# Let $\theta=\frac{2\pi}{5}$. Show $2\cos(2\theta)+2\cos(\theta)+1=0$.

I have been at this for a while. Any ideas?

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What have you learnt in school? Complex numbers? Or just trigo? –  BlackAdder Feb 17 '14 at 1:55
I know a bit of both, I have subbed in trig identities and tried to use Euler but just seem to be traveling in a circle. –  Recip Feb 17 '14 at 1:56

Hint: Consider the roots of unity of $z^5=1$.

So we have $1+z+z^2+z^3+z^4=0.$ Now using $z+z^{-1}=2\Re(z)$, see if you can get the answer.

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I don't think that's even remotely the way he's supposed to solve it in trig class. That fancy 'R' is scary. –  Shahar Feb 17 '14 at 2:07
I asked if he knew about complex numbers, and he replied in the affirmitive. In any case, it is just a different way to approach the question, a chance to broaden one's horizons. :) –  BlackAdder Feb 17 '14 at 2:08
Knowing that complex numbers exist, and having an idea about how to deal with them, are two very different things. –  Arkamis Feb 17 '14 at 3:21

Create the following diagram:

Prove that the angle $\angle BAB'= 36^\circ$. Then show that: $$\cos 2\theta = \cos \pi/5 = \frac{1+\sqrt{5}}{4}$$ And by the double angle identity: $$\cos \theta = \cos 2\pi/5 = 2\left(\frac{1+\sqrt{5}}{4}\right)^2-1$$

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So 4cos(x/2)cos(3x/2)+1=0? –  Recip Feb 17 '14 at 2:03
@Recip No, there's no $x$ in your original problem. –  Shahar Feb 17 '14 at 2:04
Sorry I meant theta. –  Recip Feb 17 '14 at 2:06

I think this is a simple way. By using De Moivre’s formula, we obtain the fifth time angle formula: $$\cos(5\theta)=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$ Set $\theta=2\pi/5$, we have: $$\cos(2\pi)=16\cos^5\theta-20\cos^3\theta+5\cos\theta\\ 16\cos^5\theta-20\cos^3\theta+5\cos\theta-1=0$$ By using factorization, we have: $$(\cos\theta-1)(4\cos^2\theta+2\cos\theta-1)^2=0$$ Obviously, $\cos\theta-1\neq0$, then the equality have been prove.

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Do you see that your left-hand side is the same as

$$\cos(2\theta)+\cos(-2\theta)+\cos(\theta)+\cos(-\theta)+\cos(0)$$

Consider five unit rays emanating from the origin at these angles: $0, \theta, 2\theta, -2\theta, -\theta$ (with $\theta=\frac{2\pi}{5}$). Consider these five rays to be force vectors, pulling away from the origin like some complicated tug-of-war in five equally spaced directions around the origin. Does any one direction win the tug-of-war? No, because there is symmetry in the pulling, and if any one direction was the winner, then there would be at least 5 total directions that win. So these force vectors must sum to the zero vector.

Now the $x$-components of the five vectors are the five terms in the sum above. And the $x$-component of what they all add up to is the $x$-component of the zero vector. This equality is what you set out to prove.

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Hint: Show that for every $\theta$ such that $\theta \not= 2k\pi$ $\left(k\in \mathbb{Z}\right)$ we have: $$1 + 2\cos (\theta) + 2\cos (2\theta) = \frac{\sin\left(\frac{5}{2}\theta \right)}{\sin\left(\frac{1}{2}\theta \right)}.$$

The trigonometry based proof is here. It hinges on $$\frac{1}{2}\cos(x)\sin(y) = \sin(x+y) - \sin(x-y)$$ and telescoping sum.

Its generalization and proofs are here: Dirichlet's kernel.

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$2×\cos(2×{{2\pi}\over 5})=-{{1+\sqrt 5}\over 2}$
$2×\cos{{2\pi}\over 5}={{-1+\sqrt 5}\over 2}$
$-{{1+\sqrt 5}\over 2}+{{-1+\sqrt 5}\over 2}+1=0$

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If $\xi=e^{i2\pi/5}$, then $0=\xi^5-1=(\xi-1)(\xi^4+\xi^3+\xi^2+\xi+1)$. Therefore, $$\xi^4+\xi^3+\xi^2+\xi+1=0\tag{1}$$ Since $2\cos(2\pi/5)=\xi+\xi^4$ and $2\cos(4\pi/5)=\xi^2+\xi^3$, $(1)$ says $$2\cos(4\pi/5)+2\cos(2\pi/5)+1=0\tag{2}$$

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If $\displaystyle5\theta=2\pi, 3\theta=2\pi-2\theta$

$\displaystyle\implies\cos3\theta=\cos(2\pi-2\theta)=\cos2\theta$

Using Triple and Double angle formulas,

$$4c^3-3c=2c^2-1\iff 4c^3-2c^2-3c+1=0\ \ \ \ (1)$$

where $c=\cos\theta$

Again, $\displaystyle\cos3\theta=\cos2\theta$

$\displaystyle\implies3\theta=2n\pi\pm2\theta$ where $n$ is any integer

Taking the '+' sign, $\displaystyle\implies3\theta=2n\pi+2\theta\iff\theta=2n\pi\implies\cos\theta=\cos2n\pi=1$

Taking the '-' sign, $\displaystyle\implies3\theta=2n\pi-2\theta\iff\theta=\frac{2n\pi}5$ where $n\equiv0,1,2,3,4\pmod5$

Again as $\displaystyle\cos(2\pi-y)=\cos y,\cos\frac{2\pi}5=\cos\left(2\pi-\frac{2\pi}5\right)=\cos\frac{8\pi}5$ and $\displaystyle\cos\frac{4\pi}5=\cos\frac{6\pi}5$

Clearly, $c-1$ is a factor of $(1)$ and $\displaystyle\cos\frac{2\pi}5,\cos\frac{4\pi}5\ne1$

So, $\displaystyle\cos\frac{2\pi}5=\cos\frac{8\pi}5$ and $\displaystyle\cos\frac{4\pi}5=\cos\frac{6\pi}5$ are the roots of $$\frac{4c^3-2c^2-3c+1}{c-1}=0\iff 4c^2+2c-1=0$$

Again use Double formula $\displaystyle\cos2\theta=2\cos^2\theta-1\implies c^2=\cos^2\theta=\frac{1+\cos2\theta}2$

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