Trigonometry (5)

It is given that:

$\begin{aligned} \sin x + \sin^2 x & = 1 \\ \sin x & = 1 - \sin^2 x \\ \implies \sin x & = \cos^2 x \end{aligned}$

Therefore,

$\begin{aligned} X & = \cos^8 x + 2\cos^6 x + \cos^4 x & \small \color{#3D99F6} \text{As }\sin x = \cos^2 x \\ & = \sin^4 x + 2\sin^3 x + \sin^2 x \\ & = \left({\color{#3D99F6}\sin^2 x + \sin x} \right)^2 & \small \color{#3D99F6} \text{Given that }\sin x + \sin^2 x = 1 \\ & = \left({\color{#3D99F6}1} \right)^2 = \boxed{1} \end{aligned}$

The given expression can be written as

$\cos^{4}(x)(\cos^{4}(x) + 2\cos^{2}(x) + 1) = (\cos^{2}(x)(\cos^{2}(x) + 1))^{2}$ , (A).

Now we are given that $\sin(x) = 1 - \sin^{2}(x) = \cos^{2}(x)$ , so (A) can be written as

$(\sin(x)(\sin(x) + 1))^{2} = (\sin^{2}(x) + \sin(x))^{2} = 1^{2} = \boxed{1}$ .