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I have been trying to express a power of $\cos(\frac {2\pi}{5})$ as a "rational number", or trying to find a "rational number" that results from a linear combination of powers of $\cos(\frac {2\pi}{5})$. Meaning $\frac {p}{q} = [\cos (\frac {2\pi}{5})]^n$ or $\frac {p}{q} = a_1[\cos (\frac {2\pi}{5})]^n + \cdots + a_n[\cos (\frac {2\pi}{5})]^m $.

I have seen in other posts (example: Find exact value of $\cos (\frac{2\pi}{5})$ using complex numbers. ) that this is possible by using the fifth roots of unity. So I would suppose that it is not possible without the aid of complex numbers? If that is or isn't the case, I do not have an idea about how to argument this.

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By well-known trigonometric formulas and a few transformations

$$\cos(5t)=\cos(4t)\cos(t)-\sin(4t)\sin(t)\\ =(2\cos^2(2t)-1)\cos(t)-2\sin(2t)\cos(2t)\sin(t)\\ =(2(2\cos^2(t)-1)^2-1)\cos(t)-4\sin(t)\cos(t)(2\cos^2(t)-1)\sin(t)\\ =16\cos^5(t)- 20\cos^3(t)+ 5\cos(t)$$

Then,

$$\cos\left(5\frac{2\pi}5\right)=16\cos^5\left(\frac{2\pi}5\right)-20\cos^3\left(\frac{2\pi}5\right)+\cos\left(\frac{2\pi}5\right)=1$$ is a rational expression.

We can solve further

$$\cos(5t)=1,$$ or

$$16x^5-20x^3+5x-1=0.$$

We have the obvious root $x=1$ and by synthetic division

$$16x^5-20x^3+5x-1=(x-1)(16x^4+16x^3-4x^2-4x+1).$$

As the four other solutions of the trigonometric equation go in pairs, the second factor must be a perfect square.

The leading term of the squared polynomial is obviously $4x^2$, and by $$(4x^2+ax+b)^2=16x^4+8ax^3+(a^2+8b)x^2+2abx+b^2$$ we get $a=2,b=-1$ and the factor is

$$(4x^2+2x-1)^2.$$

Your rational number can be

$$4\cos^2\left(\frac{2\pi}5\right)+2\cos\left(\frac{2\pi}5\right)=1.$$

Finally, the roots of the polynomial are

$$\frac{-1\pm\sqrt{5}}4$$ and we choose the positive value.