<s>[INST] Solving trigonometric identity with condition. Problem : If $\sin\theta +\sin^2\theta +\sin^3\theta=1$ Then prove $\cos^6\theta -4\cos^4\theta +8\cos^2\theta =4$
My working :
As $\sin\theta +\sin^2\theta +\sin^3\theta=1 \Rightarrow \sin\theta +\sin^3\theta = \cos^2\theta$
Now the given equation : $\cos^6\theta -4\cos^4\theta +8\cos^2\theta$ can be written as 
$(\sin\theta +\sin^3\theta)^3-4(\sin\theta+\sin^3\theta )^2+8(\sin\theta +\sin^3\theta)$
= $\sin^3\theta +\sin^6\theta +3\sin^5\theta +3\sin^7\theta -4\sin^2\theta -4\sin^6\theta -8\sin^4\theta + 8\sin\theta + 8\sin^3\theta$
But I think this is not the right way of doing this...Please suggest other alternative.. Thanks...
 [/INST] Clearly, we need to eliminate $\sin\theta$ and let us derive the proposition instead of verifying it.
As $\sin\theta +\sin^2\theta +\sin^3\theta=1$
$\implies \sin\theta(1 +\sin^2\theta) =1-\sin^2\theta$
$\implies \sin\theta(2 -\cos^2\theta) =\cos^2\theta$ using $\cos^2\theta=1-\sin^2\theta$
Squaring we get, $\sin^2\theta(2 -\cos^2\theta)^2 =\cos^4\theta$
or,  $(1-\cos^2\theta)(2 -\cos^2\theta)^2 =\cos^4\theta$
On simplification, $\cos^6\theta-4\cos^4\theta+8\cos^2\theta=4$
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