If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now, 
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^2 x$$
$$=\sin^4 x + \sin^3 x + \sin^3 x + \sin^2 x$$
$$=\sin^3 x(\sin x +1) +\sin^2 x(\sin x +1)$$
$$=(\sin x +1) (\sin^3 x +\sin^2 x)$$
How do I proceed further?
 A: Almost finished from there: 
$$(\sin x + 1)(\sin^3 x + \sin^2 x)\\ = (\sin x + 1)(\sin x + 1) \sin^2 x\\ = (\sin^2 x + \sin x) (\sin^2 x + \sin x) \\= 1$$
A: Here is a shorter way to do it. Your condition can be rewritten in the forms
$$\sin x = \cos^2(x) \implies \sin^n(x) = \cos^{2n}(x) \implies \sin^n(x)+\sin^{n-1}(x) = \sin^{n-2}(x)$$
We thus rewrite your cosines as
$$\begin{align}
\cos^8 x + 2\cos^6 x + \cos^4 x &= \sin^4(x)+2\sin^3(x)+\sin^2(x)\\ &= [\sin^4(x)+\sin^3(x)]+[\sin^3(x)+\sin^2(x)]\\ &= \sin^2(x)+\sin(x)\\&=1
\end{align}
$$
A: Even shorter:
You know $\sin x = \cos^2 x$.  Then
$\cos^8 x + 2 \cos^6 x + \cos^4 x = \sin^4 x + 2 \sin^3 x + \sin^2 x = (\sin^2 x +\sin x)^2 = 1^2 = 1$.
A: $$
\cos^8 x + 2\cos^6 x + \cos^4 x = (\cos^2 x(1+\cos^2 x))^2 
=(\sin x(1+\sin x))^2 = 1
$$
A: Hint:
$$\cos^8x+2\cos^6x+\cos^4x=(\cos^4x+\cos^2x)^2$$
Now as $\cos^2x=\sin x,\cos^4x=(\cos^2x)^2=?$
A: \begin{align}
& \cos^8 x + 2\cos^6 x + \cos^4 x \\[8pt]
= {} & \sin^4 x + 2\sin^3 x + \sin^2 x \\[8pt]
= {} & (\sin^2 x)^2 + 2(\sin^2 x)(\sin x) + \sin^2 x \\[8pt]
= {} & (1-\cos^2 x)^2 + 2(1-\cos^2 x)(\sin x) + (1-\cos^2 x) \\[8pt]
= {} & (1-\sin x)^2 + 2(1 - \sin x)(\sin x) + (1-\sin x) \\[8pt]
= {} & (1-2\sin x + \sin^2 x) + 2(\sin x - \sin^2 x) + (1-\sin x) \\[8pt]
= {} & 2 - \sin x -\sin^2 x \\[8pt]
= {} & 2 - \sin x -(1-\cos^2 x) \\[8pt]
= {} & 2 - \sin x - (1-\sin x) \\[8pt]
= {} & 1.
\end{align}
