Find the exact value of $\cos\frac{2\pi}{5}$ by solving equation There's this concept in the topic of complex numbers which I don't really understand much (and was devastated upon realising it'll appear in the topic often) - trigonometry!
I'm super lost to be honest.
We are asked to express $\cos3\theta$ and $\cos2\theta$ in terms of $\cos\theta$. 
Then show that $\cos3\theta=\cos2\theta$ can be written as $4z^3-2z^2-3z+1=0$ where $z=\cos\theta$.
Lastly, by solving the equation above for $z$, we are to find out the value of $\cos\frac{2\pi}{5}$.
I attempted the question with just subbing in $z=\cos\theta$ into the equation and I got: $$4\cos^3\theta-3\cos\theta=\cos2\theta$$ and don't really know where to go next.
 A: try this:
$\cos(2\theta)=\cos(\theta + \theta)$ then addition formula:
$=\cos^2(\theta)-\sin^2(\theta)$ 
use the same for $\cos(3\theta)=\cos(2\theta+\theta)$
A: Since $cos(\pi + x) = cos(\pi -x)$, we have $cos (\pi +\pi/5) = cos (\pi - \pi/5)$, that is $cos(6\pi/5) = cos (4\pi/5)$ or   
$cos(3\theta) = cos (2 \theta)$ for given $\theta \quad(=\frac{2\pi}{5}) \qquad$ ....$(1)$
Using $cos(3\theta) = 4cos^3(\theta) - 3 cos(\theta)$ and $cos (2\theta) = 2cos^2(\theta) - 1$,
$4z^3 - 2z^2-3z+1 =0$ , where $\qquad z=cos(\theta)$  
Now, $z=0$ is a root (you could easily see that as $\theta =0$ satisfies equation $(1)$).
So, $(z-1)(4z^2+2z-1)=0$, and you know $cos(2\pi/5) \neq 1$, so solve the remaining quadratic equation $(4z^2+2z-1)=0$ to get the positive root as the required solution.
A: Here I use the formulas
$$\cos(x+y) = \cos x \cos y - \sin x \sin y$$
$$\sin(x+y) = \sin x \cos y + \cos x \sin y.$$
So $$\cos 2\theta = \cos(\theta+\theta) = \cos^2 \theta -\sin^2 \theta = 2\cos^2 \theta - 1.$$
Similarly
\begin{align}
\cos 3\theta = \cos (2\theta+\theta) & = \cos 2\theta\cos \theta-\sin 2\theta \sin \theta \\
& = 2\cos^3 \theta-\cos \theta-2\sin^2 \theta \cos \theta \\
& = 4\cos^3 \theta -3\cos \theta.
\end{align}
Now if $\cos 3\theta = \cos 2\theta$ you get
$$4\cos^3 \theta - 2\cos^2 \theta - 3\cos \theta + 1 = 0$$
which is what you claim putting $z = \cos \theta$. It's easy to see that $z = 1$ is a solution, so 
$$4z^3-2z^2-3z+1 = (z-1)(4z^2+2z-1) = \bigg(z-1\bigg)\bigg(z+\frac{1+\sqrt{5}}{4}\bigg)\bigg(z-\frac{-1+\sqrt{5}}{4}\bigg). $$
Now $\theta = \frac{2\pi}{5}$ satisfies $\cos 3\theta = \cos 2\theta$, and $\cos \frac{2\pi}{5}$ must be positive (because $\frac{2\pi}{5} < \frac{\pi}{2}$) and different from $1$ (because $\frac{2\pi}{5} \neq 0$). Hence
$$\cos \frac{2\pi}{5} = \frac{\sqrt{5}-1}{4}.$$
A: $$\begin{align}
%
\cos(3\theta) - \cos(2\theta) &=
%
\mathcal R \left\{e^{3i\theta}\right\} - \left\{e^{2i\theta}\right\} \\ \\ &=
%
\mathcal R \left\{\left(e^{i\theta}\right)^3 - \left(e^{i\theta}\right)^2\right\} \\ \\ &=
%
\mathcal R \left\{\left(\cos(\theta) + \rm i \sin(\theta) \right)^3 - \left(\cos(\theta) + \rm i \sin(\theta) \right)^2\right\} \\ \\ &=
%
\mathcal R \left\{\begin{array} {ll}
& \left(\cos(\theta)^3 + 3~\rm i \cos(\theta)^2\sin(\theta) -
  3~\rm \cos(\theta)\sin(\theta)^2 - \rm i \sin(\theta)^2 \right)
\\ - & \left(\cos(\theta)^2 + 2~\rm i \cos(\theta) \sin(\theta) - \sin(\theta)^2\right)
\end{array}\right\} 
\\ \\ &=
%
\cos(\theta)^3 - 3\cos(\theta)\sin(\theta)^2 - \cos(\theta)^2 + \sin(\theta)^2 \\ \\ &=
%
z^3 - 3z(1 - z^2) - z^2 + 1 - z^2 \\ \\ &=
%
4z^3 - 2z^2 - 3z + 1
\end{align}$$
A: The trick here is to find $2$ representations of the complex exponential $$e^{ix} = \cos(x)+i\sin(x).$$ 
With this in mind, we have 
$$e^{2ix} = \cos(2x)+i\sin(2x)$$ and 
$$e^{3ix} = \cos(3x)+i\sin(3x) $$ along with 
$$(e^{ix})^2 = (\cos(x)+i\sin(x))^2 = (\cos(x)^2 - \sin(x)^2) + i(2\sin(x)\cos(x))$$ and 
$$(e^{ix})^3 = (\cos(x)+i\sin(x))^3 = (\cos(x)^3 -3\cos(x)\sin(x)^2) + i(3\cos(x)^2\sin(x) - sin(x)^3).$$
Equating the real part of $e^{2ix}$ yields 
$$\cos(2x) = \cos(x)^2 - \sin(x)^2.$$
Using the Pythagorean identity, we see that $\sin(x)^2 = 1-\cos(x)^2$ and so we have
$$\cos(2x) = \cos(x)^2 - \sin(x)^2 = \cos(x)^2 - (1-\cos(x)^2) = 2\cos(x)^2-1.$$
Similarly, we may equate the real part of $e^{3ix}$ to get 
$$\cos(3x) = \cos(x)^3 -3\cos(x)\sin(x)^2 = \cos(x)^3 - 3\cos(x)(1-\cos(x)^2) = 4\cos(x)^3 - 3\cos(x).$$
So, if we set $$\cos(2z) = \cos(3z)$$
then we have 
$$2\cos(z)^2-1 = 4\cos(z)^3 - 3\cos(z)$$ which implies 
$$4\cos(z)^3 - 2\cos(z)^2 -3 \cos(z) +1 =0, \text{ as desired}.$$
A: We have 
\begin{eqnarray*}
\cos( 2 \theta) =2 \cos^2(\theta)-1 \\
\cos( 4 \theta) =4 \cos^3(\theta)-3 \cos(  \theta)
\end{eqnarray*}
Let $z=\cos( \theta) $ then $\cos(3 \theta) = \cos(2 \theta)$ becomes $4z^3-2z^2-3z+1=0$. This will factorise $(z-1)(4z^2+2z-1)=0$, so $\color{red}{\cos(\frac{2 \pi}{5} )=\frac{-1+\sqrt{5}}{4}}$.
