Entangled sequences

It's easy to prove, that $a_n=\cos36n$ , $b_n=\sin36n$ . That explains the recurring cycle of 10 steps, since $10\times 36=360$ .

Now, because 2018= $201\times10$ +8, $a_{2018}=a_8=-\cos72=0.309$

Similar solution with @Emil Gueorguiev 's

We find that $a_n = \cos \frac {n\pi}5$ and $b_n = \sin \frac {n \pi}5$ . Let us prove the claim by induction that it is true for all $n \ge 1$ .

Proof: For $n=1$ , we find that $a_1 = \cos \frac \pi 5$ and $b_1 = \sin \frac \pi 5$ (see note below). Therefore the claim is true for $n=1$ . Now assuming that the claim is true for $n$ , then:

$\begin{cases} a_{n+1} = a_na_1 - b_nb_1 = \cos \frac {n\pi}5\cos \frac \pi 5 - \sin \frac {n\pi}5\sin \frac \pi 5 = \cos \frac {(n+1)\pi}5 \\ b_{n+1} = b_na_1 + a_nb_1 = \sin \frac {n\pi}5\cos \frac \pi 5 + \cos \frac {n\pi}5\sin \frac \pi 5 = \sin \frac {(n+1)\pi}5 \end{cases}$

Therefore the claim is also true for $n+1$ and hence true for all $n \ge 1$ . And we have $a_{2018} = \cos \frac {2018\pi}5 = \cos \frac {8\pi}5 \approx \boxed{0.309}$ .

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Note: Considering the following identity:

$\begin{aligned} \cos \frac \pi 5 + \color{#3D99F6} \cos \frac {3\pi}5 & = \frac 12 & \small \color{#3D99F6} \text{As }\cos (\pi - \theta) = - \cos \theta \\ \cos \frac \pi 5 - \color{#3D99F6} \cos \frac {2\pi}5 & = \frac 12 & \small \color{#3D99F6} \text{Also }\cos (2x\theta) = 2\cos^2 \theta - 1 \\ \cos \frac \pi 5 - \color{#3D99F6} 2 \cos^2 \frac \pi 5 + 1 & = \frac 12 \\ 2 \cos^2 \frac \pi 5 - \cos \frac \pi 5 - \frac 12 & = 0 & \small \color{#3D99F6} \text{Solving the quadratic} \end{aligned}$

$\begin{aligned} \implies \cos \frac \pi 5 & = \frac {1+\sqrt 5}4 \\ \implies \sin \frac \pi 5 & = \sqrt{1-\left(\frac {1+\sqrt 5}4\right)} = \sqrt{\frac {5+\sqrt 5}8} \end{aligned}$