Tri-gono-metry

Let $f(n) = {\sin}^n \theta + {\cos}^n \theta$ .
So, lets derive a general formula for $\dfrac{f(n) - f(n + 2)}{f(n - 2)}$ which is what the answer we want is like.
$\begin{aligned} \dfrac{f(n) - f(n + 2)}{f(n - 2)} & = \dfrac{{\sin}^n \theta + {\cos}^n \theta - {\sin}^{n+2} \theta - {\cos}^{n+2} \theta}{{\sin}^{n-2} \theta + {\cos}^{n-2} \theta}\\ \\ & = \dfrac{{\sin}^n \theta\left(1 - {\sin}^2 \theta\right) + {\cos}^n \theta\left(1 - {\cos}^2 \theta\right)}{{\sin}^{n-2} \theta + {\cos}^{n-2} \theta}\\ \\ & = \dfrac{{\sin}^n \theta {\cos}^2 \theta + {\cos}^n \theta {\sin}^2 \theta}{{\sin}^{n-2} \theta + {\cos}^{n-2} \theta}\\ \\ & = \dfrac{{\sin}^2 \theta {\cos}^2 \theta\left({\sin}^{n-2} \theta + {\cos}^{n-2} \theta\right)}{{\sin}^{n-2} \theta + {\cos}^{n-2} \theta}\\ \\ & = \color{#20A900}{\boxed{{\sin}^2 \theta {\cos}^2 \theta}} \end{aligned}$

So, putting the value of the angle (i.e. ${45}^°$ ) we get:-
$\begin{aligned} \dfrac{f(2016) - f(2018)}{f(2014)} & = {\sin}^2 {45}^° {\cos}^2 {45}^°\\ \\ & = \dfrac{1}{4}\\ \\ & = \color{#3D99F6}{\boxed{0.25}} \end{aligned}$

$\quad f(x) = \left(\dfrac{1}{\sqrt2}\right)^{x}+\left(\dfrac{1}{\sqrt2}\right)^{x} \implies 2^{1-\frac{x}{2}} \\ \quad \dfrac{f(2016)-f(2018)}{f(2014)} = \dfrac{2^{1-1008} - 2^{1-1009}}{2^{1-1007}} = 2^{-1}-2^{-2} \\ \quad \text{Our answer : } \dfrac{1}{4} = \boxed{0.25}$