}{100}}\right)\cos\left({\frac{3\pi}{25}}\right)$

may be this is irrational and the irrational part cancells the irrational part of $\cos\left({\frac{\pi}{4}}\right)$

we need to prove that above is not true

 

[kaliprasad]

Here I provide solution

Spoiler

as $\cos 2x =2\cos^2x-1$

so if
$\cos\, x$ is rational then $\cos 2x$ is rational

now

$\cos (n+1) x + \cos(n-1)x =2\cos\,nx\cos \, x$

hence
$\cos (n+1) x =2\cos\,nx\cos \ x- \cos(n-1)x ..\cdots(1)$

from (1) if $\cos\,x$ , $\cos\,nx$ , $\cos(n-1)x$ are rational then $\cos (n+1) x$ is rational

we have shown that if $\cos\,x$ is rational then
$\cos\,2x$ is rational and from (1) using n = 2 we can show that $\cos\,3x$ is rational so on
$\cos\,nx$ is rational for all n

now

$x=\dfrac{\pi}{100}$
letting n= 25

we get $\cos\dfrac{\pi}{4}$ is rational which is a contradiction as $\cos\dfrac{\pi}{4}$ being $\dfrac{1}{\sqrt{2}}$ is not

so $\cos\dfrac{\pi}{100}$ is irrational

 

[anemone]

I'm disappointed in myself and I apologize to the community for my temporary lack of attention and I apologize too for making the inappropriate statement. :(

However, here is a solution of other that I want to share with MHB:

Spoiler

Let $a=\cos \left(\dfrac{\pi}{100}\right)$ and $b=\sin \left(\dfrac{\pi}{100}\right)$.

Then from the Pythagorean identity, we have

$a^2+b^2=1$

$b^2=1-a^2$

Using De Moivre's Theorem and the Binomial Theorem, we have

$$
\begin{align*}\cos \left(\dfrac{\pi}{4}\right)&=\text{Re}(a+ib)^{25}\\&=\text{Re}\sum_{k=0}^{25}{25 \choose k}i^kb^ka^{25-k}\\&=\sum_{m=0}^{12}{25 \choose 2m}(-1)^m(1-a^2)^ma^{25-2m}\\&\end{align*}$$

If $a$ were rational, then each term on the RHS would be rational as well, contradicting the fact that $\cos \left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$ is irrational.

Therefore $\cos \left(\dfrac{\pi}{100}\right)$ is irrational.