Infinite fraction, square root or trig function?

Well, we need to check every option to conclude our answer. So, get ready for some bashing. :P And we'll also use the quadratic formula to solve any quadratic equations we encounter.

Option 1: Continued Fraction Option

Take the fraction as $y$ . We can formulate the following equation then:

$y=\frac{1}{1+y}\implies y^2+y-1=0\implies y=\frac{-1\pm \sqrt{5}}{2}$

Since the fraction cannot have a negative value, we will reject the negative solution of $y$ . So, we have $y=\dfrac{\sqrt{5}-1}{2}$ .

Option 2: Iterated square root option

Take the expression as $z$ . We can formulate the following equation then:

$z^2=1+z\implies z^2-z-1=0\implies z=\frac{1\pm \sqrt{5}}{2}$

Just as before, the iterated square root expression here cannot have a negative value, so we will reject the negative solution of $z$ . So, we have $z=\dfrac{1+\sqrt{5}}{2}$ .

Option 3: Trigonometric option

Let $x=\sin 18^{\circ}$ . Now,

$\sin 72^{\circ}=2\sin 36^{\circ}\cos 36^{\circ}\\ \implies \cos 18^{\circ}=4\sin 18^{\circ}\cos 18^{\circ}(1-2\sin^2 18^{\circ})\\ \implies 1=4\sin 18^{\circ}(1-2\sin^2 18^{\circ})\\ \implies 1=4x(1-2x^2)\\ \implies 1=4x-8x^3\\ \implies 8x^3-4x+1=0\\ \implies (2x-1)(4x^2+2x-1)=0\\ \implies 4x^2+2x-1=0\\ \implies x=\frac{-2\pm \sqrt{20}}{8}=\frac{-1\pm\sqrt{5}}{4}$

Here too, we cannot have a negative solution for $x$ since $\sin {18}^{\circ}$ is the sine of a first quadrant angle and is clearly $\gt 0$ . So, negative solution of $x$ is to be rejected.

Thus, $x=\dfrac{\sqrt{5}-1}{4}$ . Using this, we can easily compute the required value as,

$2\cos 36^{\circ}=2(1-2\sin^2 18^{\circ})=2(1-2x^2)=\frac{1+\sqrt{5}}{2}$

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So, we can see that both option $2$ and option $3$ are equal, so the correct answer is None of These.