## Irrational Cosine

#### Problem

Prove that $\cos(1^o)$ is irrational.

#### Solution

Using Euler's formula:

$$\cos(5x) + i \sin(5x) = e^{5xi} = (e^{xi})^5 = (cos(x) + i\sin(x))^5$$

Let $c = \cos(x)$ and $s = \sin(x)$.

\begin{align}\therefore \cos(5x) + i\sin(5x) &= (c + si)^5\\&= c^5 + 5c^4si - 10c^3s^2 - 10c^2s^3i + 5cs^4 + s^5i\end{align}

Equating real coefficients: $\cos(5x) = c^5 - 10c^3s^2 + 5cs^4$.

Using the identity $s^2 = 1 - c^2$ throughout:

\begin{align}\cos(5x) &= c^5 - 10c^3(1 - c^2) + 5c(1 - c^2)^2\\&= 11c^5 - 10c^3 + 5c(1 - 2c^2 + c^4)\\&= 16c^4 - 20c^3 + 5c\end{align}

In the same way we can show that $\cos(3x) = 4c^3 - 3c$.

Using the identity $\cos(A - B) = \cos(A) \cos(B) + \sin(A) \sin(B)$:

\begin{align}\cos(15) &= \cos(45 - 30)\\&= \cos(45)\cos(30) + \sin(45)\sin(30)\\&= \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2}\\&= \dfrac{\sqrt{3} + 1}{2\sqrt{2}}\end{align}

So clearly $\cos(15)$ is irrational.

Using $\cos(5x) = 16c^5 - 20c^3 + 5c$:

$$\cos(15) = 16\cos^5(3) - 20\cos^3(3) + 5\cos(3) = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}$$

Now if $\cos(3)$ were rational then so too would $\cos(15)$. But as we know that $\cos(15)$ is irrational it follows that $\cos(3)$ is also irrational.

In the same way by using $\cos(3x) = 4c^3 - 3c$:

$$\cos(3) = 4\cos^3(1) - 3\cos(1)$$

And as $\cos(3)$ is irrational it follows that $\cos(1)$ is indeed irrational. Q.E.D.

Problem ID: 280 (13 May 2006)     Difficulty: 4 Star

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